# Factoring Quadratics

## Quadratics Factoring - Basic

Given a quadratic equation \(ax^2+bx+c=0\), how can we factor it?

First, we need to know that the factored form of a quadratic equation is \(a(x-r_1)(x-r_2)\), where \(r_1\) and \(r_2\) are the roots of the equation and \(a\) is the coefficient of the first term.

By expanding this, we get \(a\left(x^2-(r_1+r_2)x+r_1 r_2\right)\).

Now we can try to factor the equation, but first we need to factor out \(a\), which is the coefficient of the first term. Then, we can find the roots of the equation by trying out some values, since we know \(\frac{-b}{a} = r_1+r_2\) and \(\frac{c}{a}=r_1 r_2\).

## Factorize \(x^2-7x+6\).

We don't need to factor out \(a\) since \(a=1\).

Now that we know \(r_1+r_2=-b=7\) and \(r_1 r_2=c=6\), we know that the values \(1\) and \(6\) satisfy the condition for \(r_1\) and \(r_2\).

Therefore, \[x^2-7x+6=(x-6)(x-1). \ _\square\]

## Quadratics Factoring - Intermediate

Given a quadratic equation \(ax^2 + bx + c = 0\), we can factor it easily using methods described here. This, however, may not be the best method when dealing with the general quadratic with solutions which are not real. For these cases, the quadratic formula might be more applicable.

Consider the quadratic equation \(ax^2+bx+c=0\). If we divide the whole expression by \(a\), we get \[x^2 + \frac{b}{a}x + \frac{c}{a} = 0 .\] Rearranging gives \[x^2 + \frac{b}{a}x = - \frac{c}{a} .\] Let us now add \(\frac{b^2}{4a^2}\) to both sides: \[\begin{align} x^2 + \frac{b^2}{4a^2} + \frac{b}{a}x &= \frac{b^2}{4a^2} - \frac{c}{a}\\ \left(x + \frac{b}{2a}\right)^2 &= \frac{b^2-4ac}{4a^2} . \end{align}\] Taking the square root of both sides, we have \[x + \frac{b}{2a} = \pm \sqrt{\frac{b^2-4ac}{4a^2}} .\] Rearranging one last time, we get \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} .\]

Notice that now we are able to find the roots of a quadratic even if they are not real. This allows us to factor it very easily. Thus for any quadratic \(ax^2 + bx + c = 0\), we can factor it into \(k(x+\alpha)(x+\beta) = 0 \), where \(\alpha= \frac{-b+ \sqrt{b^2 - 4ac}}{2a}, \) \( \beta= \frac{-b - \sqrt{b^2 - 4ac}}{2a}, \) and \(k\) is some constant.

## Factorize \(x^2 + x - 1 = 0.\)

Using our formula, we obtain \[\phi= \frac{-b+ \sqrt{b^2 - 4ac}}{2a} = \frac{-1 +\sqrt{5}}{2}, \Phi= \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{-1 - \sqrt{5}}{2} .\] Our quadratic is thus \[k(x-\phi)(x-\Phi) = 0\] for some constant \(k\). However, since the leading coefficient of our first term is 1, we know that the factorization must be \[(x-\phi)(x - \Phi) = \left(x - \frac{-1 +\sqrt{5}}{2} \right)\left(x- \frac{-1 - \sqrt{5}}{2}\right). \ _\square\]

This method of factorization also works when we are not dealing with quadratics with real solutions/coefficients.

## Factorize \(2x^2 +2 x + 2 = 0.\)

We first observe that this has no real solutions because its discriminant is negative: \[D = b^2 - 4ac = 2^2 - 4\times2\times2 = -12 < 0.\] We can proceed to finding its roots using \[x = \frac{-b+ \sqrt{b^2 - 4ac}}{2a} =\frac{-b+ \sqrt{D}}{2a} . \] Using our formula \[\alpha= \frac{-2+ \sqrt{-12}}{2.2} =\frac{-2+ \sqrt{4\times-1\times 3}}{4} =\frac{-2+ 2i\sqrt{3}}{4} = \frac{-1}{2} + \frac{i\sqrt{3}}{2} . \] Similarly, we get \[\beta= \frac{-2 - \sqrt{-12}}{2.2} =\frac{-2 - \sqrt{4\times-1\times 3}}{4} =\frac{-2 - 2i\sqrt{3}}{4} = \frac{-1}{2} - \frac{i\sqrt{3}}{2} .\]

We can now write our equation as \[k\left(x-\frac{-1}{2} - \frac{i\sqrt{3}}{2}\right)\left(x-\frac{-1}{2} + \frac{i\sqrt{3}}{2}\right)=0.\] Notice that \(k\) has to be \(2\) in order to make the coefficient of \(x^2\) \(2\). Thus, we finally get \[2\left(x+\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)\left(x+\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)=0. \ _\square\]

This allows us to factorize quadratics with irrational and even imaginary coefficients.

## Factorize \( \sqrt{24}x^2 + 5\sqrt{2}x + \sqrt{6} = 0.\)

Again, our discriminant is \[D = b^2 - 4\times a\times c = 50 - 4\times12 = 2 > 0. \] Using our formula we get \[\begin{align} \alpha&= \frac{-5\sqrt{2} + \sqrt{2} }{4\sqrt{6}} = \frac{ - 4\sqrt{2} }{ 4\sqrt{6}} = -\frac{\sqrt{3}}{3 } \\ \beta&= \frac{-5\sqrt{2} - \sqrt{2} }{4\sqrt{6}} = \frac{ - 6\sqrt{2} }{ 4\sqrt{6}} = -\frac{\sqrt{3}}{2}. \end{align} \]

Thus, our factorization becomes

\[k\left(x+\frac{\sqrt{3}}{3 }\right)\left(x + \frac{\sqrt{3}}{2} \right) = 0. \]

Again, by comparing this with our initial polynomial, we see that \(k = \sqrt{24} = 2\sqrt{6} \). Thus, our final factorized form is \[\sqrt{24}\left(x+\frac{\sqrt{3}}{3} \right)\left(x + \frac{\sqrt{3}}{2} \right) = 0. \ _\square \]

**Cite as:**Factoring Quadratics.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/factoring-quadratics/