Question
Use binomial theorem to expand (s2 + 3)5
Hint:
The binomial expansion is 
, here n ≥ 0 . We are asked to use binomial theorem to expand (s2 + 3)5 .
The correct answer is: 243
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:
Thus, the expansion is: 
Thus, the expansion is:
For the expansion of an expression (x + y)n , we would have n+1 terms. This is something you need to keep in mind.
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Use binomial theorem to expand (x - 3)4
For the expansion of an expression (x + y)n , we would have n+1 terms. This is something you need to keep in mind.
Use binomial theorem to expand (x - 3)4
For the expansion of an expression (x + y)n , we would have n+1 terms. This is something you need to keep in mind.
Use polynomial identities to factor the polynomials or simplify the expressions :
Use polynomial identities to factor the polynomials or simplify the expressions :
How is 
obtained from 
How is obtained from
Use Pascal triangle to expand (x + y)6
The expansion of the polynomial (x + y)6 can be found using the values of 6Cr
Use Pascal triangle to expand (x + y)6
The expansion of the polynomial (x + y)6 can be found using the values of 6Cr
Use polynomial identities to multiply the expressions ?
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Use Pascal triangle to expand
The answer cam also be found by expanding the formula of 5Cr .
Use Pascal triangle to expand
The answer cam also be found by expanding the formula of 5Cr .
Use polynomial identities to multiply the expressions ?
Use polynomial identities to multiply the expressions ?
How can you use polynomial identities to factor polynomials and simplify numerical expressions ?
How can you use polynomial identities to factor polynomials and simplify numerical expressions ?
Use polynomial identities to multiply the expressions ?
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How is ( x + y )n obtained from (x + y)n-1
You could also get the value of from by just multiplying a ( x + y) with .
How is ( x + y )n obtained from (x + y)n-1
You could also get the value of from by just multiplying a ( x + y) with .
