**Factorization allows us to convert expressions with elements that are added or subtracted into expressions with elements that are multiplied.**

**Factorization allows us to convert expressions with elements that are added or subtracted into expressions with elements that are multiplied.**

Factorization helps to solve different exercises, including 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.

Question 1

How many solutions does the equation have?

\( x^4+12x^3+36x^2=0 \)

Question 2

\( 2x^{90}-4x^{89}=0 \)

Question 3

Extract the common factor:

\( 4x^3+8x^4= \)

Question 4

Solve the following by removing a common factor:

\( 6x^6-9x^4=0 \)

Question 5

\( x^2+6x+9=0 \)

What is the value of X?

How many solutions does the equation have?

$x^4+12x^3+36x^2=0$

Let's solve the given equation:

$x^4+12x^3+36x^2=0$We note that it is possible __to factor the expression__ which is in the left side of the given equation, this is done by __taking out the common factor__ $x^2$ which is the __greatest common factor of the numbers and letters__ in the expression:

$x^4+12x^3+36x^2=0 \\
\downarrow\\
x^2(x^2+12x+36) =0$**We continue** and refer to the expression on the left side of the equation that we got in the last step which contains algebraic expressions and on the right side the number 0, therefore, **since the only way to get the result 0 from a product is to multiply by 0**, __at least__ one of the expressions in the product on the left side, must be equal to zero,

Meaning:

$x^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \boxed{x=0}$

__Or:__

$x^2+12x+36=0$** We continue** and in order to find the

__Note that the leading coefficient is 1__, so we can (try) to solve it using __the quick trinomial formula,__

However, we can factor, in this case, also using __the short multiplication formula for a binomial squared:__

$(\textcolor{red}{a}+\textcolor{blue}{b})^2 = \textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$**The motivation** for trying factoring in this approach is the fact that we can identify in the left side expression in the equation we got in the last step, that the two terms which are **in the extreme positions** (meaning the term in the first position - it is the squared term and the term in the zero position - it is the free number in the expression) can be presented (simply) as a squared term:

$x^2+12x+36=0 \\ \downarrow\\ x^2+12x+6^2=0$

Equating the expression on the left side **in the equation:**

$\downarrow\\ x^2+12x+6^2$

To the expression on the right side **in the short multiplication formula above:**

$\textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

The conclusion from this is that **what remains to check is whether the middle term in the expression in the equation** matches the middle term in the short multiplication formula above, meaning - after identifying $a$$b$which are both in the first position in the short multiplication formula above in which $a$and $b$we check if the middle term in the expression in the left side of the equation **can be presented** as $2\cdot a \cdot b$we check if so, we start by presenting the equality of the short multiplication formula to the given expression:

$\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2 \Leftrightarrow \textcolor{red}{x}^2+\underline{12x}+\textcolor{blue}{6}^2$__And__ ** indeed it holds** that:

$2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}=12x$Meaning **the middle term in the expression in the equation indeed matches the form of the middle term in the short multiplication formula** (highlighted with a line below), mathematically:

$x^2+\underline{12x}+6^2=0 \\
\textcolor{red}{x}^2+\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}}+\textcolor{blue}{6}^2=0\\
\downarrow\\
(\textcolor{red}{x}+\textcolor{blue}{6})^2=0$We can now remember that **a real root can be calculated only for a positive number or for the number zero** (since __it is not possible to get a negative number from squaring a real number itself__), and therefore for an equation there are two real solutions (or one solution) only if:

**Next** we note that if: $(x+6)^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\
x+6=0\\
\boxed{x=-6}$then the __only__ solution to the equation is:

$x^4+12x^3+36x^2=0 \\
\downarrow\\
x^2(x^2+12x+36) =0 \\
\downarrow\\
x^2=0\rightarrow\boxed{x=0} \\
x^2+12x+36=0\\
x^2+2\cdot2\cdot6+6^2=0\\
\rightarrow(x+6)^2=0\rightarrow\boxed{x=-6}\\
\downarrow\\
\boxed{x=0,-6}$Therefore, ** we can summarize what was explained in the rule**:

In the quadratic equation:

$ax^2+bx+c =0$in which the coefficients are substituted and the discriminant is calculated:

$a,b$__If it holds:__

** a.**$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$:

__There is no (real) solution__ to the equation.

** b.**$\Delta$:

__There exists a single (real) solution__ to the equation.

** c.**$x_{1,2}=\frac{-b\pm\sqrt{\textcolor{orange}{\Delta}}}{2a} \longleftrightarrow\textcolor{orange}{\Delta}=b^2-4ac$:

__There exist two (real) solutions__ to the equation.

** Now let's return to the given equation** and extract from it the

$\Delta\geq0$We continue and __calculate $\Delta=0$:__

$x=-\frac{b}{2a}$Therefore for the quadratic equation that we solved, ** one (real) solution**,

and in combination with the solution $ax^2+bx+c =0$(the additional solution we found for the given equation which is indicated in the first step after factoring using the common factor),

**Therefore we get that for the given equation:**

$\Delta=b^2-4ac$

__two real solutions.__

Two solutions

$2x^{90}-4x^{89}=0$

__The equation in the problem is:__

$2x^{90}-4x^{89}=0$__Beginning__ Let's pay attention to the left side **The expression can be broken down into factors by taking out a common factor,** The **greatest** common factor for the numbers and letters in this case is $2x^{89}$and that is because the power of 89 is the ** lowest power in the equation** and therefore

For the numbers, note that the number 4 is a multiple of the number 2, so the number 2 is the **greatest** common factor __for the numbers__ for the two terms in the expression,

**Continuing** if so and performing the factorization:

$2x^{90}-4x^{89}=0 \\
\downarrow\\
2x^{89}(x-2)=0$Let's continue and refer to the fact that on the left side of the equation that was obtained in the last step there is an algebraic expression and on the right side the number is 0, so, **since the only way to get the result 0 from a product is for at least**, __one of the factors in the product on the left side, must be equal to zero__,

Meaning:

$2x^{89}=0 \hspace{8pt}\text{/}:2\\ x^{89}=0 \hspace{8pt}\text{/}\sqrt[89]{\hspace{6pt}}\\ \boxed{x=0}$In the solution of the equation above, we first equated the two sides of the equation by moving the term with the unknown and then we took out a root of order 89 on both sides of the equation.

(In this case __taking out an odd-order root on the right side of the equation yields one possibility)__

__Or:__

$x-2=0 \\ \boxed{x=2}$

**In summary** if so the solution to the equation:

$2x^{90}-4x^{89}=0 \\
\downarrow\\
2x^{89}(x-2)=0 \\
\downarrow\\
2x^{89}=0 \rightarrow\boxed{ x=0}\\
x-2=0\rightarrow \boxed{x=2}\\
\downarrow\\
\boxed{x=0,2}$__And therefore the correct answer is answer a.__

$x=0,2$

Extract the common factor:

$4x^3+8x^4=$

First, we use the power law to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$It is necessary to keep in mind that:

$x^4=x^3\cdot x$Next, we return to the problem and extract the greatest common factor for the numbers separately and for the letters separately,

For the numbers, the greatest common factor is

$4$and for the letters it is:

$x^3$and therefore for the extraction

$4x^3$outside the parenthesis

We obtain the expression:

$4x^3+8x^4=4x^3(1+2x)$To determine what the expression inside the parentheses is, we use the power law, our knowledge of the multiplication table, and the answer to the question: "How many times do we multiply the common factor that we took out of the parenthesis to obtain each of the terms of the original expression that we factored?

__Therefore, the correct answer is: a.__

It is always recommended to review again and check that you get each and every one of the terms of the expression that is factored when opening the parentheses (through the distributive property), this can be done in the margin, on a piece of scrap paper, or by marking the factor we removed and each and every one of the terms inside the parenthesis, etc.

$4x^3(1+2x)$

Solve the following by removing a common factor:

$6x^6-9x^4=0$

First, we take out the smallest power

$6x^6-9x^4=$

$6x^4\left(x^2-1.5\right)=0$

If possible, we reduce the numbers by a common factor

Finally, we will compare the two sections with: $0$

$6x^4=0$

We divide by: $6x^3$

$x=0$

$x^2-1.5=0$

$x^2=1.5$

$x=\pm\sqrt{\frac{3}{2}}$

$x=0,x=\pm\sqrt{\frac{3}{2}}$

$x^2+6x+9=0$

What is the value of X?

3-

Question 1

\( x^2-3x+2=0 \)

What is the value of X?

Question 2

\( x^2-12x+36=0 \)

What is the value of X?

Question 3

\( -x^2+13x-14=0 \)

What is the value of X?

Question 4

\( 2x^2+4x-6=0 \)

What is the value of X?

Question 5

\( x^2-1=0 \)

$x^2-3x+2=0$

What is the value of X?

$1,2$

$x^2-12x+36=0$

What is the value of X?

6

$-x^2+13x-14=0$

What is the value of X?

$x_1=14,x_2=-1$

$2x^2+4x-6=0$

What is the value of X?

$x_1=1,x_2=-3$

$x^2-1=0$

$x=\pm1$

Question 1

\( x^2-5x-50=0 \)

Question 2

\( x^2-8x+16=0 \)

Question 3

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

Question 4

\( x^2+9x+20=0 \)

Question 5

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

$x^2-5x-50=0$

$x=10,x=-5$

$x^2-8x+16=0$

$x=4$

$x^2+x-2=0$

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

$x^2+9x+20=0$

$x=-4,x=-5$

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

$x=3,x=-1$