Chapter 1.3, Problem 72E

To determine

To calculate: The factors of the trinomial $5x^2-7x-6$ .

Answer to Problem 72E

The factors of the trinomial $5x^2-7x-6$ are $5x+3$ and $x-2$ .

Explanation of Solution

Given information:

The expression $5x^2-7x-6$ .

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 $5x^2-7x-6$ .

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 $pq=5$ , $ps+qr=-7$ and $rs=-6$ .

Here, $p=5$ , $q=1$ , $r=3$ and $s=-2$ , so, $5x^2-7x-6$ will be factorized as,

  $\begin{array}{c}5x^2-7x-6=5x^2+\left(-10+3\right)x+\left(-2\right)\cdot \left(3\right)\\ =5x^2-10x+3x-6\\ =5x\left(x-2\right)+3\left(x-2\right)\\ =\left(5x+3\right)\left(x-2\right)\end{array}$

Thus, the factors of trinomial $5x^2-7x-6$ are $5x+3$ and $x-2$ .