# How to Factor Quadratic Equations

Factoring (Factorizing in the UK) quadratic equations is one way of finding the roots of a quadratic equation. A quadratic equation in the standard form ax2 + bx + c = 0 is factored as the product of two linear factors (x – k)(x – h); here, h and k are the two roots.

There are different methods by which we can factor quadratic equations:

### Factoring out the GCF

The simplest form of factoring the quadratics is taking the common factor out of the equation. Let us consider an example to understand the concept.

Solve the quadratic equation 4y2 + 16y = 0 by factoring the GCF

Solution:

In the given equation 4y2 + 16y = 0
4 and 16 have GCF 4
Thus, the equation can be written as follows:
4(y2 + 4y) = 0
Again, y2, y have GCF y
Thus, the equation can be further written as:
4y(y + 4) = 0
Thus, (4y)(y + 4) are the factors of the given quadratic equation.

### Splitting the Middle Term

To factor a quadratic equation by splitting the middle term, follow the following steps:
Step 1: Find two numbers whose product will give ac, and their addition gives b

Thus, number 1 × number 2 = ac and number 1 + number 2

Step 2: Spilt the middle term using the two numbers selected in step 1

ax2 + (number 1)x + (number 2)x + c = 0

Step 3: Take the common factor out and simplify

Let us factor a quadratic equation by splitting its middle term.

Solve the quadratic equation x2 – 2x – 15 = 0 by splitting the middle term.

Solution:

In the quadratic equation x2 – 2x – 15 = 0, a = 1, b = -2 and c = -15
Thus, ac = 1 × (-15) = -15
Factors of 15: 1, 3, 5, 15
The two factors that have their sum -2 and product -15 are 5 and 3
Sum of the factors = (-5) + 3 = -2
Product of the factors = (-5) × 3 = -15
Splitting the middle term,
x2 – 5x + 3x – 15 = 0
=> x(x – 5) + 3(x – 5) = 0
=> (x – 5)(x + 3) = 0
Thus, (x – 5)(x + 3) are the factors of the given quadratic equation.
Thus, the two roots are:
x – 5 = 0 or x + 3 = 0
=> x = {5, -3}

Here, we will directly use the quadratic formula:

x = ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

Substituting the values of a, b, and c and simplifying the expression, we get the values of the roots. We will get the two factors of the given quadratic equation from the roots.

Let us consider an example using the above formula.

Solve the quadratic equation x2 + 7x + 12 = 0 by using the quadratic formula.

Solution:

In the quadratic equation x2 + 7x + 12 = 0, a = 1, b = 7 and c = 12
Substituting the values of a, b, and c in the formula, we get
x = ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
= ${x =${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$}$
= ${\dfrac{-7\pm \sqrt{49-48}}{2\times 1}}$
= ${\dfrac{-7\pm \sqrt{1}}{2}}$
= ${\dfrac{-7+1}{2}}$ or ${\dfrac{-7-1}{2}}$
= {-3, -4}
Thus, (x + 3)(x + 4) are the two factors of the given quadratic equation.

### Using Algebraic Identities

3 algebraic identities can be utilized to find the factors of a quadratic equation. The 3 identities are:

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

Let us factor different algebraic equations using each of the above identities.

Factorize x2 + 6x + 9 = 0

Solution:

Since the equation x2 + 6x + 9 = 0is in the form (a + b)2, let us break the given expression accordingly
x2 + 6x + 9 = 0
=> x2 + 2× x × 3 + 32 = 0
=> (x + 3)2 = 0
Thus, (a + 3) (a + 3) are the two factors of the given quadratic equation.

Factorize a2 – 14a + 49 = 0

Solution:

Since the equation a2 – 14a + 49 = 0 is in the form (a – b)2, let us break the given expression accordingly,
a2 – 14a + 49 = 0
=> a2 – 2 × a × 7 + 72 = 0
=> (a – 7)2
Thus, (a – 7) (a – 7) are the two factors of the given quadratic equation.

Factorize 49a2 – 36b2 = 0

Solution:

Since the equation 49a2 – 36b2 = 0is in the form a2 – b2, let us break the given expression accordingly,
49a2 – 36b2 = 0
=> (7 × 7 × a × a) – (6 × 6 × b × b) = 0
=> (7a)2 – (6b)2
Thus, (7a + 6) (7a – 6) are the two factors of the given quadratic equation.

### Using Symmetry

Quadratic Equations have symmetry. The left and right parts are mirror images.

The midline is fixed at ${-\dfrac{b}{2}}$ and w is calculated with the following steps:

Step 1: a will always be 1. If not 1, divide b and c by a

b = b/a and c = c/a

Step 2: midline is at ${-\dfrac{b}{2}}$

Step 3: ${W=\sqrt{mid^{2}-c}}$

Steps 4: Roots are mid + w and mid – w

Factorize p2 + 3p – 4 = 0

Solution:

In the given equation p2 + 3p – 4 = 0, a = 1, b = 3, c = -4
Since a = 1, we can go to the next step
mid = -3/2
Thus, w = ${w=\sqrt{\left[ \left( \dfrac{3}{2}\right) ^{2}-\left( -4\right) \right] }}$
= ${\sqrt{\dfrac{9}{4}+4}}$
= ${\sqrt{\dfrac{25}{4}}}$
= ${\dfrac{-3}{2}-\dfrac{5}{2}}$ and ${\dfrac{-3}{2}+\dfrac{5}{2}}$
= {-4, 1} Thus, (p + 4) (p – 1) are the two factors of the given quadratic equation.