Chapter 6.2, Problem 47WE

To determine

To calculate: The simplified form of the expression $\frac{3x^2-14x-5}{8x^2-12x}\cdot \frac{8x^2-18}{3x^2+4x+1}$.

Answer to Problem 47WE

The simplified form of the expression $\frac{3x^2-14x-5}{8x^2-12x}\cdot \frac{8x^2-18}{3x^2+4x+1}$ is $\frac{\left(x-5\right)\left(2x+3\right)}{2x\left(x+1\right)}$.

Explanation of Solution

Given information:

The expression $\frac{3x^2-14x-5}{8x^2-12x}\cdot \frac{8x^2-18}{3x^2+4x+1}$.

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 $\frac{3x^2-14x-5}{8x^2-12x}\cdot \frac{8x^2-18}{3x^2+4x+1}$.

Take the trinomial $3x^2-14x-5$.

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.

Rewrite the trinomial and take out the common terms,

  $\begin{array}{c}3x^2-14x-5=3x^2-15x+x-5\\ =3x\left(x-5\right)+1\left(x-5\right)\\ =\left(x-5\right)\left(3x+1\right)\end{array}$

Take the trinomial $3x^2+4x+1$.

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,

  $\begin{array}{c}3x^2+4x+1=3x^2+3x+x+1\\ =3x\left(x+1\right)+1\left(x+1\right)\\ =\left(x+1\right)\left(3x+1\right)\end{array}$

Take the term $4x$ common from denominator of first term and the term $2$ from numerator of second term.

Rewrite the original expression as,

  $\begin{array}{c}\frac{3x^2-14x-5}{8x^2-12x}\cdot \frac{8x^2-18}{3x^2+4x+1}=\frac{\left(x-5\right)\left(3x+1\right)}{4x\left(2x-3\right)}\cdot \frac{2\left(4x^2-9\right)}{\left(3x+1\right)\left(x+1\right)}\\ =\frac{\left(x-5\right)\left(3x+1\right)}{4x\left(2x-3\right)}\cdot \frac{2\left[{\left(2x\right)}^2-3^2\right]}{\left(3x+1\right)\left(x+1\right)}\end{array}$

Take the binomial $4x^2-9$.

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,

  $\begin{array}{c}\frac{3x^2-14x-5}{8x^2-12x}\cdot \frac{8x^2-18}{3x^2+4x+1}=\frac{\left(x-5\right)\left(3x+1\right)}{4x\left(2x-3\right)}\cdot \frac{2\left(4x^2-9\right)}{\left(3x+1\right)\left(x+1\right)}\\ =\frac{\left(x-5\right)\left(3x+1\right)}{4x\left(2x-3\right)}\cdot \frac{2\left[{\left(2x\right)}^2-3^2\right]}{\left(3x+1\right)\left(x+1\right)}\\ =\frac{\left(x-5\right)\left(3x+1\right)}{4x\left(2x-3\right)}\cdot \frac{2\left(2x-3\right)\left(2x+3\right)}{\left(3x+1\right)\left(x+1\right)}\end{array}$

Strike off the similar terms,

  $\begin{array}{c}\frac{3x^2-14x-5}{8x^2-12x}\cdot \frac{8x^2-18}{3x^2+4x+1}=\frac{\left(x-5\right)\left(3x+1\right)}{4x\left(2x-3\right)}\cdot \frac{2\left(4x^2-9\right)}{\left(3x+1\right)\left(x+1\right)}\\ =\frac{\left(x-5\right)\left(3x+1\right)}{4x\left(2x-3\right)}\cdot \frac{2\left[{\left(2x\right)}^2-3^2\right]}{\left(3x+1\right)\left(x+1\right)}\\ =\frac{\left(x-5\right)}{2x}\cdot \frac{\left(2x+3\right)}{\left(x+1\right)}\end{array}$

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

Apply it,

  $\begin{array}{c}\frac{3x^2-14x-5}{8x^2-12x}\cdot \frac{8x^2-18}{3x^2+4x+1}=\frac{\left(x-5\right)\left(3x+1\right)}{4x\left(2x-3\right)}\cdot \frac{2\left(4x^2-9\right)}{\left(3x+1\right)\left(x+1\right)}\\ =\frac{\left(x-5\right)\left(3x+1\right)}{4x\left(2x-3\right)}\cdot \frac{2\left[{\left(2x\right)}^2-3^2\right]}{\left(3x+1\right)\left(x+1\right)}\\ =\frac{\left(x-5\right)}{2x}\cdot \frac{\left(2x+3\right)}{\left(x+1\right)}\\ =\frac{\left(x-5\right)\left(2x+3\right)}{2x\left(x+1\right)}\end{array}$

Thus, the simplified form of the expression $\frac{3x^2-14x-5}{8x^2-12x}\cdot \frac{8x^2-18}{3x^2+4x+1}$ is $\frac{\left(x-5\right)\left(2x+3\right)}{2x\left(x+1\right)}$.