Solving Quadratic Equations by Factoring

SOLVING QUADRATIC EQUATIONS BY FACTORING

Solving Quadratic Equations by Factoring :

In this section, you will learn how to solve a quadratic equation by factoring.

Generally we have two types of quadratic equation.

Quadratic equation of leading coefficient 1.
Quadratic equation of leading coefficient not equal to 1.

The general form of a quadratic equation is

ax² + bx + c = 0

In a quadratic equation, leading coefficient is nothing but the coefficient of x².

Solving Quadratic Equations by Factoring with a Leading Coefficient of 1 - Procedure

(i) In a quadratic equation in the form ax² + bx + c = 0, if the leading coefficient is 1, we have to decompose the constant term "c" into two factors.

(ii) The product of the two factors must be equal to the constant term "c" and the addition of two factors must be equal to the coefficient of x, that is "b".

(iii) If p and q are the two factors of the constant term c, then we have to factor the quadratic equation using p and q as shown below.

(x + p)(x + q) = 0

(iv) Solving the above equation, we get

x = -p and x = -q

How to assign signs for the two factors ?

Quadratic Equation

Signs of Factors

ax² + bx + c = 0

Positive sign for both the factors.

ax² - bx + c = 0

Negative sign for both the factors.

ax² + bx - c = 0

Negative sign for smaller factor and positive sign for larger factor.

ax² - bx - c = 0

Positive sign for smaller factor and negative sign for larger factor.

Solving Quadratic Equations by Factoring with a Leading Coefficient of 1 - Examples

Example :

Solve :

x² + 17 x + 60 = 0

Solution :

The given quadratic equation is in the form of

ax² + bx + c = 0

Check whether the coefficient of x² is 1 or not.

Because the coefficient of x² is 1, we have to decompose 60 into two factors as shown below.

Because the constant term 60 is having positive sign, both the factors must be positive.

In the above four pairs of factors, we have to select the a pair of factors such that the product of two factors is equal to the constant term "+60" and the addition of two factors is equal to the coefficient of x, that is "+17".

Now, factor the given quadratic equation and solve for x as shown below.

(x + 12)(x + 5) = 0

x + 12 = 0 or x + 5 = 0

x = -12 or x = -5

So, the solution is {-12, -5}.

Solving Quadratic Equations by Factoring when Leading Coefficient is not 1 - Procedure

(i) In a quadratic equation in the form ax² + bx + c = 0, if the leading coefficient is not 1, we have to multiply the coefficient of x² and the constant term. That is "ac". Then, decompose "ac" into two factors.

(ii) The product of the two factors must be equal to "ac" and the addition of two factors must be equal to the coefficient of x, that is "b".

(iii) Divide the two factors by the coefficient of x² and simplify as much as possible.

(iv) Write the remaining number along with x (This is explained in the following example).

Solving Quadratic Equations by Factoring when Leading Coefficient is not 1 - Example

Example :

Solve :

2x² + x - 6 = 0

Solution :

The given quadratic equation is in the form of

ax² + bx + c = 0

Here, the coefficient of x² is 1 or not.

Multiply the coefficient of x² and the constant term "-6".

That is,

2 ⋅ (-6) = -12

Decompose -12 into two factors such that the product of two factors is equal to -12 and the addition of two factors is equal to the coefficient of x, that is 1.

Then, the two factors of -12 are

4 and -3

Now we have to divide the two factors 4 and -3 by the coefficient of x², that is 2.

Now, factor the given quadratic equation and solve for x as shown below.

(x + 2)(2x - 3) = 0

x + 2 = 0 or 2x - 3 = 0

x = -2 or x = 3/2

x = -2 or x = 1.5

So, the solution is {-2, 1.5}.

Solving Quadratic Equations by Factoring - Practice Problems

Problem 1 :

Solve the quadratic equation by factoring :

x² – 5x – 24 = 0

Solution :

In the given quadratic equation, the coefficient of x² is 1.

Decompose the constant term -24 into two factors such that the product of the two factors is equal to -24 and the addition of two factors is equal to the coefficient of x, that is 5.

Then, the two factors of -24 are

+3 and -8

Factor the given quadratic equation using and +3 and solve for x.

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

x + 3 = 0 or x - 8 = 0

x = -3 and x = 8

So, the solution is {-3, 8}.

Problem 2 :

Solve the quadratic equation by factoring :

3x² – 5x – 12 = 0

Solution :

In the given quadratic equation, the coefficient of x² is not 1.

So, multiply the coefficient of x² and the constant term "-12".

3 ⋅ (-12) = -36

Decompose -36 into two factors such that the product of two factors is equal to -36 and the addition of two factors is equal to the coefficient of x, that is -5.

Then, the two factors of -36 are

+4 and -9

Now we have to divide the two factors 4 and -3 by the coefficient of x², that is 3.

Now, factor the given quadratic equation and solve for x as shown below.

(3x + 4)(x - 3) = 0

3x + 4 = 0 or x - 3 = 0

x = -4/3 or x = 3

So, the solution is {-4/3, 3}.

Problem 3 :

Solve the quadratic equation by factoring :

(x + 3)² - 81 = 0

Solution :

Write the given quadratic equation in the form

ax² + bx + c = 0

Then,

(x + 3)² - 81 = 0

(x + 3)(x + 3) - 81 = 0

x² + 3x + 3x + 9 - 81 = 0

x² + 6x - 72 = 0

In the quadratic equation above, the coefficient of x² is 1.

Decompose the constant term -72 into two factors such that the product of the two factors is equal to -72 and the addition of two factors is equal to the coefficient of x, that is +6.

Then, the two factors of -72 are

+12 and -6

Factor the given quadratic equation using +12 and -6 and solve for x.

(x + 12)(x - 6) = 0

x + 12 = 0 or x - 6 = 0

x = -12 and x = 6

So, the solution is {-12, 6}.