# Solving equations with factorization

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If we can factor the left hand side of the equation $$ax^2+bx+c=0$$, then the solution is easily found:

Consider the quadratic equation with unknown $$x$$:

$$ax^2+bx+c=0$$,

where $$a$$, $$b$$ and $$c$$ are real numbers with $$a\ne0$$. If the left hand side can be factored in linear factors:

$$ax^2+bx+c=(x-p)\cdot(x-q)\tiny,$$,

where $$p$$ and $$q$$ are real numbers, then the solution of the equation is $$x=p\lor x=q$$.

Let’s practice!

Solve the equation for $$x$$ below using factorization.

$$x^2-14 x-16=-8 x\tiny.$$

Give your answer in the form of $$x=a\lor x=b$$.

This can be solved by means of the following deduction:

$$\begin{array}{rcl} x^2-14 x-16&=&-8 x \\ &&\phantom{xxx}\color{blue}{\text{the original equation}}\\ x^2-6 x-16&=&0 \\ &&\phantom{xxx}\color{blue}{\text{all terms to the left}}\\ (x-8)\cdot (x+2)&=&0 \\ &&\phantom{xxx}\color{blue}{\text{left hand side factored}}\\ x-8=0 &\lor& x+2=0\\ &&\phantom{xxx}\color{blue}{A\cdot B=0 \text{ if and only if }A=0\lor B=0}\\ x=8 &\lor& x=-2\\ &&\phantom{xxx}\color{blue}{\text{constant terms to the right}} \end{array}$$

$$x=8\lor x=-2$$