Question

One of the factors of ${{x}^{4}}+4$ is:
(a). ${{x}^{2}}+2$
(b). ${{x}^{2}}-2x+2$
(c). ${{x}^{2}}-2$
(d). None of these

Answer 1

Hint: To find it’s factor means that the given expression ${{x}^{4}}+4$ is divisible by that number. And with that we can also conclude that the root of given expression in the options will also be the root of ${{x}^{4}}+4$, and with that we check which option is correct and if none of them from (a) to (c) satisfies then (d) will be the answer.

Complete step-by-step solution -

So, we will start by finding all the roots of the given option and then we can put these values in ${{x}^{4}}+4$= 0 to check whether it also gives 0 or not, if it gives 0 then that option is correct otherwise it is incorrect.
Let’s find for option (a):
${{x}^{2}}+2$= 0
${{x}^{2}}=-2$
Let it be in this form because we don’t want to make it complex by showing the complex roots we can also solve in this form.
Now,
${{x}^{4}}={{\left( -2 \right)}^{2}}=4$
Now putting the value in ${{x}^{4}}+4$ we get ,
4 + 4
= 8
Hence, option (a) is incorrect.
Let’s find for option (b):
${{x}^{2}}-2x+2$ = 0
First we will use sridharacharya formula to find the roots of this quadratic equation,
The formula is:
If the equation is $a{{x}^{2}}+bx+c=0$ , then the formula for finding x is:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now,
 $\begin{align}
  & x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 2\times 1}}{2} \\
 & x=\dfrac{2\pm \sqrt{4-8}}{2} \\
 & x=\dfrac{2\pm \sqrt{-4}}{2} \\
 & x=\dfrac{2\pm 2i}{2} \\
 & x=1\pm i \\
\end{align}$
Here $i=\sqrt{-1}\text{ and }{{\text{i}}^{2}}=-1$
Now putting the value of x in ${{x}^{4}}+4$we get,
Using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$\begin{align}
  & {{\left( 1+i \right)}^{4}}+4 \\
 & ={{\left( 1+2i-1 \right)}^{2}}+4 \\
 & ={{\left( 2i \right)}^{2}}+4 \\
 & =4{{i}^{2}}+4 \\
 & =-4+4 \\
 & =0 \\
\end{align}$
Hence, from this we can conclude that option (b) is the correct answer.
Let’s find for option (c):
$\begin{align}
  & {{x}^{2}}-2 \\
 & {{x}^{2}}=2 \\
\end{align}$
Now,
${{x}^{4}}={{\left( 2 \right)}^{2}}=4$
Now putting the value in ${{x}^{4}}+4$ we get ,
4 + 4
= 8
Hence, option (c) is incorrect.

Note: We can also solve this question by dividing ${{x}^{4}}+4$by each of the options and then check which one is the factor and which one is not the factor. Another method is also possible by factoring ${{x}^{4}}+4$as ${{\left( {{x}^{2}}+2 \right)}^{2}}-4{{x}^{2}}=\left( {{x}^{2}}+2-2x \right)\left( {{x}^{2}}+2+2x \right)$, and then we can directly get the answer in just 2-3 steps.