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Quadratic Equations (Factoring and solving when a ≠ 1)

When factoring Quadratic Equations, of the form:

ax2 + 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.

### If the Coefficient Of x2 Is 1

To factorize quadratic equations of the form: x2 + 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: x2 + 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) = x2 + 5x + 2x + 10 = x2 + 7x + 10
The factors are (x + 2)(x + 5)

Step 4: Going back to the original quadratic equation

x2 + 7x + 10 = 0       Factorize the left side of the quadratic equation
(x + 2) (x + 5) = 0

x + 2 = 0 ⇒ x = -2, or
x + 5 = 0 ⇒ x = -5

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

Get the values of x for the equation: x2 + 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)= x2 + 5xx – 5 = x2 + 4x – 5

Step 4: Going back to the original quadratic equation

x2 + 4x – 5 = 0       Factorize the left hand side of the equation
(x – 1)(x + 5) = 0

x - 1 = 0 ⇒ x = 1, or
x + 5 = 0 ⇒ x = -5

Example 3: (b and c are both negative)

Get the values of x for the equation: x2 – 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) = x2 – 6 x + x – 6 = x2 – 5x – 6

Step 4: Going back to the original quadratic equation

x2 – 5x – 6 = 0       Factorize the left hand side of the equation
(x + 1) (x – 6) = 0

x + 1 = 0 ⇒ x = -1, or
x - 6 = 0 ⇒ x = 6

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

Get the values of x for the equation: x2 – 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) = x2 – 4 x – 2x + 8 = x2 – 6x + 8

Step 4: Going back to the original quadratic equation

x2 – 6x + 8 = 0       Factorize the left hand side of the equation
(x – 2) (x – 4) = 0

x - 2 = 0 ⇒ x = 2, or
x - 4 = 0 ⇒ x = 4

### 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

x - 2 = 0 ⇒ x = 2, or
x - 5 = 0 ⇒ 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 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
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