Chapter 4.3, Problem 4AGP

To determine

To calculate: The factors of the polynomial $x^2-11x+30$ .

Answer to Problem 4AGP

The factors of the polynomial $x^2-11x+30$ are $x-6$ and $x-5$ .

Explanation of Solution

Given information:

The expression $x^2-11x+30$ .

Formula used:

To find the factor of the trinomial of the form $x^2+bx+c$ , find two numbers $r$ and $s$ such that sum of the numbers is equal to coefficient of $x$

  $\left(r+s=b\right)$ and product of two numbers is equal to constant term $\left(r\cdot s=c\right)$ , such that $\left(x+r\right)$ and $\left(x+s\right)$ are the factors of $x^2+bx+c$ .

  $\begin{array}{c}{x}^2+bx+c=x^2+\left(r+s\right)x+rs\\ =x^2+rx+sx+rs\\ =x\left(x+r\right)+s\left(x+r\right)\\ =\left(x+r\right)\left(x+s\right)\end{array}$

Calculation:

Consider the given expression $x^2-11x+30$ .

Recall that to find the factor of the trinomial of the form $x^2+bx+c$ , find two numbers $r$ and $s$ such that sum of the numbers is equal to coefficient of $x$

  $\left(r+s=b\right)$ and product of two numbers is equal to constant term $\left(r\cdot s=c\right)$ , such that $\left(x+r\right)$ and $\left(x+s\right)$ are the factors of $x^2+bx+c$ .

  $\begin{array}{c}{x}^2+bx+c=x^2+\left(r+s\right)x+rs\\ =x^2+rx+sx+rs\\ =x\left(x+r\right)+s\left(x+r\right)\\ =\left(x+r\right)\left(x+s\right)\end{array}$

So, find two numbers $r$ and $s$ such that $r+s=-11$ and $rs=30$ .

Here, $r=-6$ and $s=-5$ , so, $x^2-11x+30$ will be factorized as,

  $\begin{array}{c}{x}^2-11x+30=x^2+\left(-6-5\right)x+\left(6\right)\cdot \left(5\right)\\ =x^2-6x-5x+30\\ =x\left(x-6\right)-5\left(x-6\right)\\ =\left(x-6\right)\left(x-5\right)\end{array}$

Thus, the factors of polynomial $x^2-11x+30$ are $x-6$ and $x-5$ .