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.

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.

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 $0$.

- 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 $0$.

Find the value of the parameter x.

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

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 $x=7$.

**If you are interested in this article, you might also be interested in the following articles:**

- The uses of factorization
- Factorization according to short multiplication formulas
- Factorization through the extraction of the common factor outside the parentheses
- Factorization of trinomials
- Factorization of algebraic fractions
- Addition and subtraction of algebraic fractions
- Simplification of algebraic fractions
- Multiplication and division of algebraic fractions

**At** **Tutorela** **you will find a variety of articles on mathematics.**

Find the value of the parameter x.

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

Let's open 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:

$2x=0$

$x=0$

$x=0,x=10$

Find the value of the parameter x.

$(x-5)^2=0$

$x=5$

Find the value of the parameter x.

$x^2-25=0$

$x=5,x=-5$

Find the value of the parameter x.

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

$x=1,x=2.5$

Find the value of the parameter x.

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

$x=-3,x=-4$

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.

\( 2x^2-7x+5=0 \)

Question 3

Find the value of the parameter x.

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

Related Subjects

- Algebraic Method
- Factorization
- Factoring Trinomials
- Extracting the common factor in parentheses
- Factoring using contracted multiplication
- Factorization: Common factor extraction
- The Extended Distributive Property
- Abbreviated Multiplication Formulas
- The formula for the difference of squares
- The formula for the sum of squares
- Multiplication of the sum of two elements by the difference between them
- Formulas for Cubic Expressions
- The quadratic function
- Parabola
- Symmetry in a parabola
- Plotting the Quadratic Function Using Parameters a, b and c
- Finding the Zeros of a Parabola
- Methods for solving a quadratic function
- Squared Trinomial