Question

How do you factor $x^2+4x+1?$

Answer 1

Compute the factors of the equation using direct quadratic formula:

Consider,$x^2+4x+1$

Quadratic formula:$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, $a=1,b=4,c=1$

Substitute the given values in the formula,

$\begin{array}{c}⟹\mathrm{x}=\frac{-4\pm \sqrt{16-4}}{2}\\ \Rightarrow x=\frac{-4\pm \sqrt{12}}{2}\\ \Rightarrow x=\frac{-4\pm 2\sqrt{3}}{2}\\ ⟹\mathrm{x}=-2\pm \sqrt{3}\end{array}$

Therefore, $x^2+4x+1=\left(x+2-\sqrt{3}\right)\left(x+2+\sqrt{3}\right)$

The factorized expression is,

$x^2+4x+1=\left(x+2-\sqrt{3}\right)\left(x+2+\sqrt{3}\right)$