Factoring Quadratic Equations - 3

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

If the Coefficient of x² Is Greater Than 1

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 x² 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 x² 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 2x² – 14x + 20 = 0

Step 1: Find common factors if you can.

2x² – 14x + 20 = 2(x² – 7x + 10)

Step 2: Find the factors of (x² – 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(x² – 5 x – 2x + 10)
= 2(x² – 7x + 10) = 2x² – 14x + 20

Step 5: Going back to the original equation

2x² – 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 7x² + 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) ≠ 7x² + 18x + 11
(7x + 11)(x + 1) = 7x² + 18x + 11  

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

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

Step 4: Going back to the original equation

7x² + 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 4x² + 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) ≠ 4x² + 26x + 12
(4x + 12)(x + 12) ≠ 4x² + 26x + 12
(4x + 2)(x + 6) ≠ 4x² + 26x + 12
(4x + 6)(x + 2) ≠ 4x² + 26x + 12
(4x + 3)(x + 4) ≠ 4x² + 26x + 12
(4x + 4)(x + 3) ≠ 4x² + 26x + 12
(2x + 12)(2x + 1) = 4x² + 26x + 12  
(2x + 2)(2x + 6) ≠ 4x² + 26x + 12
(2x + 3)(2x + 4) ≠ 4x² + 26x + 12

Step 3: Going back to the original quadratic equation

4x² + 26x + 12 = 0       Factorize the left side of the equation
(2x + 12)(2x + 1) = 0

We get two values for x

Answer:

The following video shows an example of factoring a quadratic:

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