Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm not the best at algebra and would be grateful if someone could explain how you can get from,

$$\frac{x^2 + x-6}{x-2}$$



share|cite|improve this question
You probably know, or once knew, how to factor $x^2+x-6$. – André Nicolas Jun 1 '14 at 20:58
up vote 2 down vote accepted

\begin{align} x^2+x-6 = x^2 + \underbrace{3x - 2x} - 6 & = \underbrace{x^2+3x}+\underbrace{{}-2x-6} \\[8pt] & = x(x+3) + (-2)(x+3) \\[8pt] & = x(\cdots\cdots) + (-2)(\cdots\cdots) \\[8pt] & = x(\cdots\cdots) -2(\cdots\cdots) \\[8pt] & = (x-2)(\cdots\cdots) \\[8pt] & = (x-2)(x+3). \end{align}

share|cite|improve this answer

Try working backwards. Notice that the denominator does not change, so let's focus on the numerator:

$$(x+3)(x-2) = x\cdot x-2x+3x-2\cdot 3 \,\,\,\,\,\,\text{FOIL}.$$

If we look at like terms (meaning equal powers of $x$), we see that we have two pieces that have $x$ that we can simplify: $-2x+3x = (-2+3)x = x.$ As for the other bits, we have $x\cdot x$ which we can simplify to $x^2$ and also $2\cdot 3 = 6$. This gives us

$$(x+3)(x-2) = x^2+x-6.$$

share|cite|improve this answer

All that was done was factoring the numerator. Notice that $$ x^2+x-6=(x+3)(x-2) $$ because $(x+3)(x-2)=x^2-2x+3x-6=x^2+x-6$. Then it follows that $$ \frac{x^2+x+6}{x-2}=\frac{(x+3)(x-2)}{x-2} $$

share|cite|improve this answer

Note that the denominator of $\frac{(x^2 + x-6)}{x-2}$ and $\frac{(x+3)(x-2)}{x-2}$ is the same, so it remains to show that the numerators are the same; i.e. that $x^2+x-6\equiv(x+3)(x-2)$.

Now, $\color{green}{(x+3)(x-2)} \equiv x^2\underbrace{-2x+3x}_{\equiv \ +x}-6 \equiv \color{green}{x^2+x-6},$ as required.

To expand brackets, use the distributive property (i.e. the fact that multiplication is distributive over addition).

share|cite|improve this answer

Key fact: Knowing the roots of a polynomial (where the polynomial equals zero), let us factor it.

So if $n$ and $m$ are two roots of the quadratic $ax^2+bx+c$, then we can factor it as $$ax^2+bx+c=a(x-n)(x-m).$$ The roots of a quadratic can be determined using the quadratic formula: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

To find out the roots of the quadratic $x^2+x-6$ use the above formula and you'll find out that they are: $-3$ and $2$. Therefore we can write our polynomial as: $$x^2+x-6=(x-(-3))(x-2)=(x+3)(x-2).$$ Hence, it follows that: $$\require{cancel}\frac{x^2 + x-6}{x-2} = \frac{(x+3)\color{red}{\cancel{\color{black}{(x-2)}}}}{\color{red}{\cancel{\color{black}{x-2}}}}=x+3.\tag{assuming $x\neq2$}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.