Question

How do you factor $36{{a}^{4}}+90{{a}^{2}}{{b}^{2}}+99{{b}^{4}}$?

Answer 1

Hint: This equation seems a little complicated but after applying factorization method to it we will get the right answer. For this we will factorize 36, 90 and 99. This will help us in finding out common numbers. We will take those factors as a common one and get to the required answer.

Complete step by step solution:
First we will understand the concept of factorization. In this process, we basically factor those numbers which are actually the coefficients of the variables. After this we simply take the common numbers out and simplify the equation.
To solve this question we will take the common factor out of the given equation. In the given equation $36{{a}^{4}}+90{{a}^{2}}{{b}^{2}}+99{{b}^{4}}$ we can factorize 36 as $2\times 2\times 3\times 3$, 90 as $3\times 3\times 2\times 5$ and 99 as $3\times 3\times 11$. So, equation $36{{a}^{4}}+90{{a}^{2}}{{b}^{2}}+99{{b}^{4}}$ gets converted into $\left( 2\times 2\times 3\times 3 \right){{a}^{4}}+\left( 2\times 3\times 3\times 5 \right){{a}^{2}}{{b}^{2}}+\left( 3\times 3\times 11 \right){{b}^{4}}$.
In this equation we can clearly see that the common numbers are 3, 3. Therefore, after taking these out we get,
$3\times 3\left[ \left( 2\times 2 \right){{a}^{4}}+\left( 2\times 5 \right){{a}^{2}}{{b}^{2}}+11{{b}^{4}} \right]$ or, $9\left[ 4{{a}^{4}}+10{{a}^{2}}{{b}^{2}}+11{{b}^{4}} \right]$.

Hence, the required answer is $9\left[ 4{{a}^{4}}+10{{a}^{2}}{{b}^{2}}+11{{b}^{4}} \right]$.

Note: One should not put the given equation equal to zero and ${{a}^{4}}={{x}^{2}}$ in $36{{a}^{4}}+90{{a}^{2}}{{b}^{2}}+99{{b}^{4}}$otherwise, we might get to the following.
The equation gets converted into $36{{x}^{2}}+90{{b}^{2}}x+99{{b}^{4}}$. After keeping this equation equals to 0 we get $36{{x}^{2}}+90{{b}^{2}}x+99{{b}^{4}}=0$.
By using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and we will substitute a as 36, b as $90{{b}^{2}}$ and c as $99{{b}^{4}}$ we get,
\[\begin{align}
  & x=\dfrac{-\left( 90{{b}^{2}} \right)\pm \sqrt{{{\left( 90{{b}^{2}} \right)}^{2}}-4\left( 36 \right)\left( 99{{b}^{4}} \right)}}{2\left( 36 \right)} \\
 & \Rightarrow x=\dfrac{-90{{b}^{2}}\pm \sqrt{8100{{b}^{4}}-14256{{b}^{4}}}}{72} \\
 & \Rightarrow x=\dfrac{-90{{b}^{2}}\pm \sqrt{-6156}}{72} \\
 & \Rightarrow x=\dfrac{-90{{b}^{2}}\pm i\sqrt{6156}}{72}\,\,\left[ \text{Since, }\sqrt{-1}=i \right] \\
 & \Rightarrow x=\dfrac{-90{{b}^{2}}\pm i78.46}{72}\,\,\,\left[ \text{Since, }\sqrt{6156}=78.46 \right] \\
\end{align}\]
This will give factors as $36{{x}^{2}}+90{{b}^{2}}x+99{{b}^{4}}=0$ are \[x=\dfrac{-90{{b}^{2}}+i78.46}{72}\] and \[x=\dfrac{-90{{b}^{2}}-i78.46}{72}\].
After substituting ${{a}^{4}}={{x}^{2}}$ again in these factors we get \[{{a}^{2}}=\dfrac{-90{{b}^{2}}+i78.46}{72}\] and \[{{a}^{2}}=\dfrac{-90{{b}^{2}}-i78.46}{72}\].