Chapter 1.3, Problem 71E

To determine

To calculate: The factors of the trinomial $3x^2-16x+5$ .

Answer to Problem 71E

The factors of the trinomial $3x^2-16x+5$ are $3x-1$ and $x-5$ .

Explanation of Solution

Given information:

The expression $3x^2-16x+5$ .

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 $3x^2-16x+5$ .

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 $pq=3$ , $ps+qr=-16$ and $rs=5$ .

Here, $p=3$ , $q=1$ , $r=5$ and $s=1$ , so, $3x^2-16x+5$ will be factorized as,

  $\begin{array}{c}3x^2-16x+5=3x^2+\left(-15-1\right)x+\left(5\right)\cdot \left(1\right)\\ =3x^2-15x-x+5\\ =3x\left(x-5\right)-1\left(x-5\right)\\ =\left(3x-1\right)\left(x-5\right)\end{array}$

Thus, the factors of trinomial $3x^2-16x+5$ are $3x-1$ and $x-5$ .