Solving Quadratic Equations using Factoring - Examples, Exercises and Solutions

Solving an equation using the distributive property is related to the need to open the parentheses as the first step to then be able to simplify similar members. When an equation contains one or more pairs of parentheses, we must start by opening them all and then proceed to the next phase. 

Below, we provide you with some examples where this method is applied.

2(X+3)=8 2\left(X+3\right)=8

In this equation, we can clearly see some parentheses. To start, we must open them (that is, apply the distributive property) and then we can proceed with the following phases of the exercise.

2X+6=8 2X+6=8

2X=2 2X=2

X=1 X=1

The result of the equation is 1 1 .

Solving equations using the distributive property


Suggested Topics to Practice in Advance

  1. Solving Equations by Adding or Subtracting the Same Number from Both Sides
  2. Solving Equations by Multiplying or Dividing Both Sides by the Same Number
  3. Solving Equations by Simplifying Like Terms

Practice Solving Quadratic Equations using Factoring

Exercise #1

Solve for x:

7(2x+5)=77 7(-2x+5)=77

Video Solution

Step-by-Step Solution

To open parentheses we will use the formula:

a(x+b)=ax+ab a(x+b)=ax+ab

(7×2x)+(7×5)=77 (7\times-2x)+(7\times5)=77

We multiply accordingly

14x+35=77 -14x+35=77

We will move the 35 to the right section and change the sign accordingly:

14x=7735 -14x=77-35

We solve the subtraction exercise on the right side and we will obtain:

14x=42 -14x=42

We divide both sections by -14

14x14=4214 \frac{-14x}{-14}=\frac{42}{-14}

x=3 x=-3

Answer

-3

Exercise #2

Solve x:

5(x+3)=0 5(x+3)=0

Video Solution

Step-by-Step Solution

We open the parentheses according to the formula:

a(x+b)=ax+ab a(x+b)=ax+ab

5×x+5×3=0 5\times x+5\times3=0

5x+15=0 5x+15=0

We will move the 15 to the right section and keep the corresponding sign:

5x=15 5x=-15

Divide both sections by 5

5x5=155 \frac{5x}{5}=\frac{-15}{5}

x=3 x=-3

Answer

3 -3

Exercise #3

Solve for x:

3(12x+4)=12 -3(\frac{1}{2}x+4)=\frac{1}{2}

Video Solution

Step-by-Step Solution

We open the parentheses on the left side by the distributive property and use the formula:

a(x+b)=ax+ab a(x+b)=ax+ab

32x12=12 -\frac{3}{2}x-12=\frac{1}{2}

We multiply all terms by 2 to get rid of the fractions:

3x12×2=1 -3x-12\times2=1

3x24=1 -3x-24=1

We will move the minus 24 to the right section and keep the corresponding sign:

3x=24+1 -3x=24+1

3x=25 -3x=25

Divide both sections by minus 3:

3x3=253 \frac{-3x}{-3}=\frac{25}{-3}

x=253 x=-\frac{25}{3}

Answer

253 -\frac{25}{3}

Exercise #4

Solve for x:

8(2x+4)=6(x4)+3 -8(2x+4)=6(x-4)+3

Video Solution

Step-by-Step Solution

To open parentheses we will use the formula:

a(x+b)=ax+ab a\left(x+b\right)=ax+ab

(8×2x)+(8×4)=(6×x)+(6×4)+3 (-8\times2x)+(-8\times4)=(6\times x)+(6\times-4)+3

We multiply accordingly:

16x32=6x24+3 -16x-32=6x-24+3

Calculate the elements on the right section:

16x32=6x21 -16x-32=6x-21

In the left section we enter the elements with the X and in the left section those without the X, remember to change the plus and minus signs as appropriate when transferring:

32+21=6x+16x -32+21=6x+16x

Calculate the elements accordingly

11=22x -11=22x

We divide the two sections by 22

1122=22x22 -\frac{11}{22}=\frac{22x}{22}

12=x -\frac{1}{2}=x

Answer

12 -\frac{1}{2}

Exercise #5

Solve for x:

8(x2)=4(x+3) 8(x-2)=-4(x+3)

Video Solution

Step-by-Step Solution

To open parentheses we will use the formula:

a(x+b)=ax+ab a(x+b)=ax+ab

(8×x)+(8×2)=(4×x)+(4×3) (8\times x)+(8\times-2)=(-4\times x)+(-4\times3)

We multiply accordingly

8x16=4x12 8x-16=-4x-12

In the left section we enter the elements with the X and in the right section those without the X, remember to change the plus and minus signs as appropriate when transferring:

8x+4x=12+16 8x+4x=-12+16

We solve accordingly

12x=4 12x=4

We divide both sections by 12

12x12=412 \frac{12x}{12}=\frac{4}{12}

We reduce and obtainx=412=13 x=\frac{4}{12}=\frac{1}{3}

Answer

13 \frac{1}{3}

Exercise #1

Solve for x:

9(2x)=(x+4)3 -9(2-x)=(x+4)\cdot3

Video Solution

Step-by-Step Solution

We open the parentheses in both sections by the distributive property and use the formula:

a(x+b)=ax+ab a(x+b)=ax+ab

18+9x=3x+12 -18+9x=3x+12

We move 3X to the left section, and 18 to the right section and maintain the corresponding signs:

9x3x=12+18 9x-3x=12+18

We add the terms:

6x=30 6x=30

We divide both sections by 6:

6x6=306 \frac{6x}{6}=\frac{30}{6}

x=5 x=5

Answer

5

Exercise #2

Solve for x:

7(2x+3)4(x+2)=5(23x) -7(2x+3)-4(x+2)=5(2-3x)

Video Solution

Step-by-Step Solution

To open parentheses we will use the formula:

a(x+b)=ax+ab a\left(x+b\right)=ax+ab

(7×2x)+(7×3)+(4×x)+(4×2)=(5×2)+(5×3x) (-7\times2x)+(-7\times3)+(-4\times x)+(-4\times2)=(5\times2)+(5\times-3x)

We multiply accordingly:

14x214x8=1015x -14x-21-4x-8=10-15x

We calculate the elements in the left section:

18x29=1015x -18x-29=10-15x

In the left section we enter the elements with the X and in the right section those without the X, remember to change the plus and minus signs as appropriate when transferring:

18x+15x=10+29 -18x+15x=10+29

We calculate the elements accordingly:

3x=39 -3x=39

We divide the two sections by -3:

3x3=393 \frac{-3x}{-3}=\frac{39}{-3}

x=13 x=-13

Answer

-13

Exercise #3

8a+2(3a7)=0 8a+2(3a-7)=0

Video Solution

Answer

a=1 a=1

Exercise #4

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

Video Solution

Answer

a=0 a=0

Exercise #5

5(3b1)=0 5-(3b-1)=0

Video Solution

Answer

b=2 b=2

Exercise #1

3x+5(x+4)=0 3x+5(x+4)=0

Video Solution

Answer

x=2.5 x=-2.5

Exercise #2

2(x+4)+8=0 2(x+4)+8=0

Video Solution

Answer

x=8 x=-8

Exercise #3

Solve for y:

2(4+y)y=0 -2(-4+y)-y=0

Video Solution

Answer

y=223 y=2\frac{2}{3}

Exercise #4

Solve for x:

2(4x)=8 2(4-x)=8

Video Solution

Answer

0

Exercise #5

3(4a+8)=27a -3(4a+8)=27a

a=? a=\text{?}

Video Solution

Answer

813 -\frac{8}{13}

Topics learned in later sections

  1. First-degree equations with one unknown