Answer:

x2 – 5x – 6

Step-by-step explanation:

x

For this reason, factorization is a fundamental step towards solving any equation in mathematics. Let’s find out.

How to Factor a Quadratic Equation?

Factoring a quadratic equation can be defined as the process of breaking the equation into the product of its factors. In other words, we can also say that factorization is the reverse of multiplying out.

To solve the quadratic equation ax 2 + bx + c = 0 by factorization, the following steps are used:

Expand the expression and clear all fractions if necessary.

Move all terms to the left-hand side of the equal to sign.

Factorize the equation by breaking down the middle term.

Equate each factor to zero and solve the linear equations

Example 1

Solve: 2(x 2 + 1) = 5x

Solution

Expand the equation and move all the terms to the left of the equal sign.

⟹ 2x 2 – 5x + 2 = 0

⟹ 2x 2 – 4x – x + 2 = 0

⟹ 2x (x – 2) – 1(x – 2) = 0

⟹ (x – 2) (2x – 1) = 0

Equate each factor equal to zero and solve

⟹ x – 2 = 0 or 2x – 1 = 0

⟹ x = 2 or x = 1212

Therefore, the solutions are x = 2, 1/2.

Example 2

Solve 3x 2 – 8x – 3 = 0

Solution

3x 2 – 9x + x – 3 = 0

⟹ 3x (x – 3) + 1(x – 3) = 0

⟹ (x – 3) (3x + 1) = 0

⟹ x = 3 or x = -13

Example 3

Solve the following quadratic equation (2x – 3)2 = 25

Solution

Expand the equation (2x – 3)2 = 25 to get;

⟹ 4x 2 – 12x + 9 – 25 = 0

⟹ 4x 2 – 12x – 16 = 0

Divide each term by 4 to get;

⟹ x 2 – 3x – 4 = 0

⟹ (x – 4) (x + 1) = 0

⟹ x = 4 or x = -1

The are many methods of factorizing quadratic equations. In this article, our emphasis will be based on how to factor quadratic equations, in which the coefficient of x2 is either 1 or greater than 1.

Therefore, we will use the trial and error method to get the right factors for the given quadratic equation.

Factoring when the Coefficient of x 2 is 1

To factorize a quadratic equation of the form x 2 + bx + c, the leading coefficient is 1. You need to identify two numbers whose product and sum are c and b, respectively.

CASE 1: When b and c are both positive

Example 4

Solve the quadratic equation: x2 + 7x + 10 = 0

List down the factors of 10:

1 × 10, 2 × 5

Identify two factors with a product of 10 and a sum of 7:

1 + 10 ≠ 7

2 + 5 = 7.

Verify the factors using the distributive property of multiplication.

(x + 2) (x + 5) = x2 + 5x + 2x + 10 = x2 + 7x + 10

The factors of the quadratic equation are:(x + 2) (x + 5)

Equating each factor to zero gives;

x + 2 = 0 ⟹x= -2

x + 5 = 0 ⟹ x = -5

Therefore, the solution is x = – 2, x = – 5

Example 5

x 2 + 10x + 25.

Solution

Identify two factors with the product of 25 and sum of 10.

5 × 5 = 25, and 5 + 5 = 10

Verify the factors.

x 2 + 10x + 25 = x 2 + 5x + 5x + 25

= x (x + 5) + 5x + 25

= x (x + 5) + 5(x + 5)

= (x + 5) (x + 5)

Therefore, x = -5 is the answer.

CASE 2: When b is positive and c is negative

Example 6

Solve x2 + 4x – 5 = 0

Solution

Write the factors of -5.

1 × –5, –1 × 5

Identify the factors whose product is – 5 and sum is 4.

1 – 5 ≠ 4

–1 + 5 = 4

Verify the factors using the distributive property.

(x – 1) (x + 5) = x2 + 5x – x – 5 = x2 + 4x – 5

(x – 1) (x + 5) = 0

x – 1 = 0 ⇒ x = 1, or

x + 5 = 0 ⇒ x = -5

Therefore, x = 1, x = -5 are the solutions.

CASE 3: When b and c are both negative

Example 7

x2 – 5x – 6

Solution

Write down the factors of – 6:

1 × –6, –1 × 6, 2 × –3, –2 × 3

Now identify factors whose product is -6 and sum is –5:

1 + (–6) = –5

Check the factors using the distributive property.

(x + 1) (x – 6) = x2 – 6 x + x – 6 = x2 – 5x – 6

Equate each factor to zero and solve to get;

(x + 1) (x – 6) = 0

x + 1 = 0 ⇒ x = -1, or

x – 6 = 0 ⇒ x = 6

Therefore, the solution is x=6, x = -1