
How do you factor $y=-2{{x}^{3}}-6{{x}^{2}}+8x+24$ ?
Answer
544.5k+ views
Hint: In this question, we have to find the factors of a cubic polynomial. Thus, we will use the hit-and-trial method, long division and splitting the middle term method to get the solution. Therefore, to solve this problem, we will first find the root of the given polynomial by taking different values of x and substitute them in the equation, whichever value gives the answer equal to 0, implies that value is the root of the polynomial. After that we will divide the polynomial by the root, to get a quotient in the form of a quadratic polynomial. Then, we will solve the quotient using splitting the middle term method, to get the required solution to the problem.
Complete step by step solution:
According to the question, we have to find the factors of the cubic polynomial.
The polynomial given to us is $y=-2{{x}^{3}}-6{{x}^{2}}+8x+24$ ----------- (1)
Now, let $x=2$ and put this value I the equation (1), we get
$\begin{align}
& \Rightarrow y=-2{{(2)}^{3}}-6.{{(2)}^{2}}+8.(2)+24 \\
& \Rightarrow y=-16-24+16+24 \\
\end{align}$
As we know, the same terms with opposite signs cancel out each other, therefore we get
$\Rightarrow 0$
This implies that $x=2$ is the solution of the given polynomial.
Therefore, one of the roots of the given polynomial is $\left( x-2 \right)$ ---- (2)
Now, we will divide equation (1) by equation (2) by long division method, we get
\[\begin{align}
& \text{ -2}{{\text{x}}^{2}}-10x-12 \\
& x-2\left| \!{\overline {\,
-2{{x}^{3}}-6{{x}^{2}}+8x+24 \,}} \right. \\
& \text{ }\underline{{}_{+}-2{{x}^{3}}+{}_{-}4{{x}^{2}}} \\
& \text{ -10}{{\text{x}}^{2}}\text{+8x+24 } \\
& \text{ }\underline{\text{ }{}_{+}\text{-10}{{\text{x}}^{2}}\text{+}{}_{-}\text{20x }}\text{ } \\
& \text{ -12x+24} \\
& \text{ }\underline{{}_{+}\text{-12x+}{}_{-}24} \\
& \text{ }\underline{\text{ 0}} \\
& \text{ } \\
\end{align}\]
Therefore, we get the remainder equals 0 and the quotient is equal to $-2{{x}^{2}}-10x-12$ .
Now, we will use the splitting the middle term method in the quotient, which is
$-2{{x}^{2}}-10x-12$
So, we will split the middle term as the sum of -12x and +2x, because $a.c=-2.-10=20$ and $a+c=-4-6=-10=b$
Therefore, we get
$\Rightarrow -2{{x}^{2}}-4x-6x-12$
Now, we take -2x common from the first two terms and take common -6 from the last two terms, we get
$\Rightarrow -2x\left( x+2 \right)-6\left( x+2 \right)$
Now, take (x+2) common from the above equation, we get
$\left( x+2 \right)\left( -2x-6 \right)$ --------- (3)
From equation (2) and (3), we get
$y=\left( x-2 \right)\left( x+2 \right)\left( -2x-6 \right)$
Therefore, for the equation $y=-2{{x}^{3}}-6{{x}^{2}}+8x+24$ , the factors are equal to $\left( x-2 \right)\left( x+2 \right)\left( -2x-6 \right)$.
Note: While solving this problem, do mention all the formulas and the methods you are using to avoid confusion and mathematical error. One of the alternative methods to solve the quadratic equation is you can use the cross multiplication method or the discriminant method to get the solution.
Complete step by step solution:
According to the question, we have to find the factors of the cubic polynomial.
The polynomial given to us is $y=-2{{x}^{3}}-6{{x}^{2}}+8x+24$ ----------- (1)
Now, let $x=2$ and put this value I the equation (1), we get
$\begin{align}
& \Rightarrow y=-2{{(2)}^{3}}-6.{{(2)}^{2}}+8.(2)+24 \\
& \Rightarrow y=-16-24+16+24 \\
\end{align}$
As we know, the same terms with opposite signs cancel out each other, therefore we get
$\Rightarrow 0$
This implies that $x=2$ is the solution of the given polynomial.
Therefore, one of the roots of the given polynomial is $\left( x-2 \right)$ ---- (2)
Now, we will divide equation (1) by equation (2) by long division method, we get
\[\begin{align}
& \text{ -2}{{\text{x}}^{2}}-10x-12 \\
& x-2\left| \!{\overline {\,
-2{{x}^{3}}-6{{x}^{2}}+8x+24 \,}} \right. \\
& \text{ }\underline{{}_{+}-2{{x}^{3}}+{}_{-}4{{x}^{2}}} \\
& \text{ -10}{{\text{x}}^{2}}\text{+8x+24 } \\
& \text{ }\underline{\text{ }{}_{+}\text{-10}{{\text{x}}^{2}}\text{+}{}_{-}\text{20x }}\text{ } \\
& \text{ -12x+24} \\
& \text{ }\underline{{}_{+}\text{-12x+}{}_{-}24} \\
& \text{ }\underline{\text{ 0}} \\
& \text{ } \\
\end{align}\]
Therefore, we get the remainder equals 0 and the quotient is equal to $-2{{x}^{2}}-10x-12$ .
Now, we will use the splitting the middle term method in the quotient, which is
$-2{{x}^{2}}-10x-12$
So, we will split the middle term as the sum of -12x and +2x, because $a.c=-2.-10=20$ and $a+c=-4-6=-10=b$
Therefore, we get
$\Rightarrow -2{{x}^{2}}-4x-6x-12$
Now, we take -2x common from the first two terms and take common -6 from the last two terms, we get
$\Rightarrow -2x\left( x+2 \right)-6\left( x+2 \right)$
Now, take (x+2) common from the above equation, we get
$\left( x+2 \right)\left( -2x-6 \right)$ --------- (3)
From equation (2) and (3), we get
$y=\left( x-2 \right)\left( x+2 \right)\left( -2x-6 \right)$
Therefore, for the equation $y=-2{{x}^{3}}-6{{x}^{2}}+8x+24$ , the factors are equal to $\left( x-2 \right)\left( x+2 \right)\left( -2x-6 \right)$.
Note: While solving this problem, do mention all the formulas and the methods you are using to avoid confusion and mathematical error. One of the alternative methods to solve the quadratic equation is you can use the cross multiplication method or the discriminant method to get the solution.
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