# How do you multiply (a^2-a-12)/(a^2-5a+4)*(a^2+2a-3)/(a^2+a-6)?

Mar 25, 2015

You start by first factorising the quatratic polynomials

$\frac{\left(a + 3\right) \left(a - 4\right)}{\left(a - 1\right) \left(a - 4\right)} \cdot \frac{\left(a - 1\right) \left(a + 3\right)}{\left(a + 3\right) \left(a - 2\right)}$

$\frac{\left(a + 3\right) \left(a - 4\right) \left(a - 1\right) \left(a + 3\right)}{\left(a - 1\right) \left(a - 4\right) \left(a + 3\right) \left(a - 2\right)}$

After cancelling out all the like terms above and below the bar, we are left with:

$\frac{a + 3}{a - 2}$ which cannot be further reduced.

Mar 25, 2015

You could just multiply the numerator and denominator polynomials separately (this would answer the question and if this is what is required please re-post)
but
it is easier if you factor your polynomials and simplify:
$\frac{{a}^{2} - a - 12}{{a}^{2} - 5 a + 4} \cdot \frac{{a}^{2} + 2 a - 3}{{a}^{2} + a - 6}$

$= \frac{\left(a - 4\right) \left(a + 3\right)}{\left(a - 4\right) \left(a - 1\right)} \cdot \frac{\left(a + 2\right) \left(a - 1\right)}{\left(a + 3\right) \left(a - 2\right)}$

$= \frac{\textcolor{red}{\cancel{\left(a - 4\right)}} \textcolor{b l u e}{\cancel{\left(a + 3\right)}}}{\textcolor{red}{\cancel{\left(a - 4\right)}} \textcolor{g r e e n}{\cancel{\left(a - 1\right)}}} \cdot \frac{\left(a + 2\right) \textcolor{g r e e n}{\cancel{\left(a - 1\right)}}}{\textcolor{b l u e}{\cancel{\left(a + 3\right)}} \left(a - 2\right)}$

$= \frac{a + 2}{a - 2}$