Question

How do you write $y=2{{x}^{3}}+10{{x}^{2}}+12x$ in factored form?

Answer 1

Hint: In this problem we have given an equation and asked to write it in factored form. We can observe that the given equation is a cubic equation without any constant. We know that for a cubic equation we have three roots, so we will get a maximum of three factors. In the given equation, we don’t have any constant and all the terms are having the variable $x$. So, we will first take $x$ as common from the equation, now we will get a quadratic equation. Now we will consider the obtained quadratic equation separately to factorize it. Now we will compare the obtained equation with the standard equation $a{{x}^{2}}+bx+c$ and, now we will calculate the value of $ac$ and write the factors of the value $ac$. From the factors of the $ac$, we will consider any two factors such that $b={{x}_{1}}+{{x}_{2}}$, $ac={{x}_{1}}\times {{x}_{2}}$. Now we will split the middle term $bx$ by using the value $b={{x}_{1}}+{{x}_{2}}$. Now we will take appropriate terms as common and simplify the equation to get the factors of the quadratic equation. Now we will write all the factors we have calculated at one place to get the required result.

Complete step by step answer:
Given the equation, $y=2{{x}^{3}}+10{{x}^{2}}+12x$.
We can observe that the above equation is a cubic equation without any constant. So, we are going to take $x$ as common from the given equation, then we will get
$\Rightarrow y=x\left( 2{{x}^{2}}+10x+12 \right)$
We have the quadratic equation $2{{x}^{2}}+10x+12$ in the above equation. Considering this quadratic equation and comparing it with the standard form of the quadratic equation $a{{x}^{2}}+bx+c$, then we will get
$a=2$, $b=10$, $c=12$.
Now the value of $ac$ will be
$\begin{align}
  & \Rightarrow ac=2\times 12 \\
 & \Rightarrow ac=24 \\
\end{align}$
Factors of the value $24$ are $1$, $2$, $3$, $4$, $6$, $8$, $12$, $24$. From the above factors we can write that
$\begin{align}
  & 6\times 4=24 \\
 & 6+4=10 \\
\end{align}$
So, we can split the middle term which is $10x$ as $6x+4x$. Now the quadratic equation is modified as
$\Rightarrow 2{{x}^{2}}+10x+12=2{{x}^{2}}+6x+4x+12$
Taking $2x$ as common from the terms $2{{x}^{2}}+6x$ and taking $4$ as common from the terms $4x+12$, then we will get
$\Rightarrow 2{{x}^{2}}+10x+12=2x\left( x+3 \right)+4\left( x+3 \right)$
Now taking $x+3$ as common from the above equation, then we will get
$\Rightarrow 2{{x}^{2}}+10x+12=\left( x+3 \right)\left( 2x+4 \right)$
From the above value we can write the given equation as
$\Rightarrow y=x\left( x+3 \right)\left( 2x+4 \right)$

Hence the factors of the given equation $y=2{{x}^{3}}+10{{x}^{2}}+12x$ are $x$, $x+3$, $2x+4$.

Note: In this problem they have only asked to calculate the factors of the given equation, so we have followed the above procedure. If they have asked to calculate the roots of the given equation, then we need to equate each factor to zero and calculate the values of $x$.