Uses of Factorization

Factorization is the main key to solving more complex exercises than those you have studied up to today.
Factorization helps to solve different exercises, among them, those that have algebraic fractions.
In exercises where the sum or difference of their terms equals zero, factorization allows us to see them as a multiplication of $0$ and thus discover the terms that lead them to this result.

For exercises composed of fractions with expressions that may seem complicated, we can break them down into factors, reduce them, and thus end up with much simpler fractions.

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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Let's look at some examples

If we are presented with an exercise like the ones we mentioned before, where the total equals $0$ :
$x^2+5x+4=$

we can factor it in one of the ways that allow us to do so, and we will immediately have the solutions.
The factorization will be as follows:
$(x+4)(x+1)=0$
and the results will be $x=-1,-4$

Another example:
$2x^2+2x=0$

If we factor it, we will obtain:
$2x(x+1)=0$
Therefore, the solutions are: $x=0, -1$

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Examples and exercises with solutions on the uses of 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.

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

Video Solution

Step-by-Step Solution

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

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

We'll divide by minus 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 #4

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 #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.

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

Question 3

Find the value of the parameter x.

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

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