Factoring Quadratic Equations

As mentioned in the previous algebra lesson, when factoring Quadratic Equations , of the form:

ax ^{2} + bx + c = 0 where a , b and c are numbers and a ≠ 0.

we try to find common factors , and then look for patterns that will help you
to factorize the quadratic equation. For example: Square of Sum ,
Square of Difference and Difference of
Two Squares .

In other cases, you will have to try out different possibilities to get the
right factors for quadratic equations.

To factorize quadratic equations of the form: x ^{2} + bx + c , you will need to find two numbers whose product is c and whose sum is b .

Example 1: (b and c are both positive)

Solve the quadratic equation: x ^{2} + 7x + 10 = 0

Step 1: List out the factors of 10:

1 × 10, 2 × 5

Step 2: Find the factors whose sum is 7:

1 + 10 ≠ 7
2 + 5 = 7

Step 3: Write out the factors and check using the distributive property .

(x + 2) (x + 5) = x ^{2} + 5x + 2x + 10 = x ^{2} + 7x + 10
The factors are (x + 2) (x + 5)

Step 4: Going back to the original quadratic equation

x ^{2} + 7x + 10 = 0 Factorize the left side of the quadratic equation
(x + 2) (x + 5) = 0

We get two values for x .

Answer: x = – 2, x = – 5

Example 2: (b is positive and c is negative)

Get the values of x for the equation: x ^{2} + 4x – 5 = 0

Step 1: List out the factors of – 5:

1 × –5, –1 × 5

Step 2: Find the factors whose sum is 4:

1 – 5 ≠ 4
–1 + 5 = 4

Step 3: Write out the factors and check using the distributive property .

(x – 1)(x + 5)= x ^{2} + 5x – x – 5 = x ^{2} + 4x – 5

Step 4: Going back to the original quadratic equation

x ^{2} + 4x – 5 = 0 Factorize the left hand side of the equation
(x – 1)(x + 5) = 0

We get two values for x .

Answer: x = 1, x = – 5

Example 3: (b and c are both negative)

Get the values of x for the equation: x ^{2} – 5x – 6

Step 1: List out the factors of – 6:

1 × –6, –1 × 6, 2 × –3, –2 × 3

Step 2 : Find the factors whose sum is –5:

1 + ( –6) = –5

Step 3: Write out the factors and check using the distributive property .

(x + 1) (x – 6) = x ^{2} – 6 x + x – 6 = x ^{2} – 5x – 6

Step 4 : Going back to the original quadratic equation

x ^{2} – 5x – 6 = 0 Factorize the left hand side of the equation
(x + 1) (x – 6) = 0

We get two values for x .

Answer: x = –1, x = 6

Example 4 : (b is negative and c is positive)

Get the values of x for the equation: x ^{2} – 6x + 8 = 0

Step 1: List out the factors of 8:

We need to get the negative factors of 8 to get a negative sum.
–1 × – 8, –2 × –4

Step 2: Find the factors whose sum is – 6:

–1 + ( –8) ≠ –6
–2 + ( –4) = –6

Step 3: Write out the factors and check using the distributive property .

(x – 2) (x – 4) = x ^{2} – 4 x – 2x + 8 = x ^{2} – 6x + 8

Step 4: Going back to the original quadratic equation

x ^{2} – 6x + 8 = 0 Factorize the left hand side of the equation
(x – 2) (x – 4) = 0

We get two values for x .

Answer: x = 2, x = 4

Sometimes the coefficient of x in quadratic equations may not be 1 but the expression can be simplified by finding common factors .

When the coefficient of x ^{2} is greater than 1 and we cannot simplify the quadratic equation by finding common factors, we would need to consider the factors of the coefficient of x ^{2} and the factors of c in order to get the numbers whose sum is b . If there are many factors to consider you may want to use the quadratic formula instead.

Example 1: Get the values of x for the equation 2x ^{2} – 14x + 20 = 0

Step 1: Find common factors if you can.

2x ^{2} – 14x + 20 = 2(x ^{2} – 7x + 10)

Step 2: Find the factors of (x ^{2} – 7x + 10)

List out the factors of 10:
We need to get the negative factors of 10 to get a negative sum.
–1 × –10, –2 × –5

Step 3: Find the factors whose sum is – 7:

1 + ( –10) ≠ –7
–2 + ( –5) = –7

Step 4: Write out the factors and check using the distributive property .

2(x – 2) (x – 5) = 2(x ^{2} – 5 x – 2x + 10)
= 2(x ^{2} – 7x + 10) = 2x ^{2} – 14x + 20

Step 5: Going back to the original equation

2x ^{2} – 14x + 20 = 0 Factorize the left hand side of the equation
2(x – 2) (x – 5) = 0

We get two values for x

Answer: x = 2, x = 5

Example 2: Get the values of x for the equation 7x ^{2} + 18x + 11 = 0

Step 1: List out the factors of 7 and 11

Factors of 7:
1 × 7

Factors of 11:
1 × 11

Since 7 and 11 are prime numbers there are only two possibilities to try out.

Step 2: Write down the different combinations of the factors and perform the distributive property to check.

(7x + 1)(x + 11) ≠ 7x ^{2} + 18x + 11
(7x + 11)(x + 1) = 7x ^{2} + 18x + 11

Step 3 : Write out the factors and check using the distributive property .

(7x + 11)(x + 1) = 7x ^{2} + 7x + 11x + 11 = 7x ^{2} + 18x + 11

Step 4: Going back to the original equation

7x ^{2} + 18x + 11= 0 Factorize the left hand side of the equation
(7x + 11)(x + 1) = 0

We get two values for x

Answer:

Example 3: Get the values of x for the equation 4x ^{2} + 26x + 12 = 0

Step 1: List out the factors of 4 & 12

Factors of 4:
1 × 4, 2 × 2

Factors of 12:
1 × 12, 2 × 6, 3 × 4

Step 2: Write down the different combinations of the factors and perform the distributive property to check. When there are many factors to check, this becomes a tedious method to solve such quadratic equations, so you may want to try the quadratic formula instead.

(4x + 12)(x + 1) ≠ 4x ^{2} + 26x + 12
(4x + 12)(x + 12) ≠ 4x ^{2} + 26x + 12
(4x + 2)(x + 6) ≠ 4x ^{2} + 26x + 12
(4x + 6)(x + 2) ≠ 4x ^{2} + 26x + 12
(4x + 3)(x + 4) ≠ 4x ^{2} + 26x + 12
(4x + 4)(x + 3) ≠ 4x ^{2} + 26x + 12
(2x + 12)(2x + 1) = 4x ^{2} + 26x + 12
(2x + 2)(2x + 6) ≠ 4x ^{2} + 26x + 12
(2x + 3)(2x + 4) ≠ 4x ^{2} + 26x + 12

Step 3: Going back to the original quadratic equation

4x ^{2} + 26x + 12 = 0 Factorize the left side of the equation
(2x + 12)(2x + 1) = 0

We get two values for x

Answer:

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