How do you factor $8{{x}^{2}}-18$ ?
Answer
615.3k+ views
Hint: For answering this question we need to factorize the given expression $8{{x}^{2}}-18$ . We will first take out two as common $8{{x}^{2}}-18=2\left( 4{{x}^{2}}-9 \right)$ . If we observe the expression carefully it is in the form of ${{a}^{2}}-{{b}^{2}}$ so a formula will tick in our mind that is ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we will use this and factorize the expression.
Complete step by step solution:
Now considering from the question we have been asked to factorize the expression $8{{x}^{2}}-18$
For that sake we will take out two as common from the expression $\Rightarrow 2\left( 4{{x}^{2}}-9 \right)$ .
By observing this expression carefully we will remember the formula of factorization mathematically given as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ .
By applying this formula we will simplify the expression as $\Rightarrow 2\left( 4{{x}^{2}}-9 \right)=2\left( {{\left( 2x \right)}^{2}}-{{3}^{2}} \right)$ .
By further simplifying we will have $\Rightarrow 2\left( 2x+3 \right)\left( 2x-3 \right)$ .
Therefore we can conclude that the factors of the given expression $8{{x}^{2}}-18$ are $2,\left( 2x+3 \right)and\left( 2x-3 \right)$.
Note: In the process of solving this question we have to take care of the calculations we perform and the concept we apply. This is a very simple and easy question that can be answered in a very less span of time and very few mistakes are possible. The proof of the formula can be given by multiplying the factors $\left( a-b \right)\times \left( a+b \right)={{a}^{2}}+ab-ab-{{b}^{2}}\Rightarrow {{a}^{2}}-{{b}^{2}}$ . Similarly this expression can be factored by using the formulae for finding the roots of any quadratic expression in the form of $a{{x}^{2}}+bx+c$ is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . If the roots are ${{x}_{1}},{{x}_{2}}$ then factors are $\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)$ . For the expression $8{{x}^{2}}-18=2\left( 4{{x}^{2}}-9 \right)$ we will have the roots as
$\begin{align}
& x=\dfrac{\pm \sqrt{-4\left( 4 \right)\left( -9 \right)}}{2\left( 4 \right)} \\
& \Rightarrow x=\dfrac{\pm \left( 4\times 3 \right)}{2\left( 4 \right)} \\
& \Rightarrow x=\dfrac{\pm 3}{2} \\
\end{align}$
Then the factors of the expression will be $\left( 2x+3 \right)\left( 2x-3 \right)$ .
Complete step by step solution:
Now considering from the question we have been asked to factorize the expression $8{{x}^{2}}-18$
For that sake we will take out two as common from the expression $\Rightarrow 2\left( 4{{x}^{2}}-9 \right)$ .
By observing this expression carefully we will remember the formula of factorization mathematically given as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ .
By applying this formula we will simplify the expression as $\Rightarrow 2\left( 4{{x}^{2}}-9 \right)=2\left( {{\left( 2x \right)}^{2}}-{{3}^{2}} \right)$ .
By further simplifying we will have $\Rightarrow 2\left( 2x+3 \right)\left( 2x-3 \right)$ .
Therefore we can conclude that the factors of the given expression $8{{x}^{2}}-18$ are $2,\left( 2x+3 \right)and\left( 2x-3 \right)$.
Note: In the process of solving this question we have to take care of the calculations we perform and the concept we apply. This is a very simple and easy question that can be answered in a very less span of time and very few mistakes are possible. The proof of the formula can be given by multiplying the factors $\left( a-b \right)\times \left( a+b \right)={{a}^{2}}+ab-ab-{{b}^{2}}\Rightarrow {{a}^{2}}-{{b}^{2}}$ . Similarly this expression can be factored by using the formulae for finding the roots of any quadratic expression in the form of $a{{x}^{2}}+bx+c$ is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . If the roots are ${{x}_{1}},{{x}_{2}}$ then factors are $\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)$ . For the expression $8{{x}^{2}}-18=2\left( 4{{x}^{2}}-9 \right)$ we will have the roots as
$\begin{align}
& x=\dfrac{\pm \sqrt{-4\left( 4 \right)\left( -9 \right)}}{2\left( 4 \right)} \\
& \Rightarrow x=\dfrac{\pm \left( 4\times 3 \right)}{2\left( 4 \right)} \\
& \Rightarrow x=\dfrac{\pm 3}{2} \\
\end{align}$
Then the factors of the expression will be $\left( 2x+3 \right)\left( 2x-3 \right)$ .
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