Chapter 1.3, Problem 74E

To determine

To calculate: The factors of the trinomial $2{\left(a+b\right)}^2+5\left(a+b\right)-3$ .

Answer to Problem 74E

The factors of the trinomial $2{\left(a+b\right)}^2+5\left(a+b\right)-3$ are $\left(2a+2b-1\right)$ and $\left(a+b+3\right)$ .

Explanation of Solution

Given information:

The expression $2{\left(a+b\right)}^2+5\left(a+b\right)-3$ .

Formula used:

To find the factor of the trinomial of the form $a{x}^2+bx+c$ , find the numbers $p,q,r$ and $s$ such that sum of product of two numbers is equal to coefficient of $x$ $\left(ps+qr=b\right)$ , product of $p$ and $q$ is equal to the coefficient of $x^2$ $pq=a$ and product of $r$ and $s$ numbers is equal to constant term $\left(r\cdot s=c\right)$ , such that $\left(px+r\right)$ and $\left(qx+s\right)$ are the factors of $a{x}^2+bx+c$ .

  $\begin{array}{c}a{x}^2+bx+c=pq{x}^2+\left(ps+qr\right)x+rs\\ =pq{x}^2+psx+qrx+rs\\ =px\left(qx+s\right)+r\left(qx+s\right)\\ =\left(px+r\right)\left(qx+s\right)\end{array}$

Calculation:

Consider the given expression $2{\left(a+b\right)}^2+5\left(a+b\right)-3$ .

Recall that to find the factor of the trinomial of the form $a{x}^2+bx+c$ , find the numbers $p,q,r$ and $s$ such that sum of product of two numbers is equal to coefficient of $x$ $\left(ps+qr=b\right)$ , product of $p$ and $q$ is equal to the coefficient of $x^2$ $\left(pq=a\right)$ and product of $r$ and $s$ numbers is equal to constant term $\left(r\cdot s=c\right)$ , such that $\left(px+r\right)$ and $\left(qx+s\right)$ are the factors of $a{x}^2+bx+c$ .

  $\begin{array}{c}a{x}^2+bx+c=pq{x}^2+\left(ps+qr\right)x+rs\\ =pq{x}^2+psx+qrx+rs\\ =px\left(qx+s\right)+r\left(qx+s\right)\\ =\left(px+r\right)\left(qx+s\right)\end{array}$

So, find $p,q,r$ and $s$ such that $pq=2$ , $ps+qr=5$ and $rs=-3$ .

Here, $p=2$ , $q=1$ , $r=-1$ and $s=3$ , so, $9x^2+36x+32$ will be factorized as,

  $\begin{array}{c}2{\left(a+b\right)}^2+5\left(a+b\right)-3=2{\left(a+b\right)}^2+\left(6-1\right)\left(a+b\right)+\left(1\right)\cdot \left(-3\right)\\ =2{\left(a+b\right)}^2+6\left(a+b\right)-\left(a+b\right)-3\\ =\left(2a+2b\right)\left(a+b+3\right)-1\left(a+b+3\right)\\ =\left(2a+2b-1\right)\left(a+b+3\right)\end{array}$

Thus, the factors of trinomial $2{\left(a+b\right)}^2+5\left(a+b\right)-3$ are $\left(2a+2b-1\right)$ and $\left(a+b+3\right)$ .