Maths-
General
Easy

Question

How are Pascal’s triangle and binomial expansion such as (a + b)5 related?

Hint:

Pascal's Triangle gives us the coefficients for an expanded binomial of the form (a + b)n , where n is the row of the triangle. The Binomial Theorem tells us we can use these coefficients to find the entire expanded binomial, with a couple extra tricks thrown in.

The correct answer is: (x + y)n



    Step 1 of 1:

    Pascal's Triangle is a method to know the binomial coefficients of terms of binomial expression (x + y)n , where n can be any positive integer and x, y are real numbers. Pascal Triangle is represented in a triangular form, it is kind of a number pattern in the form of a triangular arrangement.
    The binomial expansion of (a + b)5 is:

    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell left parenthesis a plus b right parenthesis to the power of 5 equals 5 C subscript 0 a to the power of 5 plus 5 C subscript 1 left parenthesis a right parenthesis to the power of 4 left parenthesis b right parenthesis plus 5 C subscript 2 left parenthesis a right parenthesis cubed left parenthesis b right parenthesis squared plus 5 C subscript 3 left parenthesis a right parenthesis squared left parenthesis b right parenthesis cubed plus 5 C subscript 4 left parenthesis a right parenthesis left parenthesis b right parenthesis to the power of 4 plus 5 C subscript 5 left parenthesis b right parenthesis to the power of 5 end cell row cell equals a to the power of 5 plus 5 a to the power of 4 b plus 10 a cubed b squared plus 10 a squared b cubed plus 5 a b to the power of 4 plus b to the power of 5 end cell end table
    Here, the coefficients of the expansions are the elements of the sixth row of the Pascal’s triangle.

    You can find the expansion of (x + y)n  using both Pascal’s triangle and binomial expansion.

    Related Questions to study

    General
    Maths-

    Use binomial theorem to expand .open parentheses s squared plus 3 close parentheses to the power of 5

    ANSWER:
    Hint:
    The binomial expansion is left parenthesis x plus y right parenthesis to the power of n equals sum from k equals 0 to n of   n C subscript k x to the power of n minus k end exponent y to the power of k , here n greater or equal than 0 .
    We are asked to use binomial theorem to expand  open parentheses s squared plus 3 close parentheses squared.
    Step 1 of 2:
    The given expression is  ,open parentheses s squared plus 3 close parentheses to the power of 5 text , here  end text x equals s squared straight & y equals 3. The value of n=5, hence there are 5+1=6 terms in the expressions.
    Step 2 of 2:
    Substitute the values of open parentheses s squared plus 3 close parentheses to the power of 5 in the binomial equation to get the expansion:
    open parentheses s squared plus 3 close parentheses to the power of 5 equals 5 C subscript 0 open parentheses s squared close parentheses to the power of 5 plus 5 C subscript 1 open parentheses s squared close parentheses to the power of 4 left parenthesis 3 right parenthesis plus 5 C subscript 2 open parentheses s squared close parentheses cubed left parenthesis 3 right parenthesis squared plus 5 C subscript 3 open parentheses s squared close parentheses squared left parenthesis 3 right parenthesis cubed plus 5 C subscript 4 open parentheses s squared close parentheses left parenthesis 3 right parenthesis to the power of 4 plus 5 C subscript 5 left parenthesis 3 right parenthesis to the power of 5
    equals s to the power of 10 plus 5 open parentheses s to the power of 8 close parentheses left parenthesis 3 right parenthesis plus 10 open parentheses s to the power of 6 close parentheses left parenthesis 9 right parenthesis plus 10 open parentheses s to the power of 4 close parentheses left parenthesis 27 right parenthesis plus 5 open parentheses s squared close parentheses left parenthesis 81 right parenthesis plus left parenthesis 243 right parenthesis
    equals s to the power of 10 plus 15 s to the power of 8 plus 90 s to the power of 6 plus 270 s to the power of 4 plus 405 s squared plus 243
    Thus, the expansion is:open parentheses s squared plus 3 close parentheses to the power of 5 equals s to the power of 10 plus 15 s to the power of 8 plus 90 s to the power of 6 plus 270 s to the power of 4 plus 405 s squared plus 243
    Note:
    For the expansion of an expression left parenthesis x plus y right parenthesis to the power of n , we would have n+1 terms. This is something you need to keep in mind.

    Use binomial theorem to expand .open parentheses s squared plus 3 close parentheses to the power of 5

    Maths-General
    ANSWER:
    Hint:
    The binomial expansion is left parenthesis x plus y right parenthesis to the power of n equals sum from k equals 0 to n of   n C subscript k x to the power of n minus k end exponent y to the power of k , here n greater or equal than 0 .
    We are asked to use binomial theorem to expand  open parentheses s squared plus 3 close parentheses squared.
    Step 1 of 2:
    The given expression is  ,open parentheses s squared plus 3 close parentheses to the power of 5 text , here  end text x equals s squared straight & y equals 3. The value of n=5, hence there are 5+1=6 terms in the expressions.
    Step 2 of 2:
    Substitute the values of open parentheses s squared plus 3 close parentheses to the power of 5 in the binomial equation to get the expansion:
    open parentheses s squared plus 3 close parentheses to the power of 5 equals 5 C subscript 0 open parentheses s squared close parentheses to the power of 5 plus 5 C subscript 1 open parentheses s squared close parentheses to the power of 4 left parenthesis 3 right parenthesis plus 5 C subscript 2 open parentheses s squared close parentheses cubed left parenthesis 3 right parenthesis squared plus 5 C subscript 3 open parentheses s squared close parentheses squared left parenthesis 3 right parenthesis cubed plus 5 C subscript 4 open parentheses s squared close parentheses left parenthesis 3 right parenthesis to the power of 4 plus 5 C subscript 5 left parenthesis 3 right parenthesis to the power of 5
    equals s to the power of 10 plus 5 open parentheses s to the power of 8 close parentheses left parenthesis 3 right parenthesis plus 10 open parentheses s to the power of 6 close parentheses left parenthesis 9 right parenthesis plus 10 open parentheses s to the power of 4 close parentheses left parenthesis 27 right parenthesis plus 5 open parentheses s squared close parentheses left parenthesis 81 right parenthesis plus left parenthesis 243 right parenthesis
    equals s to the power of 10 plus 15 s to the power of 8 plus 90 s to the power of 6 plus 270 s to the power of 4 plus 405 s squared plus 243
    Thus, the expansion is:open parentheses s squared plus 3 close parentheses to the power of 5 equals s to the power of 10 plus 15 s to the power of 8 plus 90 s to the power of 6 plus 270 s to the power of 4 plus 405 s squared plus 243
    Note:
    For the expansion of an expression left parenthesis x plus y right parenthesis to the power of n , we would have n+1 terms. This is something you need to keep in mind.
    General
    Maths-

    Use polynomial identities to factor the polynomials or simplify the expressions :
    216 plus 27 y to the power of 12

    ANSWER:
    Hint:
    open parentheses straight a cubed plus straight b cubed close parentheses equals left parenthesis straight a plus straight b right parenthesis open parentheses straight a squared minus ab plus straight b squared close parentheses, where a and b can be real values, variables or multiples of both.
    We are asked to use polynomial identities to factorize or simplify the expression.
    Step 1 of 2:
    The given expression is 216 plus 27 y to the power of 12.
    text  It can be written as  end text left parenthesis 6 right parenthesis cubed plus open parentheses 3 y to the power of 4 close parentheses cubed text . It is of the form  end text open parentheses a cubed plus b cubed close parentheses text  where  end text a equals 6 straight & b equals 3 y to the power of 4.
    Step 2 of 2:
    Use the polynomial identity to simplify the expression;
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 216 plus 27 y to the power of 12 equals left parenthesis 6 right parenthesis cubed plus open parentheses 3 y to the power of 4 close parentheses cubed space space space space space space space space space space space end cell row cell equals open parentheses 6 plus 3 y to the power of 4 close parentheses open parentheses 6 squared minus left parenthesis 6 right parenthesis open parentheses 3 y to the power of 4 close parentheses plus open parentheses 3 y to the power of 4 close parentheses squared close parentheses end cell row cell equals open parentheses 6 plus 3 y to the power of 4 close parentheses open parentheses 36 minus 18 y to the power of 4 plus 9 y to the power of 8 close parentheses space space space space space space space space space space end cell end table
    Hence, the factor form is 216 plus 27 straight y to the power of 12 equals open parentheses 6 plus 3 straight y to the power of 4 close parentheses open parentheses 36 minus 18 straight y to the power of 4 plus 9 straight y to the power of 8 close parentheses
    Note:
    The multiplication of algebraic expressions is a method of multiplying two given expressions consisting of variables and constants.

    Use polynomial identities to factor the polynomials or simplify the expressions :
    216 plus 27 y to the power of 12

    Maths-General
    ANSWER:
    Hint:
    open parentheses straight a cubed plus straight b cubed close parentheses equals left parenthesis straight a plus straight b right parenthesis open parentheses straight a squared minus ab plus straight b squared close parentheses, where a and b can be real values, variables or multiples of both.
    We are asked to use polynomial identities to factorize or simplify the expression.
    Step 1 of 2:
    The given expression is 216 plus 27 y to the power of 12.
    text  It can be written as  end text left parenthesis 6 right parenthesis cubed plus open parentheses 3 y to the power of 4 close parentheses cubed text . It is of the form  end text open parentheses a cubed plus b cubed close parentheses text  where  end text a equals 6 straight & b equals 3 y to the power of 4.
    Step 2 of 2:
    Use the polynomial identity to simplify the expression;
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 216 plus 27 y to the power of 12 equals left parenthesis 6 right parenthesis cubed plus open parentheses 3 y to the power of 4 close parentheses cubed space space space space space space space space space space space end cell row cell equals open parentheses 6 plus 3 y to the power of 4 close parentheses open parentheses 6 squared minus left parenthesis 6 right parenthesis open parentheses 3 y to the power of 4 close parentheses plus open parentheses 3 y to the power of 4 close parentheses squared close parentheses end cell row cell equals open parentheses 6 plus 3 y to the power of 4 close parentheses open parentheses 36 minus 18 y to the power of 4 plus 9 y to the power of 8 close parentheses space space space space space space space space space space end cell end table
    Hence, the factor form is 216 plus 27 straight y to the power of 12 equals open parentheses 6 plus 3 straight y to the power of 4 close parentheses open parentheses 36 minus 18 straight y to the power of 4 plus 9 straight y to the power of 8 close parentheses
    Note:
    The multiplication of algebraic expressions is a method of multiplying two given expressions consisting of variables and constants.
    General
    Maths-

    Explain why the middle term left parenthesis x plus 5 right parenthesis squared is 10x.


    Step 1 of 1:
    The given expression is: (x + 5)2
    Here, a=x and b=5.
    Here, the middle term is 10x because while expanding you have the form,

    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell left parenthesis x plus 5 right parenthesis squared equals 2 C subscript 0 x squared plus 2 C subscript 1 left parenthesis 5 right parenthesis left parenthesis x right parenthesis plus 2 C subscript 2 5 squared end cell row cell equals x squared plus 2 left parenthesis 5 right parenthesis left parenthesis x right parenthesis plus 5 squared end cell row cell equals x squared plus 10 x plus 25 end cell end table
     

    Explain why the middle term left parenthesis x plus 5 right parenthesis squared is 10x.

    Maths-General

    Step 1 of 1:
    The given expression is: (x + 5)2
    Here, a=x and b=5.
    Here, the middle term is 10x because while expanding you have the form,

    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell left parenthesis x plus 5 right parenthesis squared equals 2 C subscript 0 x squared plus 2 C subscript 1 left parenthesis 5 right parenthesis left parenthesis x right parenthesis plus 2 C subscript 2 5 squared end cell row cell equals x squared plus 2 left parenthesis 5 right parenthesis left parenthesis x right parenthesis plus 5 squared end cell row cell equals x squared plus 10 x plus 25 end cell end table
     

    parallel
    General
    Maths-

    Use binomial theorem to expand open parentheses s squared plus 3 close parentheses squared .

    ANSWER:
    Hint:
    The binomial expansion is left parenthesis x plus y right parenthesis to the power of n equals sum from k equals 0 to n of   n C subscript k x to the power of n minus k end exponent y to the power of k  , here n greater or equal than 0  .
    We are asked to use binomial theorem to expand open parentheses s squared plus 3 close parentheses to the power of 5 .
    Step 1 of 2:
    The given expression is open parentheses s squared plus 3 close parentheses to the power of 5, here x equals s squared straight & y equals 3 . The value of n=5, hence there are 5+1=6 terms in the expressions.
    Step 2 of 2:
    Substitute the values of open parentheses s squared plus 3 close parentheses to the power of 5 in the binomial equation to get the expansion:
    open parentheses s squared plus 3 close parentheses to the power of 5 equals 5 C subscript 0 open parentheses s squared close parentheses to the power of 5 plus 5 C subscript 1 open parentheses s squared close parentheses to the power of 4 left parenthesis 3 right parenthesis plus 5 C subscript 2 open parentheses s squared close parentheses cubed left parenthesis 3 right parenthesis squared plus 5 C subscript 3 open parentheses s squared close parentheses squared left parenthesis 3 right parenthesis cubed plus 5 C subscript 4 open parentheses s squared close parentheses left parenthesis 3 right parenthesis to the power of 4 plus 5 C subscript 5 left parenthesis 3 right parenthesis to the power of 5equals s to the power of 10 plus 5 open parentheses s to the power of 8 close parentheses left parenthesis 3 right parenthesis plus 10 open parentheses s to the power of 6 close parentheses left parenthesis 9 right parenthesis plus 10 open parentheses s to the power of 4 close parentheses left parenthesis 27 right parenthesis plus 5 open parentheses s squared close parentheses left parenthesis 81 right parenthesis plus left parenthesis 243 right parenthesis
    equals s to the power of 10 plus 15 s to the power of 8 plus 90 s to the power of 6 plus 270 s to the power of 4 plus 405 s squared plus 243
    Thus, the expansion is: open parentheses s squared plus 3 close parentheses to the power of 5 equals s to the power of 10 plus 15 s to the power of 8 plus 90 s to the power of 6 plus 270 s to the power of 4 plus 405 s squared plus 243
    Note:
    For the expansion of an expression left parenthesis x plus y right parenthesis to the power of n,we would have n+1 terms. This is something you need to keep in mind.

    Thus, the expansion is:
    Note:
    For the expansion of an expression , we would have n+1 terms. This is something you need to keep in mind.

    Use binomial theorem to expand open parentheses s squared plus 3 close parentheses squared .

    Maths-General
    ANSWER:
    Hint:
    The binomial expansion is left parenthesis x plus y right parenthesis to the power of n equals sum from k equals 0 to n of   n C subscript k x to the power of n minus k end exponent y to the power of k  , here n greater or equal than 0  .
    We are asked to use binomial theorem to expand open parentheses s squared plus 3 close parentheses to the power of 5 .
    Step 1 of 2:
    The given expression is open parentheses s squared plus 3 close parentheses to the power of 5, here x equals s squared straight & y equals 3 . The value of n=5, hence there are 5+1=6 terms in the expressions.
    Step 2 of 2:
    Substitute the values of open parentheses s squared plus 3 close parentheses to the power of 5 in the binomial equation to get the expansion:
    open parentheses s squared plus 3 close parentheses to the power of 5 equals 5 C subscript 0 open parentheses s squared close parentheses to the power of 5 plus 5 C subscript 1 open parentheses s squared close parentheses to the power of 4 left parenthesis 3 right parenthesis plus 5 C subscript 2 open parentheses s squared close parentheses cubed left parenthesis 3 right parenthesis squared plus 5 C subscript 3 open parentheses s squared close parentheses squared left parenthesis 3 right parenthesis cubed plus 5 C subscript 4 open parentheses s squared close parentheses left parenthesis 3 right parenthesis to the power of 4 plus 5 C subscript 5 left parenthesis 3 right parenthesis to the power of 5equals s to the power of 10 plus 5 open parentheses s to the power of 8 close parentheses left parenthesis 3 right parenthesis plus 10 open parentheses s to the power of 6 close parentheses left parenthesis 9 right parenthesis plus 10 open parentheses s to the power of 4 close parentheses left parenthesis 27 right parenthesis plus 5 open parentheses s squared close parentheses left parenthesis 81 right parenthesis plus left parenthesis 243 right parenthesis
    equals s to the power of 10 plus 15 s to the power of 8 plus 90 s to the power of 6 plus 270 s to the power of 4 plus 405 s squared plus 243
    Thus, the expansion is: open parentheses s squared plus 3 close parentheses to the power of 5 equals s to the power of 10 plus 15 s to the power of 8 plus 90 s to the power of 6 plus 270 s to the power of 4 plus 405 s squared plus 243
    Note:
    For the expansion of an expression left parenthesis x plus y right parenthesis to the power of n,we would have n+1 terms. This is something you need to keep in mind.

    Thus, the expansion is:
    Note:
    For the expansion of an expression , we would have n+1 terms. This is something you need to keep in mind.
    General
    Maths-

    Use polynomial identities to factor the polynomials or simplify the expressions :
    4 x squared minus y to the power of 6

    ANSWER:
    Hint:
    open parentheses a squared minus b squared close parentheses equals left parenthesis a minus b right parenthesis left parenthesis a plus b right parenthesis, where a and b can be real values, variables or multiples of both.
    We are asked to use polynomial identities to simplify the given expressions.
    Step 1 of 2:
    The given expression is 4 x squared minus y to the power of 6.text  It can be written as  end text left parenthesis 2 x right parenthesis squared minus open parentheses y cubed close parentheses squared text . This is of the form  end text open parentheses a squared minus b squared close parentheses text  where  end text a equals 2 x straight & b equals y cubed text .  end text
    Step 2 of 2:
    Use the polynomial identity to simplify the expression;
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 4 x squared minus y to the power of 6 equals left parenthesis 2 x right parenthesis squared minus open parentheses y cubed close parentheses squared end cell row cell equals open parentheses 2 x minus y cubed close parentheses open parentheses 2 x plus y cubed close parentheses space space space space space space end cell end table
    Thus, the factor is: .4 straight x squared minus straight y to the power of 6 equals open parentheses 2 straight x minus straight y cubed close parentheses open parentheses 2 straight x plus straight y cubed close parentheses
    Note:
    Polynomial identities are used to simplify or to find the prduct of expressions. It reduces space and time during solving.

    Use polynomial identities to factor the polynomials or simplify the expressions :
    4 x squared minus y to the power of 6

    Maths-General
    ANSWER:
    Hint:
    open parentheses a squared minus b squared close parentheses equals left parenthesis a minus b right parenthesis left parenthesis a plus b right parenthesis, where a and b can be real values, variables or multiples of both.
    We are asked to use polynomial identities to simplify the given expressions.
    Step 1 of 2:
    The given expression is 4 x squared minus y to the power of 6.text  It can be written as  end text left parenthesis 2 x right parenthesis squared minus open parentheses y cubed close parentheses squared text . This is of the form  end text open parentheses a squared minus b squared close parentheses text  where  end text a equals 2 x straight & b equals y cubed text .  end text
    Step 2 of 2:
    Use the polynomial identity to simplify the expression;
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 4 x squared minus y to the power of 6 equals left parenthesis 2 x right parenthesis squared minus open parentheses y cubed close parentheses squared end cell row cell equals open parentheses 2 x minus y cubed close parentheses open parentheses 2 x plus y cubed close parentheses space space space space space space end cell end table
    Thus, the factor is: .4 straight x squared minus straight y to the power of 6 equals open parentheses 2 straight x minus straight y cubed close parentheses open parentheses 2 straight x plus straight y cubed close parentheses
    Note:
    Polynomial identities are used to simplify or to find the prduct of expressions. It reduces space and time during solving.
    General
    Maths-

    How can you use polynomial identities to rewrite expressions efficiently ?


    Step 1 of 1:
    A polynomial expression can be in the expanded form majority of the times. We can use the polynomial identities to factorize them in to the standard form. This will reduce time and space and enhance the quality of writing. Moreover, the polynomial identities are not just limited form variables but for numbers as well. We can use these identities to multiply higher numbers and reduce the calculation part to an extent.
     

    How can you use polynomial identities to rewrite expressions efficiently ?

    Maths-General

    Step 1 of 1:
    A polynomial expression can be in the expanded form majority of the times. We can use the polynomial identities to factorize them in to the standard form. This will reduce time and space and enhance the quality of writing. Moreover, the polynomial identities are not just limited form variables but for numbers as well. We can use these identities to multiply higher numbers and reduce the calculation part to an extent.
     
    parallel
    General
    Maths-

    Use binomial theorem to expand (2c + d)6


    Step 1 of 2:
    The given expression is (2c + d)6  , here x = 2c & y = d . The value of n=6, hence we would have 6+1=7 terms in the expression.
    Step 2 of 2:
    Substitute the values of (2c + d)6 in the binomial expression

    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell left parenthesis 2 c plus d right parenthesis to the power of 6 equals 6 C subscript 0 left parenthesis 2 c right parenthesis to the power of 6 plus 6 C subscript 1 left parenthesis 2 c right parenthesis to the power of 5 left parenthesis d right parenthesis plus 6 C subscript 2 left parenthesis 2 c right parenthesis to the power of 4 left parenthesis d right parenthesis squared plus 6 C subscript 3 left parenthesis 2 c right parenthesis cubed left parenthesis d right parenthesis cubed plus 6 C subscript 4 left parenthesis 2 c right parenthesis squared left parenthesis d right parenthesis to the power of 4 plus 6 C subscript 5 left parenthesis 2 c right parenthesis left parenthesis d right parenthesis to the power of 5 plus 6 C subscript 6 left parenthesis d right parenthesis to the power of 6 end cell row cell equals 64 c to the power of 6 plus 6 open parentheses 32 c to the power of 5 close parentheses left parenthesis d right parenthesis plus 15 open parentheses 16 c to the power of 4 close parentheses open parentheses d squared close parentheses plus 20 open parentheses 8 c cubed close parentheses open parentheses d cubed close parentheses plus 15 open parentheses 4 c squared close parentheses open parentheses d to the power of 4 close parentheses plus 6 left parenthesis 2 c right parenthesis open parentheses d to the power of 5 close parentheses plus d to the power of 6 end cell row cell equals 64 c to the power of 6 plus 192 d c to the power of 5 plus 240 d squared c to the power of 4 plus 160 d cubed c cubed plus 60 d to the power of 4 c squared plus 12 d to the power of 5 c plus d to the power of 6 end cell end table
    Thus, the expansion is: left parenthesis 2 c plus d right parenthesis to the power of 6 equals 64 c to the power of 6 plus 192 d c to the power of 5 plus 240 d squared c to the power of 4 plus 160 d cubed c cubed plus 60 d to the power of 4 c squared plus 12 d to the power of 5 c plus d to the power of 6.

    Use binomial theorem to expand (2c + d)6

    Maths-General

    Step 1 of 2:
    The given expression is (2c + d)6  , here x = 2c & y = d . The value of n=6, hence we would have 6+1=7 terms in the expression.
    Step 2 of 2:
    Substitute the values of (2c + d)6 in the binomial expression

    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell left parenthesis 2 c plus d right parenthesis to the power of 6 equals 6 C subscript 0 left parenthesis 2 c right parenthesis to the power of 6 plus 6 C subscript 1 left parenthesis 2 c right parenthesis to the power of 5 left parenthesis d right parenthesis plus 6 C subscript 2 left parenthesis 2 c right parenthesis to the power of 4 left parenthesis d right parenthesis squared plus 6 C subscript 3 left parenthesis 2 c right parenthesis cubed left parenthesis d right parenthesis cubed plus 6 C subscript 4 left parenthesis 2 c right parenthesis squared left parenthesis d right parenthesis to the power of 4 plus 6 C subscript 5 left parenthesis 2 c right parenthesis left parenthesis d right parenthesis to the power of 5 plus 6 C subscript 6 left parenthesis d right parenthesis to the power of 6 end cell row cell equals 64 c to the power of 6 plus 6 open parentheses 32 c to the power of 5 close parentheses left parenthesis d right parenthesis plus 15 open parentheses 16 c to the power of 4 close parentheses open parentheses d squared close parentheses plus 20 open parentheses 8 c cubed close parentheses open parentheses d cubed close parentheses plus 15 open parentheses 4 c squared close parentheses open parentheses d to the power of 4 close parentheses plus 6 left parenthesis 2 c right parenthesis open parentheses d to the power of 5 close parentheses plus d to the power of 6 end cell row cell equals 64 c to the power of 6 plus 192 d c to the power of 5 plus 240 d squared c to the power of 4 plus 160 d cubed c cubed plus 60 d to the power of 4 c squared plus 12 d to the power of 5 c plus d to the power of 6 end cell end table
    Thus, the expansion is: left parenthesis 2 c plus d right parenthesis to the power of 6 equals 64 c to the power of 6 plus 192 d c to the power of 5 plus 240 d squared c to the power of 4 plus 160 d cubed c cubed plus 60 d to the power of 4 c squared plus 12 d to the power of 5 c plus d to the power of 6.

    General
    Maths-

    Use binomial theorem to expand left parenthesis x minus 3 right parenthesis to the power of 4 .

    ANSWER:
    Hint:
    The binomial expansion is left parenthesis x plus y right parenthesis to the power of n equals sum from k equals 0 to n of   n C subscript k x to the power of n minus k end exponent y to the power of k , here n greater or equal than 0.
    We are asked to use binomial theorem to expand left parenthesis x minus 3 right parenthesis to the power of 4.
    Step 1 of 2:
    Here, the value of n=4. So, the combination permutation we would use is .
    We would have 4+1=5 terms in the expansion of the expression left parenthesis x minus 3 right parenthesis to the power of 4.
    Step 2 of 2:
    Substitute the values of left parenthesis x minus 3 right parenthesis to the power of 4 in the binomial expansion:
    left parenthesis x minus 3 right parenthesis to the power of 4 equals 4 C subscript 0 x to the power of 4 plus 4 C subscript 1 left parenthesis x right parenthesis cubed left parenthesis negative 3 right parenthesis plus 4 C subscript 2 left parenthesis x right parenthesis squared left parenthesis negative 3 right parenthesis squared plus 4 C subscript 3 left parenthesis x right parenthesis left parenthesis negative 3 right parenthesis cubed plus 4 C subscript 4 left parenthesis negative 3 right parenthesis to the power of 4
    equals x to the power of 4 plus 4 left parenthesis x right parenthesis cubed left parenthesis negative 3 right parenthesis plus 6 x squared left parenthesis 9 right parenthesis plus 4 left parenthesis x right parenthesis left parenthesis negative 27 right parenthesis plus 1 left parenthesis negative 3 right parenthesis to the power of 4
    equals x to the power of 4 minus 12 x cubed plus 54 x squared minus 108 x plus 81
    Thus, the expansion is:left parenthesis x minus 3 right parenthesis to the power of 4 equals x to the power of 4 minus 12 x cubed plus 54 x squared minus 108 x plus 81
    Note:
    For the expansion of an expression left parenthesis x plus y right parenthesis to the power of n , we would have n+1 terms. This is something you need to keep in mind.

    Use binomial theorem to expand left parenthesis x minus 3 right parenthesis to the power of 4 .

    Maths-General
    ANSWER:
    Hint:
    The binomial expansion is left parenthesis x plus y right parenthesis to the power of n equals sum from k equals 0 to n of   n C subscript k x to the power of n minus k end exponent y to the power of k , here n greater or equal than 0.
    We are asked to use binomial theorem to expand left parenthesis x minus 3 right parenthesis to the power of 4.
    Step 1 of 2:
    Here, the value of n=4. So, the combination permutation we would use is .
    We would have 4+1=5 terms in the expansion of the expression left parenthesis x minus 3 right parenthesis to the power of 4.
    Step 2 of 2:
    Substitute the values of left parenthesis x minus 3 right parenthesis to the power of 4 in the binomial expansion:
    left parenthesis x minus 3 right parenthesis to the power of 4 equals 4 C subscript 0 x to the power of 4 plus 4 C subscript 1 left parenthesis x right parenthesis cubed left parenthesis negative 3 right parenthesis plus 4 C subscript 2 left parenthesis x right parenthesis squared left parenthesis negative 3 right parenthesis squared plus 4 C subscript 3 left parenthesis x right parenthesis left parenthesis negative 3 right parenthesis cubed plus 4 C subscript 4 left parenthesis negative 3 right parenthesis to the power of 4
    equals x to the power of 4 plus 4 left parenthesis x right parenthesis cubed left parenthesis negative 3 right parenthesis plus 6 x squared left parenthesis 9 right parenthesis plus 4 left parenthesis x right parenthesis left parenthesis negative 27 right parenthesis plus 1 left parenthesis negative 3 right parenthesis to the power of 4
    equals x to the power of 4 minus 12 x cubed plus 54 x squared minus 108 x plus 81
    Thus, the expansion is:left parenthesis x minus 3 right parenthesis to the power of 4 equals x to the power of 4 minus 12 x cubed plus 54 x squared minus 108 x plus 81
    Note:
    For the expansion of an expression left parenthesis x plus y right parenthesis to the power of n , we would have n+1 terms. This is something you need to keep in mind.
    General
    Maths-

    Use binomial theorem to expand (x - 1)7


    Step 1 of 2:
    The given expression is (x - 1)7  , here x = x & y = -1 . The value of n=7, hence there are 7+1=8 terms in the expressions.
    Step 2 of 2:
    Substitute the values of (x - 1)7 in the binomial equation to get the expansion:
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell left parenthesis x minus 1 right parenthesis to the power of 7 equals 7 C subscript 0 left parenthesis x right parenthesis to the power of 7 plus 7 C subscript 1 left parenthesis x right parenthesis to the power of 6 left parenthesis negative 1 right parenthesis plus 7 C subscript 2 left parenthesis x right parenthesis to the power of 5 left parenthesis negative 1 right parenthesis squared plus 7 C subscript 3 left parenthesis x right parenthesis to the power of 4 left parenthesis negative 1 right parenthesis cubed plus 7 C subscript 4 left parenthesis x right parenthesis cubed left parenthesis negative 1 right parenthesis to the power of 4 plus 7 C subscript 5 left parenthesis x right parenthesis squared left parenthesis negative 1 right parenthesis to the power of 5 plus 7 C subscript 6 left parenthesis x right parenthesis left parenthesis negative 1 right parenthesis to the power of 6 plus 7 C subscript 7 left parenthesis negative 1 right parenthesis to the power of 7 end cell row cell equals x to the power of 7 plus 7 x to the power of 6 left parenthesis negative 1 right parenthesis plus 21 x to the power of 5 left parenthesis 1 right parenthesis plus 35 x to the power of 4 left parenthesis negative 1 right parenthesis plus 35 x cubed left parenthesis 1 right parenthesis plus 21 x squared left parenthesis negative 1 right parenthesis plus 7 x left parenthesis 1 right parenthesis plus left parenthesis negative 1 right parenthesis end cell row cell equals x to the power of straight ⊤ minus 7 x to the power of 6 plus 21 x to the power of 5 minus 35 x to the power of 4 plus 35 x cubed minus 21 x squared plus 7 x minus 1 end cell end table
    Thus, the expansion is: left parenthesis x minus 1 right parenthesis to the power of 7 equals x to the power of 7 minus 7 x to the power of 6 plus 21 x to the power of 5 minus 35 x to the power of 4 plus 35 x cubed minus 21 x squared plus 7 x minus 1

    Use binomial theorem to expand (x - 1)7

    Maths-General

    Step 1 of 2:
    The given expression is (x - 1)7  , here x = x & y = -1 . The value of n=7, hence there are 7+1=8 terms in the expressions.
    Step 2 of 2:
    Substitute the values of (x - 1)7 in the binomial equation to get the expansion:
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell left parenthesis x minus 1 right parenthesis to the power of 7 equals 7 C subscript 0 left parenthesis x right parenthesis to the power of 7 plus 7 C subscript 1 left parenthesis x right parenthesis to the power of 6 left parenthesis negative 1 right parenthesis plus 7 C subscript 2 left parenthesis x right parenthesis to the power of 5 left parenthesis negative 1 right parenthesis squared plus 7 C subscript 3 left parenthesis x right parenthesis to the power of 4 left parenthesis negative 1 right parenthesis cubed plus 7 C subscript 4 left parenthesis x right parenthesis cubed left parenthesis negative 1 right parenthesis to the power of 4 plus 7 C subscript 5 left parenthesis x right parenthesis squared left parenthesis negative 1 right parenthesis to the power of 5 plus 7 C subscript 6 left parenthesis x right parenthesis left parenthesis negative 1 right parenthesis to the power of 6 plus 7 C subscript 7 left parenthesis negative 1 right parenthesis to the power of 7 end cell row cell equals x to the power of 7 plus 7 x to the power of 6 left parenthesis negative 1 right parenthesis plus 21 x to the power of 5 left parenthesis 1 right parenthesis plus 35 x to the power of 4 left parenthesis negative 1 right parenthesis plus 35 x cubed left parenthesis 1 right parenthesis plus 21 x squared left parenthesis negative 1 right parenthesis plus 7 x left parenthesis 1 right parenthesis plus left parenthesis negative 1 right parenthesis end cell row cell equals x to the power of straight ⊤ minus 7 x to the power of 6 plus 21 x to the power of 5 minus 35 x to the power of 4 plus 35 x cubed minus 21 x squared plus 7 x minus 1 end cell end table
    Thus, the expansion is: left parenthesis x minus 1 right parenthesis to the power of 7 equals x to the power of 7 minus 7 x to the power of 6 plus 21 x to the power of 5 minus 35 x to the power of 4 plus 35 x cubed minus 21 x squared plus 7 x minus 1
    parallel
    General
    Maths-

    Use polynomial identities to factor the polynomials or simplify the expressions :
    x cubed minus 27 y cubed

    ANSWER:
    Hint:
    open parentheses straight a cubed minus straight b cubed close parentheses equals left parenthesis straight a minus straight b right parenthesis open parentheses straight a squared plus ab plus straight b squared close parentheses, here a and b can be real values, variables or multiples of both.
    We are asked to use polynomial identities to factorize the given expression.
    Step 1 of 2:
    The given expression is x cubed minus 27 y cubed.text  It can be written as  end text left parenthesis x right parenthesis cubed minus left parenthesis 3 y right parenthesis cubed text . It is of the form  end text a cubed minus b cubed text  where  end text a equals x straight & b equals 3 y text .  end text
    Step 2 of 2:
    Use polynomial identities to simplify the given expression;
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell x cubed minus 27 y cubed equals left parenthesis x right parenthesis cubed minus left parenthesis 3 y right parenthesis cubed space space space space space space space space space space space end cell row cell equals left parenthesis x minus 3 y right parenthesis open parentheses x squared plus left parenthesis x right parenthesis left parenthesis 3 y right parenthesis plus left parenthesis 3 y right parenthesis squared close parentheses end cell row cell equals left parenthesis x minus 3 y right parenthesis open parentheses x squared plus 3 x y plus 9 y squared close parentheses space space space space space space space space end cell end table
    The factor of the expression is straight x cubed minus 27 straight y cubed equals left parenthesis straight x minus 3 straight y right parenthesis open parentheses straight x squared plus 3 xy plus 9 straight y squared close parentheses
    Note:
    The multiplication of algebraic expressions is a method of multiplying two given expressions consisting of variables and constants.

    Use polynomial identities to factor the polynomials or simplify the expressions :
    x cubed minus 27 y cubed

    Maths-General
    ANSWER:
    Hint:
    open parentheses straight a cubed minus straight b cubed close parentheses equals left parenthesis straight a minus straight b right parenthesis open parentheses straight a squared plus ab plus straight b squared close parentheses, here a and b can be real values, variables or multiples of both.
    We are asked to use polynomial identities to factorize the given expression.
    Step 1 of 2:
    The given expression is x cubed minus 27 y cubed.text  It can be written as  end text left parenthesis x right parenthesis cubed minus left parenthesis 3 y right parenthesis cubed text . It is of the form  end text a cubed minus b cubed text  where  end text a equals x straight & b equals 3 y text .  end text
    Step 2 of 2:
    Use polynomial identities to simplify the given expression;
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell x cubed minus 27 y cubed equals left parenthesis x right parenthesis cubed minus left parenthesis 3 y right parenthesis cubed space space space space space space space space space space space end cell row cell equals left parenthesis x minus 3 y right parenthesis open parentheses x squared plus left parenthesis x right parenthesis left parenthesis 3 y right parenthesis plus left parenthesis 3 y right parenthesis squared close parentheses end cell row cell equals left parenthesis x minus 3 y right parenthesis open parentheses x squared plus 3 x y plus 9 y squared close parentheses space space space space space space space space end cell end table
    The factor of the expression is straight x cubed minus 27 straight y cubed equals left parenthesis straight x minus 3 straight y right parenthesis open parentheses straight x squared plus 3 xy plus 9 straight y squared close parentheses
    Note:
    The multiplication of algebraic expressions is a method of multiplying two given expressions consisting of variables and constants.
    General
    Maths-

    Use Pascal triangle to expandleft parenthesis x plus y right parenthesis to the power of 6

    ANSWER:
    Hint:
    Hint:
    Pascal's Triangle is a method to know the binomial coefficients of terms of binomial expression (x + y)n, where n can be any positive integer and x,y are real numbers. Pascal Triangle is represented in a triangular form, it is kind of a number pattern in the form of a triangular arrangement.
    We are asked to find the binomial expansion of left parenthesis x plus y right parenthesis to the power of 6  using Pascal’s formula.
    Step 1 of 1:
    The Pascal’s triangle is given by:

    From the triangle, the seventh line would give the coefficients of the expansion of the polynomial left parenthesis x plus y right parenthesis to the power of 6 . Thus, we have:
    left parenthesis x plus y right parenthesis to the power of 6 equals x to the power of 6 plus 6 x to the power of 5 y plus 15 x to the power of 4 y squared plus 20 x cubed y cubed plus 15 x squared y to the power of 4 plus 6 x y to the power of 5 plus y to the power of 6
    Note:
    The expansion of the polynomial  left parenthesis x plus y right parenthesis to the power of 6 can be found using the values of 6 C subscript r

    Use Pascal triangle to expandleft parenthesis x plus y right parenthesis to the power of 6

    Maths-General
    ANSWER:
    Hint:
    Hint:
    Pascal's Triangle is a method to know the binomial coefficients of terms of binomial expression (x + y)n, where n can be any positive integer and x,y are real numbers. Pascal Triangle is represented in a triangular form, it is kind of a number pattern in the form of a triangular arrangement.
    We are asked to find the binomial expansion of left parenthesis x plus y right parenthesis to the power of 6  using Pascal’s formula.
    Step 1 of 1:
    The Pascal’s triangle is given by:

    From the triangle, the seventh line would give the coefficients of the expansion of the polynomial left parenthesis x plus y right parenthesis to the power of 6 . Thus, we have:
    left parenthesis x plus y right parenthesis to the power of 6 equals x to the power of 6 plus 6 x to the power of 5 y plus 15 x to the power of 4 y squared plus 20 x cubed y cubed plus 15 x squared y to the power of 4 plus 6 x y to the power of 5 plus y to the power of 6
    Note:
    The expansion of the polynomial  left parenthesis x plus y right parenthesis to the power of 6 can be found using the values of 6 C subscript r
    General
    Maths-

    Use binomial theorem to expand (s2 + 3)5


    Step 1 of 2:
    The given expression is (s2 + 3)5  , here x = s2 & y = 3 . The value of n=5, hence there are 5+1=6 terms in the expressions.
    Step 2 of 2:
    Substitute the values of (s2 + 3)5 in the binomial equation to get the expansion:

    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell open parentheses s squared plus 3 close parentheses to the power of 5 equals 5 C subscript 0 open parentheses s squared close parentheses to the power of 5 plus 5 C subscript 1 open parentheses s squared close parentheses to the power of 4 left parenthesis 3 right parenthesis plus 5 C subscript 2 open parentheses s squared close parentheses cubed left parenthesis 3 right parenthesis squared plus 5 C subscript 3 open parentheses s squared close parentheses squared left parenthesis 3 right parenthesis cubed plus 5 C subscript 4 open parentheses s squared close parentheses left parenthesis 3 right parenthesis to the power of 4 plus 5 C subscript 5 left parenthesis 3 right parenthesis to the power of 5 end cell row cell equals s to the power of 10 plus 5 open parentheses s to the power of 8 close parentheses left parenthesis 3 right parenthesis plus 10 open parentheses s to the power of 6 close parentheses left parenthesis 9 right parenthesis plus 10 open parentheses s to the power of 4 close parentheses left parenthesis 27 right parenthesis plus 5 open parentheses s squared close parentheses left parenthesis 81 right parenthesis plus left parenthesis 243 right parenthesis end cell row cell equals s to the power of 10 plus 15 s to the power of 8 plus 90 s to the power of 6 plus 270 s to the power of 4 plus 405 s squared plus 243 end cell end table
    Thus, the expansion is: open parentheses s squared plus 3 close parentheses to the power of 5 equals s to the power of 10 plus 15 s to the power of 8 plus 90 s to the power of 6 plus 270 s to the power of 4 plus 405 s squared plus 243
     

    Use binomial theorem to expand (s2 + 3)5

    Maths-General

    Step 1 of 2:
    The given expression is (s2 + 3)5  , here x = s2 & y = 3 . The value of n=5, hence there are 5+1=6 terms in the expressions.
    Step 2 of 2:
    Substitute the values of (s2 + 3)5 in the binomial equation to get the expansion:

    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell open parentheses s squared plus 3 close parentheses to the power of 5 equals 5 C subscript 0 open parentheses s squared close parentheses to the power of 5 plus 5 C subscript 1 open parentheses s squared close parentheses to the power of 4 left parenthesis 3 right parenthesis plus 5 C subscript 2 open parentheses s squared close parentheses cubed left parenthesis 3 right parenthesis squared plus 5 C subscript 3 open parentheses s squared close parentheses squared left parenthesis 3 right parenthesis cubed plus 5 C subscript 4 open parentheses s squared close parentheses left parenthesis 3 right parenthesis to the power of 4 plus 5 C subscript 5 left parenthesis 3 right parenthesis to the power of 5 end cell row cell equals s to the power of 10 plus 5 open parentheses s to the power of 8 close parentheses left parenthesis 3 right parenthesis plus 10 open parentheses s to the power of 6 close parentheses left parenthesis 9 right parenthesis plus 10 open parentheses s to the power of 4 close parentheses left parenthesis 27 right parenthesis plus 5 open parentheses s squared close parentheses left parenthesis 81 right parenthesis plus left parenthesis 243 right parenthesis end cell row cell equals s to the power of 10 plus 15 s to the power of 8 plus 90 s to the power of 6 plus 270 s to the power of 4 plus 405 s squared plus 243 end cell end table
    Thus, the expansion is: open parentheses s squared plus 3 close parentheses to the power of 5 equals s to the power of 10 plus 15 s to the power of 8 plus 90 s to the power of 6 plus 270 s to the power of 4 plus 405 s squared plus 243
     

    parallel
    General
    Maths-

    Use polynomial identities to factor the polynomials or simplify the expressions :
    8 x cubed plus y to the power of 9

    ANSWER:
    Hint:
    open parentheses straight a cubed plus straight b cubed close parentheses equals left parenthesis straight a plus straight b right parenthesis open parentheses straight a squared minus ab plus straight b squared close parentheses, where a and b can be real values, variables or multiples of both.
    We are asked to use polynomial identities to factorize or simplify the expression.
    Step 1 of 2:
    The given expression is 8 x cubed plus y to the power of 9text  It can be written as  end text left parenthesis 2 x right parenthesis cubed plus open parentheses y cubed close parentheses cubed text . It is of the form  end text open parentheses a cubed plus b cubed close parentheses text  where  end text a equals 2 x straight & b equals y cubed text .  end text
    Step 2 of 2:
    Use the polynomial identity open parentheses straight a cubed plus straight b cubed close parentheses equals left parenthesis straight a plus straight b right parenthesis open parentheses straight a squared minus ab plus straight b squared close parentheses to simplify the expression;
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 8 x cubed plus y to the power of 9 equals left parenthesis 2 x right parenthesis cubed plus open parentheses y cubed close parentheses cubed space space space space space space space space space space space space space space space space space space space space end cell row cell equals open parentheses 2 x plus y cubed close parentheses open parentheses left parenthesis 2 x right parenthesis squared minus left parenthesis 2 x right parenthesis open parentheses y cubed close parentheses plus open parentheses y cubed close parentheses squared close parentheses end cell row cell equals open parentheses 2 x plus y cubed close parentheses open parentheses 4 x squared minus 2 x y cubed plus y to the power of 6 close parentheses space space space space space space space space space space space space space end cell end table
    Thus, the factor is: 8 straight x cubed plus straight y to the power of 9 equals open parentheses 2 straight x plus straight y cubed close parentheses open parentheses 4 straight x squared minus 2 xy cubed plus straight y to the power of 6 close parentheses
    Note:
    Polynomial identities are equations that are true for all possible values of the variable. We can perform polynomial multiplication by applying the distributive property to the multiplication of polynomials.

    Use polynomial identities to factor the polynomials or simplify the expressions :
    8 x cubed plus y to the power of 9

    Maths-General
    ANSWER:
    Hint:
    open parentheses straight a cubed plus straight b cubed close parentheses equals left parenthesis straight a plus straight b right parenthesis open parentheses straight a squared minus ab plus straight b squared close parentheses, where a and b can be real values, variables or multiples of both.
    We are asked to use polynomial identities to factorize or simplify the expression.
    Step 1 of 2:
    The given expression is 8 x cubed plus y to the power of 9text  It can be written as  end text left parenthesis 2 x right parenthesis cubed plus open parentheses y cubed close parentheses cubed text . It is of the form  end text open parentheses a cubed plus b cubed close parentheses text  where  end text a equals 2 x straight & b equals y cubed text .  end text
    Step 2 of 2:
    Use the polynomial identity open parentheses straight a cubed plus straight b cubed close parentheses equals left parenthesis straight a plus straight b right parenthesis open parentheses straight a squared minus ab plus straight b squared close parentheses to simplify the expression;
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 8 x cubed plus y to the power of 9 equals left parenthesis 2 x right parenthesis cubed plus open parentheses y cubed close parentheses cubed space space space space space space space space space space space space space space space space space space space space end cell row cell equals open parentheses 2 x plus y cubed close parentheses open parentheses left parenthesis 2 x right parenthesis squared minus left parenthesis 2 x right parenthesis open parentheses y cubed close parentheses plus open parentheses y cubed close parentheses squared close parentheses end cell row cell equals open parentheses 2 x plus y cubed close parentheses open parentheses 4 x squared minus 2 x y cubed plus y to the power of 6 close parentheses space space space space space space space space space space space space space end cell end table
    Thus, the factor is: 8 straight x cubed plus straight y to the power of 9 equals open parentheses 2 straight x plus straight y cubed close parentheses open parentheses 4 straight x squared minus 2 xy cubed plus straight y to the power of 6 close parentheses
    Note:
    Polynomial identities are equations that are true for all possible values of the variable. We can perform polynomial multiplication by applying the distributive property to the multiplication of polynomials.
    General
    Maths-

    Use Pascal triangle to expand left parenthesis x plus y right parenthesis to the power of 6

    ANSWER:
    Hint:
    Hint:
    Pascal's Triangle is a method to know the binomial coefficients of terms of binomial expression (x + y)n, where n can be any positive integer and x,y are real numbers. Pascal Triangle is represented in a triangular form, it is kind of a number pattern in the form of a triangular arrangement.
    We are asked to find the binomial expansion of left parenthesis x plus y right parenthesis to the power of 6  using Pascal’s formula.
    Step 1 of 1:
    The Pascal’s triangle is given by:

    From the triangle, the seventh line would give the coefficients of the expansion of the polynomialleft parenthesis x plus y right parenthesis to the power of 6.
    Thus, we have:
    left parenthesis x plus y right parenthesis to the power of 6 equals x to the power of 6 plus 6 x to the power of 5 y plus 15 x to the power of 4 y squared plus 20 x cubed y cubed plus 15 x squared y to the power of 4 plus 6 x y to the power of 5 plus y to the power of 6
    Note:
    The expansion of the polynomial left parenthesis x plus y right parenthesis to the power of 6 can be found using the values of 6 C subscript r

    Use Pascal triangle to expand left parenthesis x plus y right parenthesis to the power of 6

    Maths-General
    ANSWER:
    Hint:
    Hint:
    Pascal's Triangle is a method to know the binomial coefficients of terms of binomial expression (x + y)n, where n can be any positive integer and x,y are real numbers. Pascal Triangle is represented in a triangular form, it is kind of a number pattern in the form of a triangular arrangement.
    We are asked to find the binomial expansion of left parenthesis x plus y right parenthesis to the power of 6  using Pascal’s formula.
    Step 1 of 1:
    The Pascal’s triangle is given by: