Question

How do you divide $\dfrac{{{y}^{4}}+9{{y}^{2}}+20}{{{y}^{2}}+4}$?

Answer 1

Hint: We have two quadratic equations in the denominator and the numerator. We factor the numerator using different processes. From them we eliminate the common factor if there is any. We find the simplified form the expression.

Complete step by step solution:
We have been given the division of two quadratic equations. We need to find the factor of the equations.
The two quadratic equations are ${{y}^{4}}+9{{y}^{2}}+20$ and ${{y}^{2}}+4$.
 We use the vanishing method to solve the numerator.
We find the value of ${{y}^{2}}=a$ for which the function $f\left( y \right)={{y}^{4}}+9{{y}^{2}}+20=0$.
We take ${{y}^{2}}=a=-4$. We can see $f\left( -4 \right)={{\left( -4 \right)}^{2}}+9\times \left( -4 \right)+20=16-36+20=0$.
So, the factor of the $f\left( y \right)={{y}^{4}}+9{{y}^{2}}+20$ will be the function $\left( {{y}^{2}}+4 \right)$.
Therefore, the term $\left( {{y}^{2}}+4 \right)$ is a factor of the polynomial ${{y}^{4}}+9{{y}^{2}}+20$.
So, \[{{y}^{4}}+9{{y}^{2}}+20=\left( {{y}^{2}}+4 \right)\left( {{y}^{2}}+5 \right)\].
Now we put the factored values for the two equations.
We get \[\dfrac{{{y}^{4}}+9{{y}^{2}}+20}{{{y}^{2}}+4}=\dfrac{\left( {{y}^{2}}+4 \right)\left( {{y}^{2}}+5 \right)}{\left( {{y}^{2}}+4 \right)}\].
Now we can see in the denominator and the numerator the common factor is $\left( {{y}^{2}}+4 \right)$.
We can eliminate the factor from both denominator and the numerator.
So, \[\dfrac{{{y}^{4}}+9{{y}^{2}}+20}{{{y}^{2}}+4}=\dfrac{\left( {{y}^{2}}+4 \right)\left( {{y}^{2}}+5 \right)}{\left( {{y}^{2}}+4 \right)}=\left( {{y}^{2}}+5 \right)\].
Therefore, the simplified form of $\dfrac{{{y}^{4}}+9{{y}^{2}}+20}{{{y}^{2}}+4}$ is $\left( {{y}^{2}}+5 \right)$.

Note: We can also factorise the quadratic equations using grouping method.
In case of ${{y}^{4}}+9{{y}^{2}}+20$, we break the middle term $9{{y}^{2}}$ into two parts of $5{{y}^{2}}$ and $4{{y}^{2}}$.
So, ${{y}^{4}}+9{{y}^{2}}+20={{y}^{4}}+5{{y}^{2}}+4{{y}^{2}}+20$.
Factorising we get
$\begin{align}
  & {{y}^{4}}+9{{y}^{2}}+20 \\
 & ={{y}^{4}}+5{{y}^{2}}+4{{y}^{2}}+20 \\
 & ={{y}^{2}}\left( {{y}^{2}}+5 \right)+4\left( {{y}^{2}}+5 \right) \\
 & =\left( {{y}^{2}}+5 \right)\left( {{y}^{2}}+4 \right) \\
\end{align}$