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Solving Quadratic Equations by Factoring

solving quadratic equations by factoring

The way by which you express any given polynomial as a product of its linear elements is called factoring the quadratics. It is a strategy for addressing issues by reducing quadratic equations and discovering their roots. A quadratic polynomial is ax2 + bx + c, where a, b, and c are all positive integers. The quadratic formula factoring approach is used to find the quadratic equation zeros of the equation ax2 + bx + c = 0.

What is meant by solving quadratic equations by factoring?

The process of presenting any given polynomial equation as the product of its linear roots is called the method of factoring quadratics. The given polynomial is a quadratic equation in the form of ax2 + bx + c = 0. The roots of the polynomial equation can be expressed in the form of (x – k) (x – h), where the variables h and k are the calculated roots of the equation. This type of method is frequently referred to as the quadratic equation factorization method. You can do factorization of quadratic equations in various methods, such as dividing the core term, by the application of the quadratic equation formula, using the methodology of completing the squares, and so on.

Meaning of factoring the quadratics

The factor theory connects any polynomial’s linear factors and zeros. Let’s call every quadratic equation two roots because the power of the quadratic equation is 2, or you can say its degree is 2; let’s call them α and β. They are the quadratic equation’s zeros. Assume the quadratic equation f(x) = 0, where f(x) is an order 2 quadratic. Assume that x = α is one of the equation’s roots. As a result, x = α is a 0 of the quadratic equation f(x). As a result, (x – α) is a factor of f(x).

Likewise, if x = β is another root of f(x) = 0, then x = β is a zero of f(x). As a result, (x – β) is a factor of f(x). As a result, factorization of the quadratics is a way of representing quadratic equations as a multiplication of their linear factors, f(x) = (x – α) (x – β). 

Let us solve the given quadratic equation x2 + 5x + 6 = 0.

You know if -3 and -2 are said to be the roots of the above-given equation, check if the value equals 0 by replacing the roots in the provided equation.

Let’s start by calculating factor 1: -3 is the first factor

-3 = x

(x + 3) = 0

Start with the LHS

put x = -3 in the equation x2 + 5x + 6

= (-3)2 + (5 × -3) + 6

= 9 -15 + 6 = 0

which is equal to the RHS

Let’s calculate factor 2: -2 is the second factor

-2 = x

(x + 2) = 0

Start with the LHS

put x = -2 in the equation x2 + 5x + 6

= (-2)2 + (5 × -2) + 6

= 4 -10 + 6 = 0

which is equal to the RHS

Hence, it is proved that the equation has 2 factors which are (x + 3) and (x + 2)

Consider the expression x2 – 16 = 0.

The equation’s two roots are 4 and -4. Check the value by replacing the roots in the above equation and seeing if it equals 0.

Check the value by replacing the roots in the above equation and seeing if it equals 0.

42 – 16 = 16 – 16 = 0

(-4)2 – 16 = 16 – 16 = 0

As a result, the equation contains two components (x+4) and (x-4)

How to solve quadratic equations by factoring – Quadratic Factoring Methods

Factoring quadratics yields the quadratic equation’s roots. Factoring quadratic equations can be accomplished using a variety of techniques. There are four methods for factoring quadratics:

  • Taking the GCD (greatest common division) into account
  • Breaking up the middle phrase
  • Making use of completing square methodology which uses algebraic identities 
  • Using the quadratic equation

Factoring Quadratics by Excluding The GCD

Finding the general numeric and algebraic factors held in common by the elements in the quadratic equation and then removing them is how quadratics are factored. Now let’s perform an instance to understand better factoring quadratic equations by removing and evaluating the GCD.

Consider the following quadratic equation: 3x2 + 6x = 0. In both cases, the numerical component is 3 (coefficient of x2). In both cases, the common algebraic element is x. 3 and x are the common factors. As a result, we remove them. As a result, 3x2 + 6x = 0 may be factored as 3x(x + 2) = 0.

Factoring Quadratics: Breaking the Middle Term in Two Parts

 -b/a is the summation of the solutions of the quadratic equation ax2 + bx + c = 0.

c/a denotes the multiplication of the quadratic equation’s roots ax2 + bx + c = 0.

Whenever we attempt to factorize quadratic equations, we separate the middle term b in the quadratic equation ax2 + bx + c = 0. We find the component pairings of the combination of a and c, whose total is equal to b.

For instance, f(x) = x2 + 8x + 12

Divide the middle element 8x so that the components of the product of 1 and 12 total up to 8. The factor pairs of 12 can be summarized as (1, 12), (3, 4), and (2, 6). We could see that the component pair (2, 6) meets our requirements because the summation of 6 and 2 is 8, and the pair multiplication is 12. As a result, we divide the center term and rewrite the quadratic equation as follows: x2 + 8x + 12 = 0

As a result, we divide the center term and rewrite the quadratic equation as follows: x2 + 8x + 12 = 0

Now combine the terms as follows:

(x2 + 6x) + (2x + 12) = 0

x(x + 6) + 2(x + 6) = 0

Dropping away the common element (x + 6) yields (x + 2) (x + 6) = 0.

Thus, the elements of x2 + 8x + 12 = 0 are (x + 2) and (x + 6).

Identities for Quadratic Factoring

Factoring quadratics can be accomplished by finishing the squares, which necessitates the application of algebraic identities.

The following are the key algebraic identities that are utilized to complete the squares:

(a – b)2 = a2 – 2ab + b2

(a + b)2 = a2 + 2ab + b2

The following are the steps to factoring the quadratic equation ax2 + bx + c = 0 using the finishing the squares method:

  • Step 1: Reduce both sides of the quadratic equation ax2 + bx + c = 0 by dividing the equation by a. The resulting equation is x2 + (b/a) x + c/a = 0.
  • Step 2: Take c/a from both sides of the quadratic equation x2 + (b/a) x + c/a = 0. The resulting equation is x2 + (b/a) x = -c/a.
  • Step 3: Take the square of (b/2a) and multiply it by both sides of the  equation x2 + (b/a)x = -c/a. The equation found is x2 + (b/a) x + (b/2a)2 = -c/a + (b/2a) 2
  • Step 4: The LHS of the quadratic equation x2 + (b/a) x + (b/2a)2 = -c/a + (b/2a)2 may now be represented as a full square, and the RHS can be simplified if required. Resultant equation is (x + b/2a)2 = -c/a + (b/2a)2
  • Step 5: We may acquire the provided quadratic equation’s roots and generate the equation’s components. Another algebraic identity used for quadratic factoring is a2 – b2 = (a + b) (a – b).

Important Notes on Quadratic Factoring

  • Linear factors have the formula ax + b and cannot be factored further.
  • A polynomial that is of degree two is called a quadratic polynomial.
  • -b/a is the summation of the roots of the quadratic equation ax2 + bx + c = 0.
  • c/a denotes the multiplication of the roots in the quadratic equation ax2 + bx + c = 0.

Solving quadratic equations by factoring examples

The given examples consist of solving quadratic equations by factoring answers as well. Make sure to thoroughly go through them. 

Example 1: f(x) = 16x2 – 25 (difference of 2 perfect squares)

Solution: 16x2 – 25 = (4x)2 – 52

Noticing the equation we can identify it in the form of a2 – b2 = (a + b)(a – b)

Hence, we factorize the equation 16x2 – 25 = 0 as (4x+5) (4x-5)

16x2 – 25 = (4x+5) (4x-5)

Example 2: Ram and Shyam share 45 chips. After dropping 5 chips each, the product of their current marble counts is 124. How to calculate how many marbles they had to begin with.

Solution: Assume Ram possessed x marbles.

The amount of chips Shyam had then Equaled 45 – x.

Ram’s remaining chips after losing 5 chips Equals x – 5

The amount of chips Shyam has left after losing 5 chips is 45 – x – 5 = 40 – x.

The number of chips multiplied by the number of chips equals 124.

(x – 5) (40 – x) = 124

40x – x2 – 200 + 5x = 124

– x2 + 45x – 200 = 124

x2 – 45x + 324 = 0

This is a representation of the quadratic equation. As a result of solving the given equation for x, we have;

x = 36, x = 9

So, the answer you get is Ram had 36 chips while Shyam had 9 chips or vice versa.

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