- Rafia Shabbir

# How To Factor Quadratic Equations?

Updated: Oct 31, 2020

Quadratic equations are second-degree polynomials, i.e. the highest degree of these polynomials is 2. We categorize a quadratic equation as a trinomial because it is constituted of three terms. The standard form of a quadratic equation is given below:

Here, a, b and c are constants and x is a variable. The leading coefficient of quadratics should be non zero. If we set the above quadratic equation equal to 0, we will get a quadratic formula.

Unlike linear equations, graphing quadratic equations in a graphing calculator will yield a parabola.

**Examples of quadratic equations**

Some examples of quadratic functions are given below:

If the leading coefficient in a quadratic equation is not given like in this equation below, then we assume that its coefficient is 1.

**Methods for Solving Quadratic Equations**

There are multiple methods for solving quadratic functions. It is important to choose the correct method based on the question.

**Deciding which method to use**

First, we try to solve the equation by factoring because it is the simplest method to get the values of x.

If we can’t solve by factoring, then we see if the equation is a perfect square or not. If so, we find the value of x by taking the square root on both sides of the equation. Be careful with plus-minus signs after taking the square roots.

If the leading coefficient is 1 and the middle term of the quadratic function is even, we can solve the equation by completing the square method.

If you are unable to solve the equation through any of the above methods, then try using the quadratic formula.

In this article, we will only discuss how to factor quadratic equations and find the zeros and roots through factorization.

**Process of solving quadratics by factoring**

Before proceeding to examples, let us see how to solve quadratic equations by factoring.

**Step 1** – Multiply the constant with the leading coefficient.

**Step 2** – Expand the middle term bx into two terms in such a way that the addition or subtraction of two terms is equal to the midterm and the multiplication of expanded terms is equal to the product obtained in step 1.

**Step 3 **– You will have four terms now. Factor the terms by grouping and write the final answer as a product of two binomial terms. Binomials are polynomial functions constituted of two terms only.

**Example 1**

Solve the following quadratic equation through factorization.

**Solution**

**Step 1** – Multiply first term with the constant to get -42 x ^ {2}.

**Step 2** – Expand the midterm -x into two terms in such a way that adding or subtracting two terms will be equal to -x and their product is equal to -42 x ^2.

**Step 3 **– Group the first and second, and third and fourth terms together and find the greatest common factor between them in pairs.

Hence, the factors of the quadratic equation are (x – 7) (x + 6). Let us use these factors to find zeroes or roots of the function. Set these factors equal to zero:

(x – 7) (x + 6) = 0

According to zero product property, either x – 7 = 0 or x + 6 = 0. Solve for x by taking -7 and 6 to the right hand side of the equations. Hence, x = 7 or x = - 6. The numbers 7 and -6 are the roots of the above quadratic polynomial equation.

**Example 2**

Solve the following quadratic equation by factoring.

**Solution**

Follow the following steps for solving the equation by factoring.

**Step 1** – Multiply the constant term -3 by the leading term of the equation to obtain the product -6 x ^2.

**Step 2** – Expand the midterm +5x into two terms in such a way that adding or subtracting two terms will be equal to +5x and their product is equal to -6 x ^ {2}.

Step 3 – Group the first and second, and third and fourth terms together and find the greatest common factor between them in pairs.

=2x (x +3) -1 ( x + 3)

=(x -1) (x + 3)

Now, we have the factors of the polynomial, let us find the roots of the polynomial by applying the zero product property.

(2x – 1)(x + 3) = 0

Set individual factors equal to zero and solve for x by taking constants to the right-hand side of the equation.

2x – 1 = 0, x =1/2

x + 3 = 0 , x = -3

Hence, 1/2 and -3 are the roots of the quadratic equation.

**Example 3**

**Solution**

Follow the steps below to solve it by factoring.

**Step 1 **– Multiply -2 by the first term 5x ^ 2 to obtain the product -10 x ^2.

**Step 2** – Expand the second term -9x into two terms in such a way that when we add or subtract two terms, the result will be equal to -9x and the product will be -10 x ^2.

Step 3 – Group the first and second, and third and fourth terms together and find the greatest common factor between them in pairs.

= 5x (x - 2) + 1 ( x - 2)

= (5x + 1) (x - 2)

Now, we have the factors of the polynomial, let us find the roots of the polynomial by applying the zero product property.

(5x + 1) (x – 2) = 0

5x + 1 = 0, x = -1/5

x – 2 =0, x = 2

Hence, the roots of the polynomial are -1/5 and 2.

**Example 4**

**Solution**

**Step 1** – To factorize the quadratic expression, multiply -18 by the first term x ^ 2 to obtain the product -18 x ^2.

**Step 2** – Expand the midterm -3x into two terms in such a way that adding or subtracting two terms will be equal to -3x and their product is equal to -18 x ^2.

**Step 3** – Group the first and second, and third and fourth terms together and find the greatest common factor between them in pairs.

=x (x - 6) +3 ( x - 6)

= (x - 6) (x + 3)

Now, we have the factors of the polynomial, let us find the roots of the polynomial by applying the zero product property.

(x - 6) (x + 3) = 0

x - 6 = 0, x = 6

x + 3 =0, x = -3

Hence, the roots of the polynomial are -3 and 6.

**Example 5**

**Solution**

Step 1 – Multiplying -40 by the first term x ^ 2 will give us -40 x ^2.

**Step 2** – Expand the midterm +3x into two terms in such a way that adding or subtracting two terms will be equal to +3x and their product is equal to -40 x ^2.

**Step 3** – Group the first and second, and third and fourth terms together and find the greatest common factor between them in pairs.

=x (x + 8) - 5 ( x + 8)

=(x + 8) (x - 5)

(x + 8) (x - 5) = 0

x + 8 = 0, x = -8

x - 5 =0, x = 5

Hence, the roots of the polynomial are -8 and 5.

**Example 6**

**Solution**

**Step 1** – To factorize the given equation, -1 will be multiplied by the first term 3x ^ 2 will give us -3 x ^2.

**Step 2 **– Expand the midterm -2x into two terms in such a way that adding or subtracting two terms will be equal to -2x and their product is equal to -3 x ^2.

Step 3 – Group the first and second, and third and fourth terms together and find the greatest common factor between them in pairs.

=3x (x - 1) + 1 ( x - 1)

=(x - 1) (3x + 1)

(3x+1) (x - 1) = 0

3x + 1 = 0, x = -1/3

x - 1 =0, x = 1

Hence, the roots of the polynomial are -1/3 and 1.

**Example 7**

**Solution**

**Step 1** – To solve this equation, -12 will be multiplied by the first term to get -24 x ^2.

**Step 2** – Expand the midterm -2x into two terms in such a way that adding or subtracting two terms will be equal to -2x and their product is equal to -24 x ^2.

**Step 3 **– Group the first and second, and third and fourth terms together and find the greatest common factor between them in pairs.

=2x (x - 3) + 4 ( x - 3)

=(x - 3) (2x + 4)

(2x + 4) (x - 3) = 0

2x + 4 = 0, x = -4/2 = -2

x - 3 =0, x = 3

Hence, the 3 and -2 are the roots of a quadratic equation. It means that when these roots are substituted in the original equation, the result will be zero.

**Example 8**

**Solution**

**Step 1** – To solve the equation by factoring, the first term and the constant -30 will be multiplied together to get 30 x ^ 2

**Step 2** – Expand the midterm 11x into two terms in such a way that adding or subtracting two terms will be equal to 11x and their product is equal to 30 x ^2.

**Step 3** – Group the first and second, and third and fourth terms together and find the common factors between them in pairs.

=x (x + 5) + 6 ( x + 5)

=(x + 5) (x + 6)

(x + 5) (x + 6) = 0

x + 5 = 0, x = -5

x + 6 =0, x = -6

Hence, the -5 and -6 are the roots of a quadratic equation. It means that by substituting these roots in the original polynomial function we will get the result zero.

**Example 9**

**Solution**

**Step 1** – By multiplying the first term and the constant -2 we will get -30 x ^ 2

**Step 2** – Expand the midterm +x into two terms in such a way that adding or subtracting two terms will be equal to +x and their product is equal to -30 x ^2.

**Step 3** – Group the first and second, and third and fourth terms together and find the common factors between them in pairs.

=3x (5x + 2) - 1 (5 x + 2)

=(5x + 2) (3x - 1)

(5x + 2) (3x - 1) = 0

5x + 2 = 0

Solve the terms algebraically by taking 2 on the right hand side of the equation and then dividing it by 5

x = -2/5

3x - 1 =0

Take -1 on the right-hand side and divide it by 3 to isolate x on the left-hand side of the equation:

x = 1/3

Hence, the 1/3 and -2/5 are the roots of a quadratic equation. It means that substituting these roots in the original polynomial function will yield zero.

**Example 10**

**Solution**

**Step 1** – To solve equation, the first term and the constant +2 will be multiplied together to get 8 x ^ 2

**Step 2** – Expand the second term +6x into two terms in such a way that adding or subtracting two terms will be equal to +6x and their product is equal to 8 x ^2.

**Step 3** – Group the first and second, and third and fourth terms together and find the common factors between them in pairs.

=4x (x + 1) + 2 ( x + 1)

=(x + 1) (4x + 2)

Now, we have the equation in factored form, let us find the roots of the polynomial by applying the zero product property.

(4x + 2) (x + 1) = 0

4x + 2 = 0, x = -1/2

x + 1 =0, x = -1

Hence, - 1 and -1/2 are roots of a quadratic equation. It means that substituting these roots in the original polynomial function will yield zero.

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