Solving Equations by Factoring - Examples, Exercises and Solutions

To solve equations through factorization, we must transpose all the elements to one side of the equation and leave 0 0 on the other side.
Why? Because after factoring, we will have 0 0 as the product.

Let's remember the following property

The product of two numbers equals 0 0 when, at least, one of them is 0 0 .
If  x×y=0x\times y=0
then
either: x=0x=0
or: y=0y=0
or both are equal to 0 0 .

Steps to carry out to solve equations through factorization

  • Let's move all the elements to one side of the equation and leave 0 0 on the other.
  • Let's factor using one of the methods we have learned: by taking out the common factor, with shortcut multiplication formulas, or with trinomials.
  • Let's find out when the elements achieve a product equivalent to 0 0 .

Suggested Topics to Practice in Advance

  1. Factoring using contracted multiplication
  2. Factorization
  3. Extracting the common factor in parentheses
  4. Factorization: Common factor extraction
  5. Factoring Trinomials
  6. Algebraic Fractions
  7. Simplifying Algebraic Fractions
  8. Factoring Algebraic Fractions
  9. Addition and Subtraction of Algebraic Fractions
  10. Multiplication and Division of Algebraic Fractions

Practice Solving Equations by Factoring

Examples with solutions for Solving Equations by Factoring

Exercise #1

Solve for x.

x27x12=0 -x^2-7x-12=0

Video Solution

Step-by-Step Solution

First, factor using trinomials and remember that there might be more than one solution for the value of x x :

x27x12=0 -x^2-7x-12=0

Divide by -1:

x2+7x+12=0 x^2+7x+12=0

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

Therefore:

x+4=0 x+4=0

x=4 x=-4

Or:

x+3=0 x+3=0

x=3 x=-3

Answer

x=3,x=4 x=-3,x=-4

Exercise #2

Find the value of the parameter x.

(x5)2=0 (x-5)^2=0

Video Solution

Step-by-Step Solution

We will factor using the shortened multiplication formulas:

a2b2=(ab)(a+b) a^2-b^2=(a-b)(a+b) (ab)2=a22ab+b2 (a-b)^2=a^2-2ab+b^2

(a+b)2=a2+2ab+b2 (a+b)^2=a^2+2ab+b^2

Let's remember that there might be more than one solution for the value of x.

According to one solution, we'll take the square root:

(x5)2=0 (x-5)^2=0

x5=0 x-5=0

x=5 x=5

According to the second solution, we'll use the shortened multiplication formula:

(x5)2=x210x+25=0 (x-5)^2=x^2-10x+25=0

We'll use the trinomial:

(x5)(x5)=0 (x-5)(x-5)=0

x5=0 x-5=0

x=5 x=5

or

x5=0 x-5=0

x=5 x=5

Therefore, according to all calculations, x=5 x=5

Answer

x=5 x=5

Exercise #3

Find the value of the parameter x.

x225=0 x^2-25=0

Video Solution

Step-by-Step Solution

We will factor using the shortened multiplication formulas:

a2b2=(ab)(a+b) a^2-b^2=(a-b)(a+b) (ab)2=a22ab+b2 (a-b)^2=a^2-2ab+b^2

(a+b)2=a2+2ab+b2 (a+b)^2=a^2+2ab+b^2

Remember that there might be more than one solution for the value of x.

According to the first formula:

x2=a2 x^2=a^2

We'll take the square root:

x=a x=a

25=b2 25=b^2

We'll take the square root:

b=5 b=5

We'll use the first shortened multiplication formula:

a2b2=(ab)(a+b) a^2-b^2=(a-b)(a+b)

x225=(x5)(x+5)=0 x^2-25=(x-5)(x+5)=0

Therefore:

x+5=0 x+5=0

x=5 x=-5

Or:

x5=0 x-5=0

x=5 x=5

Answer

x=5,x=5 x=5,x=-5

Exercise #4

Find the value of the parameter x.

2x27x+5=0 2x^2-7x+5=0

Video Solution

Step-by-Step Solution

We will factor using trinomials, remembering that there is more than one solution for the value of X:

2x27x+5=0 2x^2-7x+5=0

We will factor -7X into two numbers whose product is 10:

2x25x2x+5=0 2x^2-5x-2x+5=0

We will factor out a common factor:

2x(x1)5(x1)=0 2x(x-1)-5(x-1)=0

(2x5)(x1)=0 (2x-5)(x-1)=0

Therefore:

x1=0 x-1=0 x=1 x=1

Or:

2x5=0 2x-5=0

2x=5 2x=5

x=2.5 x=2.5

Answer

x=1,x=2.5 x=1,x=2.5

Exercise #5

Find the value of the parameter x.

(x4)2+x(x12)=16 (x-4)^2+x(x-12)=16

Video Solution

Step-by-Step Solution

Let's open the parentheses, remembering that there might be more than one solution for the value of X:

(x4)2+x(x12)=16 (x-4)^2+x(x-12)=16

x28x+16+x212x=16 x^2-8x+16+x^2-12x=16

2x220x=0 2x^2-20x=0

2x(x10)=0 2x(x-10)=0

Therefore:

x10=0 x-10=0

x=10 x=10

Or:

2x=0 2x=0

x=0 x=0

Answer

x=0,x=10 x=0,x=10

Exercise #6

Find the value of the parameter x.

2x(3x)+(x3)2=9 -2x(3-x)+(x-3)^2=9

Video Solution

Step-by-Step Solution

To solve the equation 2x(3x)+(x3)2=9-2x(3-x) + (x-3)^2 = 9, follow these steps:

  • Step 1: Expand each term:
    2x(3x)=6x+2x2-2x(3-x) = -6x + 2x^2 and (x3)2=x26x+9(x-3)^2 = x^2 - 6x + 9.
  • Step 2: Substitute the expanded terms into the equation:
    6x+2x2+x26x+9=9-6x + 2x^2 + x^2 - 6x + 9 = 9.
  • Step 3: Combine like terms:
    (2x2+x2)+(6x6x)+9=9(2x^2 + x^2) + (-6x - 6x) + 9 = 9.
  • Step 4: Simplify further:
    3x212x+9=93x^2 - 12x + 9 = 9.
  • Step 5: Move all terms to one side to form a quadratic equation:
    3x212x+99=03x^2 - 12x + 9 - 9 = 0.
  • Step 6: Simplify the expression:
    3x212x=03x^2 - 12x = 0.
  • Step 7: Factor out the common term:
    3x(x4)=03x(x - 4) = 0.
  • Step 8: Solve for x x :
    Since 3x=03x = 0 or x4=0x - 4 = 0, we find x=0x = 0 or x=4x = 4.

Therefore, the values of x x that satisfy the equation are x=0\mathbf{x = 0} and x=4\mathbf{x = 4}.

Answer

x=0,x=4 x=0,x=4

Exercise #7

Find the value of the parameter x.

(x+5)2=0 (x+5)^2=0

Video Solution

Step-by-Step Solution

To solve the equation (x+5)2=0(x+5)^2 = 0, we will use the fact that a perfect square is zero only when the quantity being squared is zero itself.

  • Step 1: Set x+5=0 x+5 = 0 since it is the term being squared.
  • Step 2: Solve for x x by subtracting 5 from both sides: x+5=0 x+5 = 0 x=5 x = -5

Therefore, the solution to the equation is x=5 x = -5 .

Answer

x=5 x=-5

Exercise #8

Find the value of the parameter x.

12x39x23x=0 12x^3-9x^2-3x=0

Video Solution

Step-by-Step Solution

To solve the problem, we follow these steps:

  • Step 1: Identify the given cubic equation, 12x39x23x=0 12x^3 - 9x^2 - 3x = 0 .
  • Step 2: Look for common factors in the terms of the equation.
  • Step 3: Apply the zero-product property to factor and solve the equation.

Let's work through the solution:

Step 1: Observe that each term in the equation 12x39x23x 12x^3 - 9x^2 - 3x has a common factor of 3x 3x . So, we can factor 3x 3x out of the equation, giving us:

3x(4x23x1)=0 3x (4x^2 - 3x - 1) = 0

Step 2: Having factored out 3x 3x , we now have a product of terms equaling zero. According to the zero-product property, at least one of the factors must be zero:

3x=0or4x23x1=0 3x = 0 \quad \text{or} \quad 4x^2 - 3x - 1 = 0

This gives us one solution directly:

x=0 x = 0

Step 3: Solve the quadratic equation 4x23x1=0 4x^2 - 3x - 1 = 0 using the quadratic formula, where a=4 a = 4 , b=3 b = -3 , and c=1 c = -1 :

The quadratic formula is:

x=b±b24ac2a x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}

Applying it to our equation:

x=(3)±(3)244(1)24 x = \frac{{-(-3) \pm \sqrt{{(-3)^2 - 4 \cdot 4 \cdot (-1)}}}}{2 \cdot 4}

x=3±9+168 x = \frac{{3 \pm \sqrt{{9 + 16}}}}{8}

x=3±258 x = \frac{{3 \pm \sqrt{25}}}{8}

x=3±58 x = \frac{{3 \pm 5}}{8}

This gives us two solutions:

When 25=5 \sqrt{25} = 5 , x=3+58=1 x = \frac{{3 + 5}}{8} = 1 .

When 25=5 \sqrt{25} = -5 , x=358=14 x = \frac{{3 - 5}}{8} = -\frac{1}{4} .

Therefore, the solutions to the equation 12x39x23x=0 12x^3 - 9x^2 - 3x = 0 are x=0 x = 0 , x=1 x = 1 , and x=14 x = -\frac{1}{4} .

Verifying against the provided choices, the correct choice is choice 2: x=0,x=1,x=14 x=0,x=1,x=-\frac{1}{4} .

Answer

x=0,x=1,x=14 x=0,x=1,x=-\frac{1}{4}

Exercise #9

A right triangle is shown below.

x>1


Calculate the lengths of the sides of the triangle.

x+9x+9x+9x+2x+2x+2x+10x+10x+10

Video Solution

Step-by-Step Solution

To find the lengths of the sides of the right triangle, we will apply the Pythagorean theorem, which states a2+b2=c2a^2 + b^2 = c^2 for a right triangle, where cc is the hypotenuse.

Given the side lengths are x+9x + 9, x+2x + 2, and x+10x + 10, we assume x+10x + 10 is the hypotenuse because it is the largest value and confirm it by checking with the theorem.

Substitute into the Pythagorean theorem:
(x+2)2+(x+9)2=(x+10)2(x + 2)^2 + (x + 9)^2 = (x + 10)^2

Let's expand each side and solve for xx:
(x+2)2=x2+4x+4(x + 2)^2 = x^2 + 4x + 4
(x+9)2=x2+18x+81(x + 9)^2 = x^2 + 18x + 81
(x+10)2=x2+20x+100(x + 10)^2 = x^2 + 20x + 100

Combine these into a single equation:
x2+4x+4+x2+18x+81=x2+20x+100x^2 + 4x + 4 + x^2 + 18x + 81 = x^2 + 20x + 100

Simplify and combine like terms:
2x2+22x+85=x2+20x+1002x^2 + 22x + 85 = x^2 + 20x + 100

Rearrange to form a quadratic equation:
2x2+22x+85x220x100=02x^2 + 22x + 85 - x^2 - 20x - 100 = 0
x2+2x15=0x^2 + 2x - 15 = 0

Factor the quadratic equation:
(x+5)(x3)=0(x + 5)(x - 3) = 0

Solve for xx:
x+5=0x=5x + 5 = 0 \Rightarrow x = -5 (Not valid as x>1x > 1)
x3=0x=3x - 3 = 0 \Rightarrow x = 3

Therefore, substituting x=3x = 3 will provide the side lengths:
<begin><begin> - Short side: x+2=3+2=5x + 2 = 3 + 2 = 5
<sametime><same time> - Other side: x+9=3+9=12x + 9 = 3 + 9 = 12
<sell><sell> - Hypotenuse: x+10=3+10=13x + 10 = 3 + 10 = 13

These side lengths 55, 1212, and 1313 form a well-known Pythagorean triple. Therefore, the solution to the problem is 5,12,135, 12, 13.

Answer

5,12,13 5,12,13

Exercise #10

A right triangle is shown below.

x>1

Find the lengths of the sides of the triangle.

x+2x+2x+2xxxx+4x+4x+4

Video Solution

Step-by-Step Solution

To solve this problem, we begin by using the Pythagorean theorem, as the triangle is right-angled. Let's identify the hypotenuse:

  • The side lengths are given as x x , x+2 x + 2 , and x+4 x + 4 .
  • Since x+4 x + 4 is the largest, it will serve as the hypotenuse, so we'll denote the sides as follows: a=x a = x , b=x+2 b = x + 2 , and c=x+4 c = x + 4 .

Using the Pythagorean theorem, we write:

x2+(x+2)2=(x+4)2 x^2 + (x + 2)^2 = (x + 4)^2

Let's expand and simplify the equation:

x2+(x2+4x+4)=x2+8x+16 x^2 + (x^2 + 4x + 4) = x^2 + 8x + 16

Simplifying further:

2x2+4x+4=x2+8x+16 2x^2 + 4x + 4 = x^2 + 8x + 16

Rearrange all terms to one side:

2x2+4x+4x28x16=0 2x^2 + 4x + 4 - x^2 - 8x - 16 = 0

Simplifying gives:

x24x12=0 x^2 - 4x - 12 = 0

This is a standard quadratic equation, which we can solve using factoring. By factoring, we find:

(x6)(x+2)=0 (x - 6)(x + 2) = 0

Setting each factor equal to zero gives solutions x=6 x = 6 and x=2 x = -2 . Since x>1 x > 1 , we discard x=2 x = -2 .

The valid solution is x=6 x = 6 .

Now, substitute x=6 x = 6 back into the expressions for the side lengths:

  • The first side: x=6 x = 6
  • The second side: x+2=6+2=8 x + 2 = 6 + 2 = 8
  • The hypotenuse: x+4=6+4=10 x + 4 = 6 + 4 = 10

Therefore, the lengths of the sides of the triangle are 6 6 , 8 8 , and 10 10 , which matches choice 4.

Therefore, the correct answer is choice 4: 6,8,10 6, 8, 10 .

Answer

6,8,10 6,8,10

Exercise #11

In front of you is a square.

The expressions listed next to the sides describe their length.

( x>-4 length measurements in cm).

Since the area of the square is 36.

Find the lengths of the sides of the square.

363636x+4x+4x+4

Video Solution

Step-by-Step Solution

To solve for the side length of the square, we follow these steps:

  • Step 1: Given each side of the square as x+4 x+4 cm, use the area formula for a square: Area=(side length)2 \text{Area} = (\text{side length})^2 .
  • Step 2: Since the area is 36 cm2^2, set up the equation:
  • (x+4)2=36(x+4)^2 = 36.

  • Step 3: Solve for x+4 x+4 :
  • Taking the square root of both sides,

    x+4=6x+4 = 6 or x+4=6x+4 = -6.

  • Step 4: Solve each equation for x x :
    • From x+4=6x+4 = 6, we get x=2x = 2.
    • From x+4=6x+4 = -6, we get x=10x = -10.
  • Step 5: Apply the condition x>4 x > -4 :
  • Only x=2x = 2 satisfies the condition x>4x > -4.

  • Step 6: Calculate the side length using x=2 x = 2 :
  • Side length = x+4=2+4=6 x+4 = 2+4 = 6 cm.

Therefore, the length of the sides of the square is 6 6 cm.

Answer

6

Exercise #12

In front of you is a square.

The expressions listed next to the sides describe their length.

( x>-2 length measurements in cm).

Since the area of the square is 16.

Find the lengths of the sides of the square.

161616x+2x+2x+2

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Set up the equation using the area of the square.
  • Step 2: Solve for x x using algebraic methods.
  • Step 3: Validate the solution against the given condition.

Now, let's work through each step:
Step 1: Given that the area of the square is 16, we use the area formula: (x+2)2=16(x+2)^2 = 16.
Step 2: Solving the equation, take the square root of both sides:
x+2=±4 x + 2 = \pm 4 This produces two solutions: x+2=4orx+2=4 x + 2 = 4 \quad \text{or} \quad x + 2 = -4 Step 3: Solve each equation for x x :
For x+2=4 x + 2 = 4 , we have: x=42=2 x = 4 - 2 = 2 For x+2=4 x + 2 = -4 , we have: x=42=6 x = -4 - 2 = -6 Since x>2 x > -2 , we discard the solution x=6 x = -6 because it does not satisfy the condition.
Thus, the acceptable value is x=2 x = 2 .

The length of the sides of the square is x+2=2+2=4 x + 2 = 2 + 2 = 4 cm.

Therefore, the solution to the problem is 4, and it matches with choice 1.

Answer

4

Exercise #13

In front of you is an isosceles right triangle.

The expressions listed next to the sides describe their length.

( x>-5 length measurements in cm).

Since the area of the triangle is 12.5.

Find the lengths of the sides of the triangle.

12.512.512.5x+5x+5x+5

Video Solution

Step-by-Step Solution

The problem involves finding the side lengths of an isosceles right triangle given its area. Let's proceed with the solution.

Since the triangle is an isosceles right triangle, the two legs are equal, and the area AA is provided by the formula: A=12baseheight A = \frac{1}{2} \cdot \text{base} \cdot \text{height} For this triangle, both the base and the height are x+5x + 5.

The area is given as 12.5, so we set up the equation: 12.5=12(x+5)(x+5) 12.5 = \frac{1}{2} \cdot (x + 5) \cdot (x + 5) 12.5=12(x+5)2 12.5 = \frac{1}{2} \cdot (x + 5)^2

Multiply both sides by 2 to solve for (x+5)2(x + 5)^2: 25=(x+5)2 25 = (x + 5)^2

Take the square root of both sides: x+5=25 x + 5 = \sqrt{25} x+5=5 x + 5 = 5

Solve for xx: x=55 x = 5 - 5 x=0 x = 0

Therefore, the length of each leg of the triangle is x+5=5x + 5 = 5 cm.

For the hypotenuse cc, use the Pythagorean theorem c2=a2+a2c^2 = a^2 + a^2, where a=x+5=5a = x + 5 = 5: c2=52+52=25+25=50 c^2 = 5^2 + 5^2 = 25 + 25 = 50 c=50=52 c = \sqrt{50} = 5\sqrt{2}

Thus, the lengths of the sides of the triangle are 55, 55, and 525\sqrt{2}.

Therefore, the correct solution is 5,5,52 5, 5, 5\sqrt{2} .

Answer

5,5,52 5,5,5\sqrt{2}

Exercise #14

In front of you is an isosceles right triangle.

The expressions listed next to the sides describe their length.

( x>-8 length measurements in cm).

Since the area of the triangle is 32.

Find the lengths of the sides of the triangle.

323232x+8x+8x+8

Video Solution

Step-by-Step Solution

To solve this problem, we'll utilize the properties of an isosceles right triangle and the area formula:

  • Step 1: Identify the expression for the legs of the triangle as x+8 x + 8 .
  • Step 2: Use the area formula for a right triangle, 12×(leg)2=32 \frac{1}{2} \times (\text{leg})^2 = 32 .
  • Step 3: Given that the triangle is isosceles, solve for x x by substituting and expanding the terms.
  • Step 4: Derive the hypotenuse using the known leg value and calculate leg×2 \text{leg} \times \sqrt{2} .

Now, let's work through each step:

Step 1: Recognize both legs of the isosceles triangle are x+8 x + 8 .

Step 2: Apply the area formula for right triangles:
We know 12×(leg)×(leg)=32 \frac{1}{2} \times (\text{leg}) \times (\text{leg}) = 32 . Therefore, the equation is:

12×(x+8)2=32 \frac{1}{2} \times (x + 8)^2 = 32

Step 3: Simplify and solve for x x .

(x+8)2=64(x + 8)^2 = 64 x+8=64x + 8 = \sqrt{64} x+8=8orx+8=8x + 8 = 8 \quad \text{or} \quad x + 8 = -8

Given the constraint x>8x > -8, we discard x+8=8x + 8 = -8 since it violates the condition. Therefore,

x+8=8x + 8 = 8 x=0x = 0

Recalculate x+8 x + 8 , which states the leg is 8.

Step 4: Determine the hypotenuse.

hypotenuse=82hypotenuse = 8\sqrt{2}

Therefore, the side lengths of the triangle are 8,8,82 8, 8, 8\sqrt{2} . Match this to the choices given, which is option 3.

The lengths of the sides of the triangle are 8,8,82 8, 8, 8\sqrt{2} .

Answer

8,8,82 8,8,8\sqrt{2}

Topics learned in later sections

  1. Uses of Factorization