# HOW TO SOLVE QUADRATIC EQUATIONS BY FACTORING

A quadratic polynomial is of the form of

ax2 + bx + c = 0

where a, b, c are real numbers.

Factoring quadratics is a method of expressing the quadratic equation ax2 + bx + c = 0 as a product of its linear factors as (x - k)(x - h),

where h, k are the roots of the quadratic equation ax2 + bx + c = 0.

Factorise :

Problem 1 :

x2–7x – 8 = 0

Solution :

Step 1 :

Product of factors = -8

Sum of factors = -7

Step 2 :

Finding the factors of product and sum.

2 · 4 = 8 and 2 + 4 ≠ 6

-8 · 1 = -8 and -8 + 1 = -7

So, the factors are -8 and +1

Step 3 :

x2 + 1x – 8x – 8 = 0

By grouping,

(x2 + 1x) + (-8x – 8) = 0

Step 4 :

By taking the common factor, we get

x(x + 1) - 8(x + 1) = 0

(x – 8)(x + 1) = 0

Step 5 :

Equating each factors to zero.

x – 8 = 0 and x + 1 = 0

x = 8 and x = -1

Problem 2 :

x2 + 4x – 21 = 0

Solution :

Step 1 :

Product of factors = -21

Sum of factors = 4

Step 2 :

Finding the factors of product and sum.

1 · (-21) = -21 and 1 + (-21) ≠ -20

7 · (-3) = -21 and 7 + (-3) = 4

So, the factors are +7 and -3

Step 3 :

x2 + 7x – 3x – 21 = 0

By grouping,

(x2 + 7x) + (-3x – 21) = 0

Step 4 :

By taking the common factor, we get

x(x + 7) -3(x + 7) = 0

(x – 3)(x + 7) = 0

Step 5 :

Equating each factor to zero

x – 3 = 0 and x + 7 = 0

x = 3 and x = -7

Problem 3 :

x2 - x – 2 = 0

Solution :

Step 1 :

Product of factors = -2

Sum of factors = -1

Step 2 :

Finding the factors of product and sum.

2 · (-1) = -2 and 2 + (-1) ≠ 1

1 · (-2) = -2 and 1 + (-2) = -1

So, the factors are -2 and 1

Step 3 :

x2 + x – 2x – 2 = 0

By grouping,

(x2 + x) + (-2x – 2) = 0

Step 4 :

By taking the common factor, we get

x(x + 1) -2(x + 1) = 0

(x – 2)(x + 1) = 0

Step 5 :

Equating each factors to zero.

x – 2 = 0 and x + 1 = 0

x = 2 and x = -1

Problem 4 :

x2 - 2x – 8 = 0

Solution :

Step 1 :

Product of factors = -8

Sum of factors = -2

Step 2 :

Finding the factors of product and sum.

(-1) · 8 = -8 and -1 + 8 ≠ 7

(-4) · 2 = -8 and (-4) + 2 = -2

So, the factors are -4 and 2.

Step 3 :

x2 - 4x + 2x – 8 = 0

By grouping,

(x2 - 4x) + (2x – 8) = 0

Step 4 :

By taking the common factor, we get

x(x - 4) + 2(x - 4) = 0

(x + 2)(x - 4) = 0

Step 5 :

Equating each factors to zero.

x + 2 = 0 and x - 4 = 0

x = -2 and x = 4

Problem 5 :

x2 + 5x – 24 = 0

Solution :

Step 1 :

Product of factors = -24

Sum of factors = 5

Step 2 :

Finding the factors of product and sum.

(-6) · 4 = -24 and (-6) + 4 ≠ -2

8 · (-3) = -24 and 8 + (-3) = 5

So, the factors are 8 and -3

Step 3 :

x2 + 8x – 3x – 24 = 0

By grouping,

(x2 + 8x) + (-3x – 24) = 0

Step 4 :

By taking the common factor, we get

x(x + 8) -3(x + 8) = 0

(x – 3)(x + 8) = 0

Step 5 :

Equating each factors to zero.

x – 3 = 0 and x + 8 = 0

x = 3 and x = -8

Problem 6 :

x2 - 3x – 10 = 0

Solution :

Step 1 :

Product of factors = -10

Sum of factors = -3

Step 2 :

Finding the factors of product and sum.

(-1) · 10 = -10 and -1 + 10 ≠ 9

(-5) · 2 = -10 and (-5) + 2 = -3

So, the factors are -5 and +2

Step 3 :

x2 - 5x + 2x – 10 = 0

By grouping,

(x2 - 5x) + (2x – 10) = 0

By taking the common factor, we get

x(x - 5) + 2(x - 5) = 0

(x – 5)(x + 2) = 0

Equating each factors to zero.

x – 5 = 0 and x + 2 = 0

x = 5 and x = -2

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