Chapter 6.2, Problem 45WE

To determine

To calculate: The simplified form of the expression $\frac{t^2-2t-8}{4-t^2}\cdot \frac{t^2-5t+6}{t^2-t-12}$.

Answer to Problem 45WE

The simplified form of the expression $\frac{t^2-2t-8}{4-t^2}\cdot \frac{t^2-5t+6}{t^2-t-12}$ is $\frac{\left(3-t\right)}{\left(3+t\right)}$.

Explanation of Solution

Given information:

The expression $\frac{t^2-2t-8}{4-t^2}\cdot \frac{t^2-5t+6}{t^2-t-12}$.

Formula used:

Fractions are multiplied as $\frac{a}{b}\cdot \frac{c}{d}=\frac{ab}{cd}$.

To factor a polynomial split the middle term in such a way that sum of two numbers is the middle term and product of two numbers is same is product of first and last term.

Calculation:

Consider the expression $\frac{t^2-2t-8}{4-t^2}\cdot \frac{t^2-5t+6}{t^2-t-12}$.

Take the trinomial $t^2-2t-8$.

Recall that to factor a polynomial split the middle term in such a way that sum of two numbers is the middle term and product of two numbers is same is product of first and last term.

For the above polynomial $-4$ and $2$ are two numbers such that their sum is $-2$ and product is $-8$.

Therefore, the trinomial is rewritten as,

  $t^2-2t-8=t^2-4t+2t-8$

Take out the common terms,

  $\begin{array}{c}{t}^2-2t-8=t^2-4t+2t-8\\ =t\left(t-4\right)+2\left(t-4\right)\\ =\left(t-4\right)\left(t+2\right)\end{array}$

Take the trinomial $t^2-5t+6$.

Recall that to factor a polynomial split the middle term in such a way that sum of two numbers is the middle term and product of two numbers is same is product of first and last term.

Factor and take out the common terms,

  $\begin{array}{c}{t}^2-5t+6=t^2-2t-3t+6\\ =t\left(t-2\right)-3\left(t-2\right)\\ =\left(t-2\right)\left(t-3\right)\end{array}$

Take the trinomial $t^2-t-12$.

Recall that to factor a polynomial split the middle term in such a way that sum of two numbers is the middle term and product of two numbers is same is product of first and last term.

Factor and take out the common terms,

  $\begin{array}{c}{t}^2-t-12=t^2-4t+3t-12\\ =t\left(t-4\right)+3\left(t-4\right)\\ =\left(t-4\right)\left(t+3\right)\end{array}$

Take the binomial $4-t^2$.

Recall that difference of square of two numbers a and bis $a^2-b^2=\left(a-b\right)\left(a+b\right)$.

Apply it,

  $\begin{array}{c}4-t^2=2^2-t^2\\ =\left(2+t\right)\left(2-t\right)\end{array}$

Rewrite the original expression as,

  $\frac{t^2-2t-8}{4-t^2}\cdot \frac{t^2-5t+6}{t^2-t-12}=\frac{\left(t-4\right)\left(t+2\right)}{\left(2+t\right)\left(2-t\right)}\cdot \frac{\left(t-2\right)\left(t-3\right)}{\left(t-4\right)\left(t+3\right)}$

Strike off the similar terms,

  $\begin{array}{c}\frac{t^2-2t-8}{4-t^2}\cdot \frac{t^2-5t+6}{t^2-t-12}=\frac{\left(t-4\right)\left(t+2\right)}{\left(2+t\right)\left(2-t\right)}\cdot \frac{\left(t-2\right)\left(t-3\right)}{\left(t-4\right)\left(t+3\right)}\\ \frac{t^2-2t-8}{4-t^2}\cdot \frac{t^2-5t+6}{t^2-t-12}=-\frac{1}{1}\cdot \frac{\left(t-3\right)}{\left(t+3\right)}\end{array}$

Recall that fractions are multiplied as $\frac{a}{b}\cdot \frac{c}{d}=\frac{ab}{cd}$.

Apply it,

  $\begin{array}{c}\frac{t^2-2t-8}{4-t^2}\cdot \frac{t^2-5t+6}{t^2-t-12}=\frac{\left(t-4\right)\left(t+2\right)}{\left(2+t\right)\left(2-t\right)}\cdot \frac{\left(t-2\right)\left(t-3\right)}{\left(t-4\right)\left(t+3\right)}\\ \frac{t^2-2t-8}{4-t^2}\cdot \frac{t^2-5t+6}{t^2-t-12}=-\frac{1}{1}\cdot \frac{\left(t-3\right)}{\left(t+3\right)}\\ =-\frac{\left(t-3\right)}{\left(t+3\right)}\\ =\frac{\left(3-t\right)}{\left(3+t\right)}\end{array}$

Thus, the simplified form of the expression $\frac{t^2-2t-8}{4-t^2}\cdot \frac{t^2-5t+6}{t^2-t-12}$ is $\frac{\left(3-t\right)}{\left(3+t\right)}$.