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

The Uses of Factorization

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 00 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.

Practice Factorization

Exercise #1

How many solutions does the equation have?

x4+12x3+36x2=0 x^4+12x^3+36x^2=0

Video Solution

Step-by-Step Solution

Let's solve the given equation:

x4+12x3+36x2=0 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 x2 x^2 which is the greatest common factor of the numbers and letters in the expression:

x4+12x3+36x2=0x2(x2+12x+36)=0 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:

x2=0/x=0 x^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \boxed{x=0}

Or:

x2+12x+36=0 x^2+12x+36=0 We continue and in order to find the additional solutions to the equation we solve the equation that we got in the last step:

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:

(a+b)2=a2+2ab+b2 (\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:

x2+12x+36=0x2+12x+62=0 x^2+12x+36=0 \\ \downarrow\\ x^2+12x+6^2=0

Equating the expression on the left side in the equation:

x2+12x+62 \downarrow\\ x^2+12x+6^2

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

a2+2ab+b2 \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 a b b which are both in the first position in the short multiplication formula above in which a a and b b we check if the middle term in the expression in the left side of the equation can be presented as 2ab 2\cdot a \cdot b we check if so, we start by presenting the equality of the short multiplication formula to the given expression:

a2+2ab+b2x2+12x+62 \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:

2x6=12x 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:

x2+12x+62=0x2+2x6+62=0(x+6)2=0 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/x+6=0x=6 (x+6)^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ x+6=0\\ \boxed{x=-6} then the only solution to the equation is:

x4+12x3+36x2=0x2(x2+12x+36)=0x2=0x=0x2+12x+36=0x2+226+62=0(x+6)2=0x=6x=0,6 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:

ax2+bx+c=0 ax^2+bx+c =0 in which the coefficients are substituted and the discriminant is calculated:

a,b a,b If it holds:

a.x1,2=b±b24ac2a x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} :

There is no (real) solution to the equation.

b.Δ \Delta :

There exists a single (real) solution to the equation.

c.x1,2=b±Δ2aΔ=b24ac 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 coefficients:

Δ0 \Delta\geq0 We continue and calculate Δ=0 \Delta=0 :

x=b2a x=-\frac{b}{2a} Therefore for the quadratic equation that we solved, one (real) solution,

and in combination with the solution ax2+bx+c=0 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:

Δ=b24ac \Delta=b^2-4ac

two real solutions.

Answer

Two solutions

Exercise #2

2x904x89=0 2x^{90}-4x^{89}=0

Video Solution

Step-by-Step Solution

The equation in the problem is:

2x904x89=0 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 2x89 2x^{89} and that is because the power of 89 is the lowest power in the equation and therefore included both in the term where the power is 90 and in the term where the power is 89, any power higher than that is not included in the term where the power of 89 is the lowest, and therefore it is the term with the highest power that can be taken out of all the terms in the expression as a common factor for the letters,

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:

2x904x89=02x89(x2)=0 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:

2x89=0/:2x89=0/89x=0 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:

x2=0x=2 x-2=0 \\ \boxed{x=2}

In summary if so the solution to the equation:

2x904x89=02x89(x2)=02x89=0x=0x2=0x=2x=0,2 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.

Answer

x=0,2 x=0,2

Exercise #3

Extract the common factor:

4x3+8x4= 4x^3+8x^4=

Video Solution

Step-by-Step Solution

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

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

x4=x3x 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 4 and for the letters it is:

x3 x^3 and therefore for the extraction

4x3 4x^3 outside the parenthesis

We obtain the expression:

4x3+8x4=4x3(1+2x) 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.

Answer

4x3(1+2x) 4x^3(1+2x)

Exercise #4

Solve the following by removing a common factor:

6x69x4=0 6x^6-9x^4=0

Video Solution

Step-by-Step Solution

First, we take out the smallest power

6x69x4= 6x^6-9x^4=

6x4(x21.5)=0 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 0

6x4=0 6x^4=0

We divide by: 6x3 6x^3

x=0 x=0

x21.5=0 x^2-1.5=0

x2=1.5 x^2=1.5

x=±32 x=\pm\sqrt{\frac{3}{2}}

Answer

x=0,x=±32 x=0,x=\pm\sqrt{\frac{3}{2}}

Exercise #5

x2+6x+9=0 x^2+6x+9=0

What is the value of X?

Video Solution

Answer

3-

Exercise #1

x23x+2=0 x^2-3x+2=0

What is the value of X?

Video Solution

Answer

1,2 1,2

Exercise #2

x212x+36=0 x^2-12x+36=0

What is the value of X?

Video Solution

Answer

6

Exercise #3

x2+13x14=0 -x^2+13x-14=0

What is the value of X?

Video Solution

Answer

x1=14,x2=1 x_1=14,x_2=-1

Exercise #4

2x2+4x6=0 2x^2+4x-6=0

What is the value of X?

Video Solution

Answer

x1=1,x2=3 x_1=1,x_2=-3

Exercise #5

x21=0 x^2-1=0

Video Solution

Answer

x=±1 x=\pm1

Exercise #1

x25x50=0 x^2-5x-50=0

Video Solution

Answer

x=10,x=5 x=10,x=-5

Exercise #2

x28x+16=0 x^2-8x+16=0

Video Solution

Answer

x=4 x=4

Exercise #3

x2+x2=0 x^2+x-2=0

Video Solution

Answer

(x1)(x+2)=0 (x-1)(x+2)=0

Exercise #4

x2+9x+20=0 x^2+9x+20=0

Video Solution

Answer

x=4,x=5 x=-4,x=-5

Exercise #5

x22x3=0 x^2-2x-3=0

Video Solution

Answer

x=3,x=1 x=3,x=-1