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 $0$ 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.

Examples with solutions for Factorization

Exercise #1

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

What is the value of X?

Step-by-Step Solution

The equation in the problem is:

$x^2+6x+9=0$We want to solve this equation using factoring,

First, we'll check if we can factor out a common factor, but this is not possible, since there is no common factor for all three terms on the left side of the equation, we can identify that we can factor the expression on the left side using the quadratic formula for a trinomial squared, however, we prefer to factor it using the factoring method according to trinomials, let's refer to the search for Factoring by trinomials:

Let's note that the coefficient of the squared term (the term with the second power) is 1, so we can try to perform factoring according to the quick trinomial method: (This factoring is also called "automatic trinomial"),

But before we do this in the problem - let's recall the general rule for factoring by quick trinomial method:

The rule states that for the algebraic quadratic expression of the general form:

$x^2+bx+c$We can find a factorization in the form of a product if we can find two numbers $m,\hspace{4pt}n$such that the following conditions are met (conditions of the quick trinomial method):

$\begin{cases} m\cdot n=c\\ m+n=b \end{cases}$If we can find two such numbers $m,\hspace{4pt}n$then we can factor the general expression mentioned above into the form of a product and present it as:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$which is its factored form (product factors) of the expression,

Let's return now to the equation in the problem that we received in the last stage after arranging it:

$x^2+6x+9=0$Let's note that the coefficients from the general form we mentioned in the rule above:

$x^2+bx+c$are:$\begin{cases} c=9 \\ b=6 \end{cases}$where we didn't forget to consider the coefficient together with its sign,

Let's continue, we want to factor the expression on the left side into factors according to the quick trinomial method, above, so we'll look for a pair of numbers $m,\hspace{4pt}n$ that satisfy:

$\begin{cases} m\cdot n=9\\ m+n=6 \end{cases}$We'll try to identify this pair of numbers through logical thinking and using our knowledge of the multiplication table, we'll start from the multiplication between the two required numbers $m,\hspace{4pt}n$ that is - from the first row of the pair of requirements we mentioned in the last stage:

$m\cdot n=9$We identify that their product needs to give a positive result, and therefore we can conclude that their signs are identical,

Next, we'll consider the factors (integers) of the number 9, and from our knowledge of the multiplication table we can know that there are only two possibilities for such factors: 3 and 3, or 9 and 1, as we previously concluded that their signs must be identical, a quick check of the two possibilities regarding the fulfillment of the second condition:

$m+n=6$ will lead to a quick conclusion that the only possibility for fulfilling both of the above conditions together is:

$3,\hspace{4pt}3$That is - for:

$m=3,\hspace{4pt}n=3$(It doesn't matter which one we call m and which one we call n)

It is satisfied that:

$\begin{cases} \underline{3}\cdot \underline{3}=9\\ \underline{3}+\underline{3}=6 \end{cases}$ From here - we understood what the numbers we are looking for are and therefore we can factor the expression on the left side of the equation in question and present it as a product:

$x^2+6x+9 \\ \downarrow\\ (x+3)(x+3)$

In other words, we performed:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

If so, we have factored the quadratic expression on the left side of the equation into factors using the quick trinomial method, and the equation is:

$x^2+6x+9=0 \\ \downarrow\\ (x+3)(x+3)=0\\ (x+3)^2=0\\$Where in the last stage we noticed that in the expression on the left side the term:

$(x+3)$

multiplies itself and therefore the expression can be written as a squared term:

$(x+3)^2$

Now that the expression on the left side has been factored into a product form (in this case not just a product but actually a power form) we will continue to quickly solve the equation we received:

$(x+3)^2=0$

Let's pay attention to a simple fact, on the left side there is a term raised to the second power, and on the right side the number 0,

and only 0 squared (to the second power) will give the result 0, so we get that the equation equivalent to this equation is the equation:

$x+3=0$(In the same way we could have operated algebraically in a pure form and taken the square root of both sides of the equation, we'll discuss this in a note at the end)

We'll solve this equation by moving the free number to the other side and we'll get that the only solution is:

$x=-3$Let's summarize then the stages of solving the quadratic equation using the quick trinomial factoring method, we got that:

$x^2+6x+9=0 \\ \downarrow\\ (x+3)(x+3)=0\\ (x+3)^2=0\\ \downarrow\\ x+3=0\\ x=-3$Therefore, the correct answer is answer C.

Note:

We could have reached the final equation by taking the square root of both sides of the equation, however - taking a square root involves considering two possibilities: positive and negative (it's enough to consider this only on one side, as described in the calculation below), that is, we could have performed:

$(x+3)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{8pt}}\\ \downarrow\\ \sqrt{(x+3)^2}=\pm\sqrt{0} \\ x+3=\pm0\\ x+3=0$Where on the left side the root (which is a half power) and the second power canceled each other out (this follows from the law of powers for power over power), and on the right side the root of 0 is 0, and we considered two possibilities positive and negative (this is the plus-minus sign indicated) except that the sign (which is actually multiplication by one or minus one) does not affect 0 which remains 0 in both cases, and therefore we reached the same equation we reached with logical and unambiguous thinking earlier - in the solution above,

In any other case where on the right side was a number different from 0, we could have solved only by taking the root etc. and considering the two positive and negative possibilities which would then give two different possibilities for the solution.

3-

Exercise #2

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

Step-by-Step Solution

Let's observe that the given equation:

$x^2-5x-50=0$is a quadratic equation that can be solved using quick factoring:

$x^2-5x-50=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=-50\\ \underline{?}+\underline{?}=-5\end{cases}\\ \downarrow\\ (x-10)(x+5)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-10)(x+5)=0 \\ \downarrow\\ x-10=0\rightarrow\boxed{x=10}\\ x+5=0\rightarrow\boxed{x=-5}\\ \boxed{x=10,-5}$Therefore, the correct answer is answer C.

$x=10,x=-5$

Exercise #3

$x^2-19x+60=0$

Step-by-Step Solution

Let's observe that the given equation:

$x^2-19x+60=0$is a quadratic equation that can be solved using quick factoring:

$x^2-19x+60=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=60\\ \underline{?}+\underline{?}=-19\end{cases}\\ \downarrow\\ (x-4)(x-15)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-4)(x-15)=0 \\ \downarrow\\ x-4=0\rightarrow\boxed{x=4}\\ x-15=0\rightarrow\boxed{x=15}\\ \boxed{x=4,15}$Therefore, the correct answer is answer A.

$x=15,x=4$

Exercise #4

$x^2+10x-24=0$

Step-by-Step Solution

Let's observe that the given equation:

$x^2+10x-24=0$is a quadratic equation that can be solved using quick factoring:

$x^2+10x-24=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=-24\\ \underline{?}+\underline{?}=10\end{cases}\\ \downarrow\\ (x+12)(x-2)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x+12)(x-2)=0 \\ \downarrow\\ x+12=0\rightarrow\boxed{x=-12}\\ x-2=0\rightarrow\boxed{x=2}\\ \boxed{x=-12,2}$Therefore, the correct answer is answer B.

$x=2,x=-12$

Exercise #5

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

Step-by-Step Solution

Let's observe that the given equation:

$x^2-2x-3=0$is a quadratic equation that can be solved using quick factoring:

$x^2-2x-3=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=-3\\ \underline{?}+\underline{?}=-2\end{cases}\\ \downarrow\\ (x-3)(x+1)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-3)(x+1)=0 \\ \downarrow\\ x-3=0\rightarrow\boxed{x=3}\\ x+1=0\rightarrow\boxed{x=-1}\\ \boxed{x=-1,3}$Therefore, the correct answer is answer B.

$x=3,x=-1$

Exercise #6

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

Step-by-Step Solution

Let's observe that the given equation:

$x^2-7x+12=0$is a quadratic equation that can be solved using quick factoring:

$x^2-7x+12=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=12\\ \underline{?}+\underline{?}=-7\end{cases}\\ \downarrow\\ (x-3)(x-4)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-3)(x-4)=0 \\ \downarrow\\ x-3=0\rightarrow\boxed{x=3}\\ x-4=0\rightarrow\boxed{x=4}\\ \boxed{x=3,4}$Therefore, the correct answer is answer A.

$x=3,x=4$

Exercise #7

$x^2-3x-18=0$

Step-by-Step Solution

Let's observe that the given equation:

$x^2-3x-18=0$is a quadratic equation that can be solved using quick factoring:

$x^2-3x-18=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=-18\\ \underline{?}+\underline{?}=-3\end{cases}\\ \downarrow\\ (x-6)(x+3)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-6)(x+3)=0 \\ \downarrow\\ x-6=0\rightarrow\boxed{x=6}\\ x+3=0\rightarrow\boxed{x=-3}\\ \boxed{x=6,-3}$Therefore, the correct answer is answer A.

$x=-3,x=6$

Exercise #8

$x^2-3x-18=0$

Step-by-Step Solution

Let's observe that the given equation:

$x^2-3x-18=0$is a quadratic equation that can be solved using quick factoring:

$x^2-3x-18=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=-18\\ \underline{?}+\underline{?}=-3\end{cases}\\ \downarrow\\ (x-6)(x+3)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-6)(x+3)=0 \\ \downarrow\\ x-6=0\rightarrow\boxed{x=6}\\ x+3=0\rightarrow\boxed{x=-3}\\ \boxed{x=6,-3}$Therefore, the correct answer is answer A.

$x=-3,x=6$

Exercise #9

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

Step-by-Step Solution

Let's observe that the given equation:

$x^2+9x+20=0$is a quadratic equation that can be solved using quick factoring:

$x^2+9x+20=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=20\\ \underline{?}+\underline{?}=9\end{cases}\\ \downarrow\\ (x+5)(x+4)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x+5)(x+4)=0 \\ \downarrow\\ x+5=0\rightarrow\boxed{x=-5}\\ x+4=0\rightarrow\boxed{x=-4}\\ \boxed{x=-4,-5}$Therefore, the correct answer is answer A.

$x=-4,x=-5$

Exercise #10

$x^2+10x+16=0$

Step-by-Step Solution

Let's observe that the given equation:

$x^2+10x+16=0$is a quadratic equation that can be solved using quick factoring:

$x^2+10x+16=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=16\\ \underline{?}+\underline{?}=10\end{cases}\\ \downarrow\\ (x+2)(x+8)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x+2)(x+8)=0 \\ \downarrow\\ x+2=0\rightarrow\boxed{x=-2}\\ x+8=0\rightarrow\boxed{x=-8}\\ \boxed{x=-2,-8}$Therefore, the correct answer is answer B.

$x=-8,x=-2$

Exercise #11

$x^2-1=0$

Step-by-Step Solution

Let's solve the given equation:

$x^2-1=0$ We will do this simply by isolating the unknown on one side and taking the square root of both sides:

$x^2-1=0 \\ x^2=1\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \boxed{x=\pm1}$

$x=\pm1$

Exercise #12

Complete the equation:

$(x+3)(x+\textcolor{red}{☐})=x^2+5x+6$

Step-by-Step Solution

Let's simplify the expression given in the left side:

$(x+3)(x+\textcolor{purple}{\boxed{?}})$ For ease of calculation we will replace the square with the question mark (indicating the missing part that needs to be completed) with the letter $\textcolor{purple}{k}$, meaning we will perform the substitution:

$(x+3)(x+\textcolor{purple}{\boxed{?}})=x^2+5x+6 \\ \downarrow\\ (x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\$Next, we will expand the parentheses using the expanded distribution law:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$Let's note that in the formula template for the distribution law mentioned we assume by default that the operation between the terms inside the parentheses is addition, so we won't forget of course that the sign preceding the term is an integral part of it, and we will also apply the laws of sign multiplication and thus we can represent any expression in parentheses, which we expand using the aforementioned formula, first, as an expression where addition is performed between all terms (if necessary),

Therefore, we will first represent each of the expressions in parentheses in the multiplication on the left side as an expression where addition exists:

$(x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ \big(x+(+3)\big)\big(x+(\textcolor{purple}{+k})\big)=x^2+5x+6 \\$Now for convenience, let's write down again the expanded distribution law mentioned earlier:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$And we'll apply it to our problem:

$\big (\textcolor{red}{x}+\textcolor{blue}{(+3)}\big)\big(x+(+\textcolor{purple}{k})\big)=x^2+5x+6 \\ \downarrow\\ \textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (+\textcolor{purple}{k})+\textcolor{blue}{(+3)}\cdot x +\textcolor{blue}{(+3)}\cdot (+\textcolor{purple}{k})=x^2+5x+6 \\$We'll continue and apply the laws of multiplication signs, remembering that multiplying expressions with identical signs will yield a positive result, and multiplying expressions with different signs will yield a negative result:

$\textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (+\textcolor{purple}{k})+\textcolor{blue}{(+3)}\cdot x +\textcolor{blue}{(+3)}\cdot (+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ x^2+\textcolor{purple}{k}x+3x+3\textcolor{purple}{k}=x^2+5x+6 \\$Now, we want to present the expression on the left side in a form identical to the expression on the right side, that is - as a sum of three terms with different exponents: second power (squared), first power, and zero power (i.e., the free number - not dependent on x). To do this - we will factor out the part of the expression on the left side where the terms are in the first power:

$x^2+\underline{\textcolor{purple}{k}x+3x}+3\textcolor{purple}{k}=x^2+5x+6 \\ \downarrow\\ x^2+\underline{(\textcolor{purple}{k}+3)x}+3\textcolor{purple}{k}=x^2+5x+6$

Now in order for equality to hold - we require that the coefficient of the first-power term on both sides of the equation be identical and at the same time - we require that the free term on both sides of the equation be identical as well:

$x^2+\underline{\underline{(\textcolor{purple}{k}+3)}}x+\underline{\underline{\underline{3\textcolor{purple}{k}}}}=x^2+\underline{\underline{5}}x+\underline{\underline{\underline{6}}}$In other words, we require that:

$\begin{cases} \textcolor{purple}{k}+3=5\rightarrow\boxed{\textcolor{purple}{k}=2}\\ 3\textcolor{purple}{k}=6\rightarrow\boxed{\textcolor{purple}{k}=2} \end{cases}$

Let's summarize the solution steps:

$(x+3)(x+\textcolor{purple}{\boxed{?}})=x^2+5x+6 \leftrightarrow\textcolor{red}{\boxed{\textcolor{purple}{\boxed{?}}=\textcolor{purple}{k}}} \\ \downarrow\\ (x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ x^2+\underline{\textcolor{purple}{k}x+3x}+3\textcolor{purple}{k}=x^2+5x+6 \\ \downarrow\\ x^2+\underline{(\textcolor{purple}{k}+3)x}+3\textcolor{purple}{k}=x^2+5x+6\\ \downarrow\\ x^2+\underline{\underline{(\textcolor{purple}{k}+3)}}x+\underline{\underline{\underline{3\textcolor{purple}{k}}}}=x^2+\underline{\underline{5}}x+\underline{\underline{\underline{6}}} \\ \begin{cases} \textcolor{purple}{k}+3=5\rightarrow\boxed{\textcolor{purple}{k}=2}\\ 3\textcolor{purple}{k}=6\rightarrow\boxed{\textcolor{purple}{k}=2} \end{cases} \\ \textcolor{red}{\bm{\rightarrow}\boxed{\textcolor{purple}{\boxed{?}}=\textcolor{purple}{2}}}$Therefore, the missing expression is the number $2$meaning - the correct answer is a'.

2

Exercise #13

How many solutions does the equation have?

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

Step-by-Step Solution

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 will focus on the left side of the equation and then on the right side (the number 0).

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:

$x^2+12x+36=0$In order to find the additional solutions to the equation we must solve the equation:

Note that the first coefficient is 1, so we can try to solve it using the trinomial formula.

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

$(\textcolor{red}{a}+\textcolor{blue}{b})^2 = \textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$The reason for trying factoring in this approach is that we can identify in the left side of the equation we got in the last step, that the two terms which are in the far sides (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 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 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 $2\cdot a \cdot b$So, we start by presenting the equation of the short 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 using the following:

$ax^2+bx+c =0$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 coefficients:

$\Delta\geq0$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

Exercise #14

How many solutions does the equation have?

$x^2+10x+9=0$

Step-by-Step Solution

Let's observe that the given equation:

$x^2+10x+9=0$is a quadratic equation that can be solved using quick factoring:

$x^2+10x+9=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=9\\ \underline{?}+\underline{?}=10\end{cases}\\ \downarrow\\ (x+1)(x+9)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x+1)(x+9)=0\\ \downarrow\\ x+1=0\rightarrow\boxed{x=-1}\\ x+9=0\rightarrow\boxed{x=-9}\\ \boxed{x=-1,-9}$and therefore the given equation has two solutions,

Two solutions

Exercise #15

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

Step-by-Step Solution

The equation in the problem is:

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

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 and performing the factorization:

$2x^{90}-4x^{89}=0 \\ \downarrow\\ 2x^{89}(x-2)=0$Let's continue and remember 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.

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 to be equal to zero,

Meaning:

$2x^{89}=0 \hspace{8pt}\text{/}:2\\ x^{89}=0 \hspace{8pt}\text{/}\sqrt[89]{\hspace{6pt}}\\ \boxed{x=0}$

Or:

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

In summary:

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

$x=0,2$