Question

The real factors of $ {{x}^{2}}+4 $ are
A. $ \left( {{x}^{2}}+2 \right)\left( {{x}^{2}}-2 \right) $
A. $ \left( x+2 \right)\left( x-2 \right) $
C. does not exist
D. none of these

Answer 1

Hint: We take all the variables and the constants all together. Then we form the equation according to the identity $ {{a}^{2}}-{{b}^{2}} $ to form the factorisation of $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ . We place values $ a=x;b=2i $ . The multiplied polynomials give value 0 individually. From that we find the factors of $ {{x}^{2}}+4 $ .

Complete step-by-step answer:
We need to find the factors of the given equation $ {{x}^{2}}+4 $ .
Now we have a quadratic equation $ {{x}^{2}}+4=0 $ which gives $ {{x}^{2}}-\left( -4 \right)={{x}^{2}}-{{\left( 2i \right)}^{2}}=0 $ .
Now we find the factorisation of the equation \[{{x}^{2}}-{{\left( 2i \right)}^{2}}=0\] using the identity of $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ .
Therefore, we get
 $
  {{x}^{2}}-{{\left( 2i \right)}^{2}}=0 \\
  \Rightarrow \left( x+2i \right)\left( x-2i \right)=0 \;
$
The factorization of $ {{x}^{2}}+4 $ is $ \left( x+2i \right)\left( x-2i \right) $ .
Therefore, the real factors of $ {{x}^{2}}+4 $ do not exist.
So, the correct answer is “Option C”.

Note: The highest power of the variable or the degree of a polynomial decides the number of roots or the solution of that polynomial. Quadratic equations have 2 roots. Cubic polynomials have 3. It can be both real and imaginary roots.
We can also apply the quadratic equation formula to solve the equation $ {{x}^{2}}+4=0 $ .
We know for a general equation of quadratic $ a{{x}^{2}}+bx+c=0 $ , the value of the roots of x will be $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ .
In the given equation we have $ {{x}^{2}}+4=0 $ . The values of a, b, c are $ 1,0,4 $ respectively.
We put the values and get x as $ x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 1\times 4}}{2\times 1}=\dfrac{\pm \sqrt{-16}}{2}=\pm 2i $ . Therefore, factors can’t be real as the roots are imaginary.