Solving Equations by Factoring

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

Let's remember the following property

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

Steps to carry out to solve equations through factorization

Let's move all the elements to one side of the equation and leave $0$ 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 $0$ .

Detailed explanation

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Test yourself on solving equations by factoring!

Find the value of the parameter x.

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

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

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Examples and exercises with solutions for solving equations through factorization

Exercise #1

Find the value of the parameter x.

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

Video Solution

Step-by-Step Solution

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

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

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

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

We will factor out a common factor:

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

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

Therefore:

$x-1=0$ $x=1$

Or:

$2x-5=0$

$2x=5$

$x=2.5$

Answer

$x=1,x=2.5$

Exercise #2

Find the value of the parameter x.

$x^2-25=0$

Video Solution

Step-by-Step Solution

We will factor using the shortened multiplication formulas:

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

$(a+b)^2=a^2+2ab+b^2$

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

According to the first formula:

$x^2=a^2$

We'll take the square root:

$x=a$

$25=b^2$

We'll take the square root:

$b=5$

We'll use the first shortened multiplication formula:

$a^2-b^2=(a-b)(a+b)$

$x^2-25=(x-5)(x+5)=0$

Therefore:

$x+5=0$

$x=-5$

Or:

$x-5=0$

$x=5$

Answer

$x=5,x=-5$

Exercise #3

Find the value of the parameter x.

$(x-5)^2=0$

Video Solution

Step-by-Step Solution

We will factor using the shortened multiplication formulas:

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

$(a+b)^2=a^2+2ab+b^2$

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:

$(x-5)^2=0$

$x-5=0$

$x=5$

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

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

We'll use the trinomial:

$(x-5)(x-5)=0$

$x-5=0$

$x=5$

$x-5=0$

$x=5$

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

Answer

$x=5$

Exercise #4

Solve for x.

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

Video Solution

Step-by-Step Solution

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

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

Divide by -1:

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

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

Therefore:

$x+4=0$

$x=-4$

Or:

$x+3=0$

$x=-3$

Answer

$x=-3,x=-4$

Exercise #5

Find the value of the parameter x.

$(x-4)^2+x(x-12)=16$

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:

$(x-4)^2+x(x-12)=16$

$x^2-8x+16+x^2-12x=16$

$2x^2-20x=0$

$2x(x-10)=0$

Therefore:

$x-10=0$

$x=10$

Or:

$2x=0$

$x=0$

Answer

$x=0,x=10$

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Test your knowledge

Question 1

Find the value of the parameter x.

$$ x^2-25=0 $$

Question 2

Find the value of the parameter x.

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

Question 3

Solve for x.

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

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