
How do you factor the expression $3{{x}^{3}}-12{{x}^{2}}-45x$?
Answer
451.8k+ views
Hint: Now to solve the given expression we will first take x common from the whole expression. Now we will solve the quadratic by using the formula for roots which is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Now we know that if $\alpha $ is the root of the equation then $x-\alpha $ is the factor of the given equation.
Complete step by step solution:
Now the given expression is a polynomial in x of degree 3.
Now since there is no constant term in the equation we have x = 0 is the root of the expression.
Now taking x common from the whole expression we get, $x\left( 3{{x}^{2}}-12x-45 \right)$ .
Now consider the quadratic expression $3{{x}^{2}}-12x-45$ .
Now we will first find the roots of this expression.
Now the quadratic is in the form of $a{{x}^{2}}+bx+c$ where a = 3, b = -12 and c = -45 and we know that the roots for such quadratic expression is given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Hence substituting the values of a, b and c we get,
$\begin{align}
& \Rightarrow \dfrac{-\left( -12 \right)\pm \sqrt{{{\left( -12 \right)}^{2}}-4\left( 3 \right)\left( -45 \right)}}{2\left( 3 \right)} \\
& \Rightarrow \dfrac{12\pm \sqrt{144+540}}{6} \\
& \Rightarrow \dfrac{12\pm \sqrt{684}}{6} \\
& \Rightarrow \dfrac{12\pm 6\sqrt{19}}{6} \\
& \Rightarrow 2\pm \sqrt{19} \\
\end{align}$
Now the roots of the given expression are $2+\sqrt{19}$ and $2-\sqrt{19}$ .
Now if $\alpha $ and $\beta $ are the roots of the quadratic then $x-\alpha $ and $x-\beta $ are the factors of the expression.
Hence we get, $\left( x-\left( 2+\sqrt{19} \right) \right)$ and $\left( x-\left( 2-\sqrt{19} \right) \right)$ as the factors of the given quadratic.
Now we have, $x\left( x-\left( 2+\sqrt{19} \right) \right)\left( x-\left( 2-\sqrt{19} \right) \right)$.
Note: Now note that to factorize the quadratic we can either factorize the expression by splitting the middle term such that the product of the two terms is multiplication of first terms and last term. Now if we cannot split the middle terms in this way find the roots of the quadratic expression and then find the corresponding fraction.
Complete step by step solution:
Now the given expression is a polynomial in x of degree 3.
Now since there is no constant term in the equation we have x = 0 is the root of the expression.
Now taking x common from the whole expression we get, $x\left( 3{{x}^{2}}-12x-45 \right)$ .
Now consider the quadratic expression $3{{x}^{2}}-12x-45$ .
Now we will first find the roots of this expression.
Now the quadratic is in the form of $a{{x}^{2}}+bx+c$ where a = 3, b = -12 and c = -45 and we know that the roots for such quadratic expression is given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Hence substituting the values of a, b and c we get,
$\begin{align}
& \Rightarrow \dfrac{-\left( -12 \right)\pm \sqrt{{{\left( -12 \right)}^{2}}-4\left( 3 \right)\left( -45 \right)}}{2\left( 3 \right)} \\
& \Rightarrow \dfrac{12\pm \sqrt{144+540}}{6} \\
& \Rightarrow \dfrac{12\pm \sqrt{684}}{6} \\
& \Rightarrow \dfrac{12\pm 6\sqrt{19}}{6} \\
& \Rightarrow 2\pm \sqrt{19} \\
\end{align}$
Now the roots of the given expression are $2+\sqrt{19}$ and $2-\sqrt{19}$ .
Now if $\alpha $ and $\beta $ are the roots of the quadratic then $x-\alpha $ and $x-\beta $ are the factors of the expression.
Hence we get, $\left( x-\left( 2+\sqrt{19} \right) \right)$ and $\left( x-\left( 2-\sqrt{19} \right) \right)$ as the factors of the given quadratic.
Now we have, $x\left( x-\left( 2+\sqrt{19} \right) \right)\left( x-\left( 2-\sqrt{19} \right) \right)$.
Note: Now note that to factorize the quadratic we can either factorize the expression by splitting the middle term such that the product of the two terms is multiplication of first terms and last term. Now if we cannot split the middle terms in this way find the roots of the quadratic expression and then find the corresponding fraction.
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