Factoring Trinomials: How many solutions exist to a problem?

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,

Thus, the correct answer is answer B.

In the given equation:

$x^3+1=0$The simplest and fastest way to find the number of its solutions,

will be simply to solve it, we will do this by moving terms to isolate the unknown, then we will take the cube root of both sides of the equation, while remembering that an odd root preserves the sign of the expression inside the root (meaning - the minus sign can be taken out of an odd root):

$x^3+1=0 \\ x^3=-1\hspace{6pt}\text{/}\sqrt[3]{\hspace{4pt}}\\ \downarrow\\ \sqrt[3]{x^3}=\sqrt[3]{-1}\\ x=-\sqrt[3]{1}\\ \boxed{x=-1}$meaning the given equation has a single solution,

therefore the correct answer is answer A.

Let's solve the given equation:

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

First, we identify that we can factor the expression on the left side using factoring out the common factor:

$$$x^3-2x^2+x=0 \\ \downarrow\\ x(x^2-2x+1)=0$

We'll continue and factor the expression in parentheses, we can identify that it can be factored using the perfect square trinomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$

Let's do it:

$x(x^2-2x+1)=0 \\ x(x^2 \textcolor{blue}{-2x}+1^2)=0 \\ x(x^2\textcolor{blue}{-2\cdot x \cdot 1}+1^2)=0 \\ \downarrow\\ x(x-1)^2=0$

We should emphasize that this factoring using the mentioned formula was possible only because the middle term in the expression (which is in first power in this case and highlighted in blue in the previous calculation) indeed matched the middle term in the perfect square trinomial formula,

We'll continue and get two simpler equations and solve them:

$x(x-1)^2=0 \\ \boxed{x=0}\\ (x-1)^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\rightarrow x-1=0\rightarrow \boxed{x=1}\\ \downarrow\\ \boxed{x=0,1}$L

et's summarize the equation solving steps:

$$$x^3-2x^2+x=0 \\ \downarrow\\ x(x^2-2x+1)=0 \\ \downarrow\\ x(x-1)^2=0 \\ \downarrow\\ \boxed{x=0}\\ (x-1)^2=0\rightarrow \boxed{x=1}\\ \downarrow\\ \boxed{x=0,1}$

Therefore, the given equation has two different solutions,

Which means - the correct answer is answer B.

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:

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 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.

Let's solve the given equation:

$x^4-5x^2+4=0$

We identify that this is a bi-quadratic equation that can be easily solved using substitution of a new variable,

That is, let's notice that:

$(x^2)^2=x^4$

Therefore, we can write the given equation in the following form:$\textcolor{blue}{ x^4}-5x^2+4=0 \\ \downarrow\\ \textcolor{blue}{ (x^2)^2}-5x^2+4=0$

Now let's define a new variable, $t$, such that:

$t=x^2$

Therefore, if we substitute this new variable, $t$, in the given equation instead of $x^2$ we'll get an equation that depends only on $t$:

$(\textcolor{red}{x^2})^2-5\textcolor{red}{x^2}+4=0 \\ \downarrow\downarrow \boxed{\textcolor{red}{(x^2=t)}}\\ t^2-5t+4=0$

Now we'll continue and solve the new equation we got for variable $t$, after we find the values of variable t for which the equation holds, we'll go back and substitute each of them in the definition of t that we mentioned before and find the value of x,

We identify that the equation we got in the last step for t is a quadratic equation that can be solved using quick trinomial factoring:

$t^2-5t+4=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=4\\ \underline{?}+\underline{?}=-5\end{cases}\\ \downarrow\\ (t-4)(t-1)=0$

Therefore we'll get two simpler equations from which we'll extract the solution for t:

$(t-4)(t-1)=0 \\ \downarrow\\ t-4=0\rightarrow\boxed{t=4}\\ t-1=0\rightarrow\boxed{t=1}\\ \boxed{t=1,4}$

Now let's go back to the definition of t that was mentioned before, let's recall it:

$t=x^2$And let's notice that since the power of x is even, the variable t can get only non-negative values (meaning positive or zero),

Therefore the two values we got for t from solving the quadratic equation are indeed valid,

We'll continue and substitute each of the two values we got for t in the definition of t mentioned before to solve the equation and then extract the corresponding value of x by solving the resulting equation using square root on both sides:

$\boxed{x^2=t}\\ \downarrow\\ t=1\textcolor{blue}{\rightarrow} x^2=1\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\textcolor{blue}{\rightarrow} \boxed{x=\pm1}\\ t=4\textcolor{blue}{\rightarrow} x^2=4\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\textcolor{blue}{\rightarrow} \boxed{x=\pm2}\\ \downarrow\\ \boxed{x=\pm1,\pm2}$

Let's summarize the steps of solving the equation:

$x^4-5x^2+4=0 \\ \downarrow\\ (\textcolor{red}{x^2})^2-5\textcolor{red}{x^2}+4=0 \\ \downarrow\downarrow \boxed{\textcolor{red}{(x^2=t)}}\\ t^2-5t+4=0 \\ \downarrow\\ (t-4)(t-1)=0 \\ \downarrow\\ t-4=0\rightarrow\boxed{t=4}\\ t-1=0\rightarrow\boxed{t=1}\\ \boxed{t=1,4}\\ \updownarrow\updownarrow\\ \boxed{x^2=t}\\ \downarrow\\ t=1\textcolor{blue}{\rightarrow} x^2=1\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\textcolor{blue}{\rightarrow} \boxed{x=\pm1}\\ t=4\textcolor{blue}{\rightarrow} x^2=4\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\textcolor{blue}{\rightarrow} \boxed{x=\pm2}\\ \downarrow\\ \boxed{x=\pm1,\pm2}$

Therefore the given equation has 4 different solutions,

Which means the correct answer is answer D.