How do you factor $2{{x}^{2}}-11x-21$?
Answer
615.9k+ views
Hint: For this problem we need to calculate the factors of the given equation. We can observe that the given equation is a quadratic equation. For this we will first find the roots of the given quadratic equation which is in form of $a{{x}^{2}}+bx+c$ by using the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. After getting the roots of the quadratic equation as $\alpha $, $\beta $. We can write the factors of the given equation as $x-\alpha $, $x-\beta $.
Complete step by step solution:
Given equation, $2{{x}^{2}}-11x-21$.
Comparing the above equation with standard form of the quadratic equation which is $a{{x}^{2}}+bx+c$, then we will get
$a=2$, $b=-11$, $c=-21$.
We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c$ are given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting $a=2$, $b=-11$, $c=-21$ in the above formula to find the roots of the given equation, then we will have
$\Rightarrow x=\dfrac{-\left( -11 \right)\pm \sqrt{{{\left( -11 \right)}^{2}}-4\left( 2 \right)\left( -21 \right)}}{2\left( 2 \right)}$
When we multiplied a negative sign with the negative sign, we will get positive sign. Applying this rule in the above equation, then the above equation is modified as
$\begin{align}
& \Rightarrow x=\dfrac{11\pm \sqrt{121+168}}{4} \\
& \Rightarrow x=\dfrac{11\pm \sqrt{289}}{4} \\
\end{align}$
In the above equation we have the value $\sqrt{289}$. On prime factoring the value $289$ we will get $289=17\times 17$. From this we can write $\sqrt{289}=\sqrt{{{17}^{2}}}=17$. Substituting this value in the above equation, then we will get
$\Rightarrow x=\dfrac{11\pm 17}{4}$
Simplifying the above equation, then we will have
$\begin{align}
& \Rightarrow x=\dfrac{11+17}{4}\text{ or }\dfrac{11-17}{4} \\
& \Rightarrow x=\dfrac{28}{4}\text{ or }\dfrac{-6}{4} \\
& \Rightarrow x=7\text{ or }-\dfrac{3}{2} \\
\end{align}$
We have the roots of the given equation as $7$, $-\dfrac{3}{2}$.
Hence the factors of the given equation will be $x-7$, $x-\left( -\dfrac{3}{2} \right)=x+\dfrac{3}{2}$.
Note: We can also check whether the calculated solution is correct or not. When we multiply the calculated factors, it must be equal to the given equation otherwise the calculated solution is not correct.
Complete step by step solution:
Given equation, $2{{x}^{2}}-11x-21$.
Comparing the above equation with standard form of the quadratic equation which is $a{{x}^{2}}+bx+c$, then we will get
$a=2$, $b=-11$, $c=-21$.
We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c$ are given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting $a=2$, $b=-11$, $c=-21$ in the above formula to find the roots of the given equation, then we will have
$\Rightarrow x=\dfrac{-\left( -11 \right)\pm \sqrt{{{\left( -11 \right)}^{2}}-4\left( 2 \right)\left( -21 \right)}}{2\left( 2 \right)}$
When we multiplied a negative sign with the negative sign, we will get positive sign. Applying this rule in the above equation, then the above equation is modified as
$\begin{align}
& \Rightarrow x=\dfrac{11\pm \sqrt{121+168}}{4} \\
& \Rightarrow x=\dfrac{11\pm \sqrt{289}}{4} \\
\end{align}$
In the above equation we have the value $\sqrt{289}$. On prime factoring the value $289$ we will get $289=17\times 17$. From this we can write $\sqrt{289}=\sqrt{{{17}^{2}}}=17$. Substituting this value in the above equation, then we will get
$\Rightarrow x=\dfrac{11\pm 17}{4}$
Simplifying the above equation, then we will have
$\begin{align}
& \Rightarrow x=\dfrac{11+17}{4}\text{ or }\dfrac{11-17}{4} \\
& \Rightarrow x=\dfrac{28}{4}\text{ or }\dfrac{-6}{4} \\
& \Rightarrow x=7\text{ or }-\dfrac{3}{2} \\
\end{align}$
We have the roots of the given equation as $7$, $-\dfrac{3}{2}$.
Hence the factors of the given equation will be $x-7$, $x-\left( -\dfrac{3}{2} \right)=x+\dfrac{3}{2}$.
Note: We can also check whether the calculated solution is correct or not. When we multiply the calculated factors, it must be equal to the given equation otherwise the calculated solution is not correct.
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