Do girlfriend have any kind of idea about the factorization the polynomials? since you now have some basic information about polynomials, we will learn exactly how to deal with quadratic polynomials by factorization.

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First the all, let’s take it a quick review of the quadratic equation. A quadratic equation is a polynomial of a second degree, normally in the kind of f(x) = ax2 + bx + c where a, b, c, ∈ R, and a ≠ 0. The hatchet ‘a’ is described as the leading coefficient, if ‘c’ is the absolute term the f (x).

Every quadratic equation has two values of the unknown variable, usually well-known as the root of the equation (α, β). We can achieve the roots of a quadratic equation by factoring the equation.

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

How to aspect a Quadratic Equation?

Factoring a quadratic equation deserve to be characterized as the process of break the equation into the product the its factors. In various other words, us can additionally say that factorization is the turning back of multiplying out.

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

Expand the expression and also clear every fractions if necessary.Move every terms to the left-hand side of the equal to sign.Factorize the equation through breaking down the middle term.Equate each factor to zero and solve the direct equations

Example 1

Solve: 2(x 2 + 1) = 5x

Solution

Expand the equation and move every the terms to the left that the same 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 remedies 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 adhering to 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 every term by 4 to get;

⟹ x 2 – 3x – 4 = 0

⟹ (x – 4) (x + 1) = 0

⟹ x = 4 or x = -1

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

Therefore, we will use the trial and also error technique to acquire the right factors for the given quadratic equation.

Factoring once the Coefficient of x 2 is 1

To factorize a quadratic equation that the type x 2 + bx + c, the leading coefficient is 1. You need to determine two numbers whose product and also sum room c and also b, respectively.

CASE 1: once b and also c space both positive

Example 4

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

List under the components of 10:

1 × 10, 2 × 5

Identify two determinants with a product that 10 and a sum of 7:

1 + 10 ≠ 72 + 5 = 7.

Verify the factors using the distributive property of multiplication.

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

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

Equating each aspect to zero gives;

x + 2 = 0 ⟹x= -2

x + 5 = 0 ⟹ x = -5

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

Example 5

x 2 + 10x + 25.

Solution

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

5 × 5 = 25, and also 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: once b is positive and c is negative

Example 6

Solve x2 + 4x – 5 = 0

Solution

Write the determinants 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 components 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, orx + 5 = 0 ⇒ x = -5

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

CASE 3: as soon as b and also c space both negative

Example 7

x2 – 5x – 6

Solution

Write down the factors of – 6:

1 × –6, –1 × 6, 2 × –3, –2 × 3

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

1 + (–6) = –5

Check the determinants using the distributive property.

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

Equate each aspect to zero and solve to get;(x + 1) (x – 6) = 0

x + 1 = 0 ⇒ x = -1, orx – 6 = 0 ⇒ x = 6

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

CASE 4: once b is negative and c is positive

Example 8

x2 – 6x + 8 = 0

Solution

Write under all components of 8.

–1 × – 8, –2 × –4

Identify components whose product is 8 and sum is -6–1 + (–8) ≠ –6–2 + (–4) = –6

Check the components using the distributive property.

(x – 2) (x – 4) = x2 – 4 x – 2x + 8 = x2 – 6x + 8

Now equate each aspect to zero and also solve the expression come get;

(x – 2) (x – 4) = 0

x – 2 = 0 ⇒ x = 2, orx – 4 = 0 ⇒ x = 4

Example 9

Factorize x2 +8x+12.

Solution

Write under the factors of 12;

12 = 2 × 6 or = 4 × 3Find factors whose sum is 8:

2 + 6 = 82 × 6 ≠ 8

Use distributive property to check the factors;

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

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

Equate each factor to zero to get;

(x + 6) (x + 2)

x = -6, -2

Factoring as soon as the coefficient that x 2 is higher than 1

Sometimes, the top coefficient the a quadratic equation might be greater than 1. In this case, we can not resolve the quadratic equation through the usage of usual factors.

Therefore, we require to take into consideration the coefficient the x2 and also the factors of c to discover numbers whose amount is b.

Example 10

Solve 2x2 – 14x + 20 = 0

Solution

Determine the common factors of the equation.

2x2 – 14x + 20 ⇒ 2(x2 – 7x + 10)

Now us can find the factors of (x2 – 7x + 10). Therefore, write down factors of 10:

–1 × –10, –2 × –5

Identify factors whose amount is – 7:

1 + (–10) ≠ –7–2 + (–5) = –7

Check the factors by applying distributive property.

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2(x – 2) (x – 5) = 2(x2 – 5 x – 2x + 10)= 2(x2 – 7x + 10) = 2x2 – 14x + 20

Equate each aspect to zero and solve;2(x – 2) (x – 5) = 0

x – 2 = 0 ⇒ x = 2, orx – 5 = 0 ⇒ x = 5

Example 11

Solve 7x2 + 18x + 11 = 0

Solution

Write under the factors of both 7 and 11.

7 = 1 × 7

11 = 1 × 11

Apply distributive property to check the factors as displayed below:

(7x + 1) (x + 11) ≠ 7x2 + 18x + 11

(7x + 11) (x + 1) = 7x2 + 7x + 11x + 11 = 7x2 + 18x + 11

Now equate each variable to zero and solve to get;

7x2 + 18x + 11= 0(7x + 11) (x + 1) = 0

x = -1, -11/7

Example 12

Solve 2x2 − 7x + 6 = 3

Solution

2x2 − 7x + 3 = 0

(2x − 1) (x − 3) = 0

x=1/2​ or x=3

Example 13

Solve 9x 2 +6x+1=0

Solution

Factorize come give:

(3x + 1) (3x + 1) = 0

(3x + 1) = 0,

Therefore, x = −1​/3

Example 14

Factorize 6x2– 7x + 2 = 0

Solution

6x2 – 4x – 3x + 2 = 0

Factorize the expression;

⟹ 2x (3x – 2) – 1(3x – 2) = 0

⟹ (3x – 2) (2x – 1) = 0

⟹ 3x – 2 = 0 or 2x – 1 = 0

⟹ 3x = 2 or 2x = 1

⟹ x = 2/3 or x = ½

Example 15

Factorize x2 + (4 – 3y) x – 12y = 0

Solution

Expand the equation;

x2 + 4x – 3xy – 12y = 0

Factorize;

⟹ x (x + 4) – 3y (x + 4) = 0

x + 4) (x – 3y) = 0

⟹ x + 4 = 0 or x – 3y = 0

⟹ x = -4 or x = 3y

Thus, x = -4 or x = 3y

Practice Questions

Solve the complying with quadratic equations by factorization:

3x 2– 20 = 160 – 2x 2(2x – 3) 2 = 4916x 2 = 25(2x + 1) 2 + (x + 1) 2 = 6x + 472x 2+ x – 6 = 03x 2 = x + 4(x – 7) (x – 9) = 195x 2– (a + b) x + abdominal muscle = 0x2+ 5x + 6 = 0x2− 2x − 15 = 0

Answers

6, -6-2, 5– 5/4, 5/4-3, 3-2, 3/2-1, 4/3-6, 22a, b–3, –25, − 3Previous Lesson | Main web page | next Lesson