# How to Solve a Quadratic Equation by Factoring?

Factoring quadratics is a method of expressing the polynomial as a product of its linear factors. In this step-by-step guide, you learn about the fascinating concept of factoring quadratics.

A quadratic polynomial is of the form \(ax^2+bx+c\), where \(a, b, c\) are real numbers. Factoring quadratics is a method that helps us to find the zeros of the quadratic equation \(ax^2+bx+c=0\).

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## A step-by-step guide to factoring quadratics

Factoring quadratics is a method of expressing the quadratic equation \(ax^2+bx+c = 0\) as a product of its linear factors as \((x – k)(x – h)\), where \(h, k\) are the roots of the quadratic equation. This method is also called the method of factorization of quadratic equations.

### Methods of factoring quadratics

Different methods can be used for factoring quadratic equations. Factoring quadratics is done in four ways:

**Factoring out the GCD:**

Factoring quadratics can be done by finding the common numeric factor and the algebraic factors shared by the expressions in the quadratic equation and then taking them out. Let’s solve an example to understand the factoring quadratic equations by taking the GCD out.

**Splitting the middle term**

The sum of the roots of the quadratic equation \(ax^2+bx+c=0\) is given by \(α+β=-\frac{b}{a}\). And the product of the roots in the quadratic equation \(ax^2+bx+c=0\) is given by \(αβ = \frac{c}{a}\).

When we want to factorize quadratic equations, we split the middle term b of the quadratic equation \(ax^2+bx+c=0\). We determine the factor pairs of the product of \(a\) and \(c\) so that their sum is equal to \(b\).

**Using algebraic identities**

The quadratic factorization process can be done by completing squares that require the use of algebraic identities. The main algebraic identities used to complete squares are:

- \(\color{blue}{(a+b)^2=a^2+2ab+b^2}\)

- \(\color{blue}{(a-b)^2=a^2-2ab+b^2}\)

- \(\color{blue}{(a^2-b^2)=(a+b)(a-b)}\)

The steps of factorization of the quadratic equation \(ax^2+bx+c=0\) using the squares method are:

- Divide both sides of the quadratic equation \(ax^2+bx+c=0\) by \(a\). Now, the obtained equation is \(x^2+ (\frac{b}{a}) x + \frac{c}{a}= 0\).
- Subtract \(\frac{c}{a}\) from both the sides of quadratic equation \(x^2 + (\frac{b}{a})x + \frac{c}{a} =0\). Obtained equation is \(x^2+ (\frac{b}{a})x = -\frac{c}{a}\).
- Add the square of \((\frac{b}{2a}\) to both the sides of quadratic equation \(x^2+ (\frac{b}{a})x = -\frac{c}{a}\). Obtained equation is \(x^2+ (\frac{b}{a})x + (\frac{b}{2a})^2= -\frac{c}{a} + (\frac{b}{2a})^2\).
- Now the LHS of the quadratic equation \(x^2 + (\frac{b}{a})x + (\frac{b}{2a})^2= -\frac{c}{a}+ (\frac{b}{2a})^2\) can be written as a complete square and simplify the RHS, if necessary. Obtained equation is \((x + \frac{b}{2a})^2= -\frac{c}{a}+(\frac{b}{2a})^2\).
- The roots of the given quadratic equation can be obtained and, we can form the factors of the equation.

**Using quadratic formula**

Quadratic factorization is also done using a formula that gives us the roots of the quadratic equation and thus the factors of the equation. If \(ax^2+bx+c\) is a quadratic equation, \(a\) is the coefficient of \(x^2\), \(b\) is the coefficient of \(x\) and \(c\) is the constant term. Then we find the value of \(x\) by using the formula:

\(\color{blue}{x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}\)

### Factoring Quadratics – Example 1:

Find the factors of the quadratic equation \(x^2+ x- 12= 0\).

**Solution**:

Split the middle term of the quadratic equation \(x^2+ x- 12= 0\) to determine its factors.

\(X^2+ x -12= 0\)

\(x^2+4x-3x-12= 0\)

\(x(x + 4)- 3(x + 4)=0\)

\((x – 3)(x + 4)=0\)

Now, the factors of \(x^2+x-12 = 0\) are \((x-3)\) and \((x + 4)\).

## Exercises for Factoring Quadratics

### Find the factors of each quadratic equation using the factoring quadratics method.

- \(\color{blue}{x^2-14x+40}\)
- \(\color{blue}{6x^2+37x+6}\)
- \(\color{blue}{3x^2+16x-12}\)
- \(\color{blue}{6x^2-7x-20}\)

- \(\color{blue}{(x-4)(x-10)}\)
- \(\color{blue}{(6x+1)(x+6)}\)
- \(\color{blue}{(3x-2)(x+6)}\)
- \(\color{blue}{(3x+4)(2x-5)}\)

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lalalavigne28–The answer to #1 should be positive 10 & positive 4, not -10/-4

lalalavigne28–I apologize, I’m wrong