Solving Equations by Factoring

🏆Practice solving equations by factoring

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 .
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Test yourself on solving equations by factoring!

einstein

Solve for x.

\( -x^2-7x-12=0 \)

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Example to solve equations through factorization
x2+49=14xx^2+49=14x
First, we will transpose all the terms to one side of the equation. On the other side, we will leave 0 0 .
We will obtain:
x2+4914x=0x^2+49-14x=0
We see that it is a trinomial. Let's arrange the equation to clearly see the trinomial:
x214x+49=0x^2-14x+49=0
Now, let's factorize and we will obtain:
(x7)(x7)=0(x-7)(x-7)=0
We can easily realize that the equation equals zero when x=7x=7
Therefore, the solution to the exercise is x=7x=7.


Examples and exercises with solutions for solving equations through factorization

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.

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 #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

Let's 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.

(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 #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

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