# Factoring Quadratic Equations – Methods & Examples

Do you have any idea about the **factorization of polynomials** ? Since you now have some basic data about polynomials, we will learn how to solve quadratic polynomials by factorization. First of all, let ’ s take a **quick review of the quadratic equation**. A quadratic equation is a polynomial of a second gear degree, normally in the form of farad ( x ) = ax2 + bx + c where a, bel, degree centigrade, ∈ R, and a ≠ 0. The term ‘ a ’ is referred to as the leading coefficient, while ‘ c ’ is the absolute terminus of fluorine ( ten ). Every quadratic equation equation has **two values of the unknown variable,** normally known as the roots of the equation ( α, β ). We can obtain the roots of a quadratic equation equality by factoring the equality. For this reason, **factorization is a fundamental step** towards solving any equality in mathematics. Let ’ s find out.

## How to Factor a Quadratic Equation?

Factoring a quadratic equation can be defined as the work of breaking the equation into the product of its factors. In other words, we can besides say that factorization is the turn back 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 leave of the equal gestural. ⟹ 2x 2 – 5x + 2 = 0 ⟹ 2x 2 – 4x – x + 2 = 0 ⟹ 2x ( ten – 2 ) – 1 ( adam – 2 ) = 0 ⟹ ( ten – 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 ( adam – 3 ) + 1 ( ten – 3 ) = 0 ⟹ ( x – 3 ) ( 3x + 1 ) = 0 ⟹ x = 3 or x = -13 **Example 3** Solve the take after quadratic equation equality ( 2x – 3 ) 2 = 25 solution Expand the equality ( 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 vehemence will be based on how to factor quadratic equations, in which the coefficient of x2 is either 1 or greater than 1. consequently, we will use the trial and error method to get the right factors for the given quadratic equality .

## Factoring when the Coefficient of x 2 is 1

To factorize a quadratic equation of the form x 2 + bx + speed of light, the leading coefficient is 1. You need to identify two numbers whose merchandise and union are c and b, respectively. case 1 : When bel and c are both incontrovertible **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 merchandise of 10 and a sum of 7 : 1 + 10 ≠ 7

2 + 5 = 7. Verify the factors using the distributive property of multiplication. ( ten + 2 ) ( ten + 5 ) = x2 + 5x + 2x + 10 = x2 + 7x + 10 The factors of the quadratic equation equation are : ( x + 2 ) ( ten + 5 ) Equating each factor to zero gives ; x + 2 = 0 ⟹x= -2 ten + 5 = 0 ⟹ x = -5 therefore, the solution is ten = – 2, x = – 5 **Example 5** adam 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. ten 2 + 10x + 25 = x 2 + 5x + 5x + 25 = x ( x + 5 ) + 5x + 25 = x ( x + 5 ) + 5 ( ten + 5 ) = ( ten + 5 ) ( adam + 5 ) consequently, x = -5 is the answer. case 2 : When bacillus is plus and degree centigrade is minus **Example 6** Solve x2 + 4x – 5 = 0 solution Write the factors of -5. 1 × –5, –1 × 5 Identify the factors whose intersection is – 5 and sum is 4. 1 – 5 ≠ 4

–1 + 5 = 4 Verify the factors using the distributive property. ( ten – 1 ) ( ten + 5 ) = x2 + 5x – x – 5 = x2 + 4x – 5

( ten – 1 ) ( ten + 5 ) = 0 ten – 1 = 0 ⇒ x = 1, or

ten + 5 = 0 ⇒ x = -5 therefore, x = 1, x = -5 are the solutions. lawsuit 3 : When b-complex vitamin 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 intersection is -6 and union is –5 : 1 + ( –6 ) = –5 Check the factors using the distributive property. ( adam + 1 ) ( adam – 6 ) = x2 – 6 x + x – 6 = x2 – 5x – 6 Equate each factor to zero and solve to get ;

( x + 1 ) ( ten – 6 ) = 0 adam + 1 = 0 ⇒ x = -1, or

ten – 6 = 0 ⇒ x = 6

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therefore, the solution is x=6, x = -1 case 4 : When b is negative and hundred is convinced **Example 8** x2 – 6x + 8 = 0 solution Write down all factors of 8. –1 × – 8, –2 × –4 Identify factors whose intersection is 8 and union is -6

–1 + ( –8 ) ≠ –6

–2 + ( –4 ) = –6 Check the factors using the distributive property. ( ten – 2 ) ( adam – 4 ) = x2 – 4 ten – 2x + 8 = x2 – 6x + 8 now equate each divisor to zero and solve the construction to get ; ( adam – 2 ) ( ten – 4 ) = 0 x – 2 = 0 ⇒ x = 2, or

x – 4 = 0 ⇒ x = 4 **Example 9** Factorize x2 +8x+12. solution Write down the factors of 12 ; 12 = 2 × 6 or = 4 × 3

Find factors whose sum is 8 : 2 + 6 = 8

2 × 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 ( adam + 6 ) = ( adam + 6 ) ( x + 2 ) Equate each factor to zero to get ; ( ten + 6 ) ( ten + 2 ) ten = -6, -2

### Factoring when the coefficient of x 2 is greater than 1

sometimes, the leading coefficient of a quadratic equation may be greater than 1. In this case, we can not solve the quadratic equation equation by the habit of common factors. consequently, we need to consider the coefficient of x2 and the factors of speed of light to find numbers whose union is b. **Example 10** Solve 2×2 – 14x + 20 = 0 solution Determine the common factors of the equation. 2×2 – 14x + 20 ⇒ 2 ( x2 – 7x + 10 ) now we can find the factors of ( x2 – 7x + 10 ). consequently, write down factors of 10 : –1 × –10, –2 × –5 Identify factors whose sum is – 7 : 1 + ( –10 ) ≠ –7

–2 + ( –5 ) = –7 Check the factors by applying distributive property. 2 ( x – 2 ) ( ten – 5 ) = 2 ( x2 – 5 x – 2x + 10 )

= 2 ( x2 – 7x + 10 ) = 2×2 – 14x + 20 Equate each factor to zero and solve ;

2 ( adam – 2 ) ( ten – 5 ) = 0 ten – 2 = 0 ⇒ x = 2, or

ten – 5 = 0 ⇒ x = 5 **Example 11** Solve 7×2 + 18x + 11 = 0 solution Write down the factors of both 7 and 11. 7 = 1 × 7 11 = 1 × 11 Apply distributive place to check the factors as shown below : ( 7x + 1 ) ( ten + 11 ) ≠ 7×2 + 18x + 11 ( 7x + 11 ) ( adam + 1 ) = 7×2 + 7x + 11x + 11 = 7×2 + 18x + 11 nowadays equate each factor to zero and solve to get ; 7×2 + 18x + 11= 0

( 7x + 11 ) ( ten + 1 ) = 0 x = -1, -11/7 **Example 12** Solve 2×2 − 7x + 6 = 3 solution 2×2 − 7x + 3 = 0 ( 2x − 1 ) ( ten − 3 ) = 0 x=1/2 or x=3 **Example 13** Solve 9x 2 +6x+1=0 solution Factorize to give : ( 3x + 1 ) ( 3x + 1 ) = 0 ( 3x + 1 ) = 0, therefore, x = −1/3 **Example 14** Factorize 6×2– 7x + 2 = 0 solution 6×2 – 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 ( ten + 4 ) = 0 ten + 4 ) ( ten – 3y ) = 0

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⟹ x + 4 = 0 or x – 3y = 0 ⟹ x = -4 or x = 3y therefore, x = -4 or x = 3y