Chapter 1, Problem 55RE

To determine

To simplify $\frac{x^2-2x-3}{x^2+8x+16}.\frac{3x+12}{x-1}$ by using appropriate operations.

Answer to Problem 55RE

The answer is $\frac{3(x+3)}{(x+4)}$ .

Explanation of Solution

Given information:

  $\frac{x^2-2x-3}{x^2+8x+16}.\frac{3x+12}{x-1}$

Formula used:

Factoring Trinomials:- To factor the form $x^2+bx+c$

  $(x+r)(x+s)=x^2(r+s)x+rs$ where $r+s=b$ and $rs=c$ .

Property of Fractions: $\frac{A}{B}.\frac{C}{D}=\frac{AC}{BD}$

Calculations:

By factoring both numerator and denominator we get,

First Term $\frac{x^2-2x-3}{x^2+8x+16}$

Numerator = $x^2+2x-3$

where $r=3$ and $s=-1$ .

  $\begin{array}{l}r+s=3-1=2\\ rs=(3)(-1)=-3\\ (x+r)(x+s)=(x+3)(x-1)\end{array}$  (i)

Denominator = $x^2+8x+16$

where $r=4$ and $s=4$ .

  $\begin{array}{l}r+s=4+4=8\\ rs=(4)(4)=16\\ (x+r)(x+s)=(x+4)(x+4)\end{array}$  (ii)

Second Term $\frac{3x+12}{x-1}$

  $\begin{array}{l}=\frac{3(x+4)}{x-1}\\ \end{array}$ (iii)

Now, dividing equation (i) and (ii) and multiplying equation (iii) we get,

  $=\frac{(x+3)(x-1)}{(x+4)(x+4)}.\frac{3(x+4)}{(x-1)}$

  $=\frac{3(x+3)(x-1)(x+4)}{(x+4)(x+4)(x-1)}$

  $=\frac{3(x+3)}{(x+4)}$ ( by cancelling $(x-1)(x+4)$ )

After simplifying we get $\frac{3(x-3)}{(x+4)}$ .