Chapter 4.3, Problem 4CGP

To determine

To calculate: The factors of the polynomial $15x^2-8x+1$ .

Answer to Problem 4CGP

The factors of the polynomial $15x^2-8x+1$ are $5x-1$ and $3x-1$ .

Explanation of Solution

Given information:

The expression $15x^2-8x+1$ .

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 $15x^2-8x+1$ .

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 $pq=15$ , $ps+qr=-8$ and $rs=1$ .

Here, $p=-5$ , $q=-3$ , $r=1$ and $s=1$ , so, $15x^2-8x+1$ will be factorized as,

  $\begin{array}{c}15x^2-8x+1=15x^2-5x-3x+1\\ =5x\left(3x-1\right)-1\left(3x-1\right)\\ =\left(5x-1\right)\left(3x-1\right)\end{array}$

Thus, the factors of polynomial $15x^2-8x+1$ are $5x-1$ and $3x-1$ .