We encounter quadratic expressions and equations in a variety of problems in maths, physics, and engineering. In almost every case, we have to factorize it. The most common approach is to split the middle term. Another way is to find its root using the quadratic formula. In this article, we factor it by writing the middle term as a sum of two numbers. The key is to find these numbers. We will break the process into four simple steps. But let us first define a quadratic expression for our use.

**✩** Standard Form of a Quadratic Expression

$Ax_{2}+Bx+CA=0$

where A, B, C are real numbers

Let’s factor 2x^{2} − x − 6 by splitting the middle term.

## Step 1: Identify A, B, and C

To identify A, B and C, convert it into the form : Ax^{2} + Bx + C

The quadratic expression now looks like this:

2x^{2} + ( − 1)x + ( − 6)

We can see that A = 2, B = − 1, C = − 6

Multiply A with C to get the product AC.

AC = 2 × ( − 6) = − 12

Do not forget the sign during multiplication.

## Step 2: List pairs of numbers whose product = AC

We find the factors of AC = 12 (ignoring the sign for the moment). These are 1, 2, 3, 4, 6, and 12.

Next, we use factors of AC to create pairs of numbers whose product is -12 (=AC). We use negative numbers as required. Let us list these pairs:

- (1, -12) as 1 x (-12) = -12
- (-1, 12) as (-1) X 12 = -12
- (2, -6) as 2 X (-6) = -12
- (-2, 6) as (-2) x 6 = -12
- (3, -4) as 3 X (-4) = -12
- (-3, 4) as (-3) X 4 = -12

## Step 3: Choose a pair whose sum = B

We choose a pair whose numbers add up to B. Let us calculate the sum of the numbers of each pair:

- 1 + (-12) = -11
- (-1) + 12 = 11
- 2 + (-6) = -4
- (-2) + 6 = 4
**3 + (-4) = -1 ( = B )**- (-3) + 4 = 1

As you can see, (3, -4) satisfies this condition. We split B (or -1 in our case) with the sum of this factor pair. The rest is simple algebra, as you will see in a minute.

## Step 4: Split the middle term (B) and factor

Original expression:

2x^{2} + ( − 1)x + ( − 6)

Split B ( = -1 ) by the sum of Factor Pair [ 3 + (- 4) ]. *The order of 3 and -4 does not matter. Try to factorize the quadratic by reversing it.*

= 2x^{2} + [3 + ( − 4)]x − 6

= 2x^{2} + 3x + ( − 4)x − 6

= 2x^{2} + 3x − 4x − 6

Taking out x as a common factor of the first two terms:

= x(2x + 3) − 4x − 6

Taking out -2 as the common factor of the last two terms:

= x(2x + 3) − 2(2x + 3)

Taking out (2x+3) as a common factor:

= (2x + 3)(x − 2)

You have the answer!

## Example

Factorize: x^{2} + x − 30

**Step1: Identify A, B, and C**

x^{2} + x − 30

Writing in the standard form we get:

(1)x^{2} + (1)x + ( − 30)

Here A = 1, B = 1, C = -30

**Step 2: List pairs of numbers whose product = AC**

AC = 1 x (-30) = -30

The factors of 30 : 1, 2, 3, 5, 6, 10, 15, 30

Pairs with the product as AC (=-30):

- (1, -30) and (-1, 30)
- (2, -15) and (-2, 15)
- (3, -10) and (-1, 10)
- (5, -6) and (-5, 6)

**Step 3: Choose a pair whose sum = B**

The only pair whose numbers add up to B is ( 6, -5 ) as 6 + (-5) = 1 = B.

**Step 4: Split the middle term (B) and factor**

x^{2} + x − 30

We split the middle term using the pair above:

= x^{2} + [6 + ( − 5)]x − 30

= x^{2} + [6 − 5]x − 30

= x^{2} + 6x − 5x − 30

= x(x + 6) − 5x − 30

= x(x + 6) − 5(x + 6)

= (x + 6)(x − 5)

There it is!

## Practice Questions

### 1 Question

Factorize 7x^{2} − 2x − 5

#### Answer

Can you identify coefficients A, B, and C?

### 2 Question

Factorize 4x^{2} − 16

#### Answer

Can you identify A, B, and C?

### 3 Question

Factorize 3x^{2} − 4x − 4

#### Answer

Can you identify A, B, and C?

### 4 Question

Factorize −2x^{2} + 3x + 9

#### Answer

Can you identify the coefficients A, B, and C?

### 5 Question

Factorize −x^{2} + 7x + 18

#### Answer

Can you find the coefficients A, B, and C?

## Answers

### 1 Answer

The standard form of Quadratic expression is Ax^{2} + Bx + C.

We can rewrite original expression as: 7x^{2} + ( − 2)x + ( − 5)

Therefore A = 7, B = -2, C = -5

AC = 7 × ( − 5) = − 35

Ignoring the sign, we get 35. Factors of 35 are 1, 5, 7, 35.

We want the product to equal AC (=-35). This is done by making one number of the pair negative. We get the following pairs:

( − 1, 35)( − 1) × 35 = − 35

(1, − 35)1 × ( − 35) = − 35

( − 5, 7)( − 5) × 7 = − 35

(5, − 7)5 × ( − 7) = − 35

The sum of 5 and -7 is -2 (=B). We choose the pair ( 5, -7 ) for splitting the coefficient B.

7x^{2} + ( − 2)x + ( − 5)

We split the middle term using the selected pair.

= 7x^{2} + ( − 7 + 5)x + ( − 5)

= 7x^{2} − 7x + 5x − 5

Taking 7x and 5 as common:

= 7x(x − 1) + 5(x − 1)

Taking out (x − 1) as common:

= (x − 1)(7x + 5)

### 2 Answer

Rewrite original expression as: 4x^{2} + (0)x + ( − 16)

So, A = 4, B = 0, C = -16

AC = 4 × ( − 16) = − 64

Factors of 64, ignoring the sign are: 1, 2, 4, 8, 16, 32, 64

List of factors pairs whose product = -64:

( − 1, 64)[ − 1 × 64 = − 64]

( − 2, 32)[ − 2 × 32 = − 64]

( − 4, 16)[ − 4 × 16 = − 64]

( − 8, 8)[ − 8 × 8 = − 64]

(1, − 64)[1 × − 64 = − 64]

(2, − 32)[2 × − 32 = − 64]

(4, − 16)[4 × − 16 = − 64]

(8, − 8)[8 × − 8 = − 64]

The sum of 8 and -8 is 0 (=B). So we choose this pair.

4x^{2} + (0)x − 16

We split the middle term (0x) using the selected pair.

= 4x^{2} + (8 − 8)x − 16

= 4x^{2} + 8x − 8x − 16

= 4x(x + 2) − 8(x + 2)4x and −8 are common

= (x + 2)(4x − 8)(x + 2) is common

Solved!

### 3 Answer

The standard form of Quadratic expression is Ax^{2} + Bx + C.

Original expression in standard form: 3x^{2} + ( − 4)x + ( − 4)

So, A = 3, B = -4, C = -4

AC = 3 × ( − 4) = − 12

Factors of 12, ignoring the sign are: 1, 2, 3, 4, 6, 12

List of Factors Pairs whose product = -12:

( − 1, 12)[ − 1 × 12 = − 12]

( − 2, 6)[ − 2 × 6 = − 12]

( − 3, 4)[ − 3 × 4 = − 12]

(1, − 12)[1 × − 12 = − 12]

(2, − 6)[2 × − 6 = − 12]

(3, − 4)[3 × − 4 = − 12]

Sum of 2 and -6 is -4 = (B). So we choose this pair.

3x^{2} + ( − 4)x − 4

We split the middle term using this pair and factorize.

= 3x^{2} + [2 + ( − 6)]x − 4

= 3x^{2} + 2x + ( − 6)x − 4

= 3x^{2} + 2x − 6x − 4

= x(3x + 2) − 2(3x + 2)x and −2 are common

= (x − 2)(3x + 2)(x − 2) is common

Solved!

### 4 Answer

The standard form of Quadratic expression is Ax^{2} + Bx + C.

Original expression in standard form: ( − 2)x^{2} + 3x + 9

A = -2, B = 3, C = 9

AC = ( − 2) × 9 = − 18

Factors of 18 (ignoring the negative sign) are: 1, 2, 3, 6, 9, 18.

List of factor pair with product = -18:

( − 1, 18)[ − 1 × 18 = − 18]

( − 2, 9)[ − 2 × 9 = − 18]

( − 3, 6)[ − 3 × 6 = − 18]

( − 6, 3)[ − 6 × 3 = − 18]

( − 9, 2)[ − 9 × 2 = − 18]

(1, − 18)[1 × − 18 = − 18]

The sum of -3 and 6 is 3 (=B). So we choose the pair (-3, 6).

−2x^{2} + 3x + 9

Splitting the middle term (3x) using selected numbers we get:

= − 2x^{2} + [( − 3) + 6]x + 9

= − 2x^{2} + [ − 3x + 6x] + 9

= − 2x^{2} − 3x + 6x + 9

= − x(2x + 3) + 3(2x + 3) − x and 3 are common

= (2x + 3)( − x + 3)(2x + 3) is common

Solved!

### 5 Answer

Step 1 – Identify A, B, and C

The standard form of Quadratic expression is Ax^{2} + Bx + C.

Original expression in standard form: −x^{2} + 7x + 18

So, A = -1, B = 7, C = 18

Value of A × C:

AC = ( − 1) × 18 = − 18

Factors of 18, ignoring the sign are:

1, 2, 3, 6, 9, 18

Factors Pairs with product = AC:

( − 1, 18)[ − 1 × 18 = − 18]

( − 2, 9)[ − 2 × 9 = − 18]

( − 3, 6)[ − 3 × 6 = − 18]

( − 6, 3)[ − 6 × 3 = − 18]

( − 9, 2)[ − 9 × 2 = − 18]

(1, − 18)[1 × − 18 = − 18]

The sum of -2 and 9 is 7 (=B). So we choose the pair ( -2, 9 ).

−x^{2} + 7x + 18

We split the middle term (7x) using the selected pair.

= − x^{2} + [( − 2) + 9]x + 18

= − x^{2} + ( − 2)x + 9x + 18

= − x^{2} − 2x + 9x + 18

= − x(x + 2) + 9(x + 2) − x and 9 are common

= (x + 2)( − x + 9)(x + 2) is common

## Related

**Finding Roots of Quadratic (Equation with Examples, Graphs) ➤**

**Quadratic Function Graph (Illustrations) ➤**