Question

How do you factor $$x^2+4$$?

Answer 1

Hint: We need to factorize $$x^2+4$$. To factorize the numbers, the numbers are split as their multiplication. For example the number $$12$$ is factored as its multiple that is $$12=2\times 2\times 3$$. The factorization of the polynomial is done in the same manner. This is done by first determining the terms are multiplied to obtain the given polynomial and then factorize the term. The process is continued until there is no simplification possible.

Complete step by step solution:
The given term is $$x^2+4$$.
Here, $$x^2+4$$ is a binomial as there are only $$2$$ terms.
It is clear in the binomial $$x^2+4$$that there are no common factors other than $$1$$.
Since no factor with more than one term can be factored further so there are no real number factors.
This concludes that the factors are Prime.
The roots of the quadratic function depend on the discriminant that is $$b^2-4ac$$, if the discriminant is $$0$$ then the roots are real, if the discriminant is rational then the factors are rational too. For the positive discriminant, the roots are real and for the negative one, the roots are complex.
Calculate the discriminant of $$x^2+4$$ as,
$$\Rightarrow b^2-4ac=0^2-4\left(1\right)\left(4\right)$$
$$\Rightarrow b^2-4ac=-16$$
The discriminant is negative this means that the roots are complex.
Then, the factorization of $$x^2+4$$ is done as,
$$\Rightarrow x^2+4=x^2-\left(-4\right)$$
$$\Rightarrow x^2+4=x^2-{\left(2i\right)}^2$$
Here, $$i$$ is the imaginary root and its value is $$i=\sqrt{-1}$$.
Consider the algebraic identity $$\left(a+b\right)\left(a-b\right)=a^2-b^2$$
Then, the binomial is,
$$\Rightarrow x^2+4=\left(x-2i\right)\left(x+2i\right)$$

Thus, the factored form is $$\left(x-2i\right)\left(x+2i\right)$$.

Note:
The quadratic equation is the equation that is of the standard from $$a{x}^2+bx+c$$. Here, a and b are the coefficients and c are the constant. In the general equation, the highest power of the x is $$2$$ so the equation is called quadratic.