Chapter 1.3, Problem 100E

To determine

To calculate: The factor of the expression $2x^2+7x-4$ .

Answer to Problem 100E

The factor of the expression $2x^2+7x-4$ is $\left(2x-1\right)\left(x+4\right)$ .

Explanation of Solution

Given information:

The expression $2x^2+7x-4$ .

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 $2x^2+7x-4$ .

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 $pq=2$ , $ps+qr=7$ and $rs=-4$ .

Here, $p=2$ , $q=1$ , $r=-1$ and $s=4$ , so, $2x^2+7x-4$ will be factorized as,

  $\begin{array}{c}2x^2+7x-4=2x^2+\left(8-1\right)x+\left(-1\right)\cdot \left(4\right)\\ =2x^2+8x-x-4\\ =2x\left(x+4\right)-1\left(x+4\right)\\ =\left(2x-1\right)\left(x+4\right)\end{array}$

Thus, the factor of expression $2x^2+7x-4$ is $\left(2x-1\right)\left(x+4\right)$ .