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.

Question Types:

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 $0$.

Question 1

Find the value of the parameter x.

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

Question 2

Find the value of the parameter x.

\( x^2-25=0 \)

Question 3

Find the value of the parameter x.

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

Question 4

Find the value of the parameter x.

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

Question 5

Find the value of the parameter x.

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

Find the value of the parameter x.

$(x-5)^2=0$

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:

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

$x-5=0$

$x=5$

or

$x-5=0$

$x=5$

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

$x=5$

Find the value of the parameter x.

$x^2-25=0$

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:

$x-5=0$

$x=5$

$x=5,x=-5$

Find the value of the parameter x.

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

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:

$2x-5=0$

$2x=5$

$x=2.5$

$x=1,x=2.5$

Find the value of the parameter x.

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

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:

$x+3=0$

$x=-3$

$x=-3,x=-4$

Find the value of the parameter x.

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

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:

$2x=0$

$x=0$

$x=0,x=10$

Question 1

Find the value of the parameter x.

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

Question 2

Find the value of the parameter x.

\( 12x^3-9x^2-3x=0 \)

Question 3

Find the value of the parameter x.

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

Question 4

A right triangle is shown below.

\( x>1 \)

Calculate the lengths of the sides of the triangle.

Question 5

A right triangle is shown below.

\( x>1 \)

Find the lengths of the sides of the triangle.

Find the value of the parameter x.

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

$x=0,x=4$

Find the value of the parameter x.

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

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

Find the value of the parameter x.

$(x+5)^2=0$

$x=-5$

A right triangle is shown below.

x>1

Calculate the lengths of the sides of the triangle.

$5,12,13$

A right triangle is shown below.

x>1

Find the lengths of the sides of the triangle.

$6,8,10$

Question 1

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.

Question 2

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.

Question 3

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.

Question 4

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.

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.

6

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.

4

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.

$8,8,8\sqrt{2}$

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.

$5,5,5\sqrt{2}$