Chapter 4.6, Problem 14WE

To determine

To Find: Factor of polynomial $s^2-6s-27$

Answer to Problem 14WE

Factor of polynomial $s^2-6s-27=\left(s-9\right)\left(s+3\right)$

Explanation of Solution

Given information: polynomial $s^2-6s-27$

Formula used: If the quadratic trinomial $a{x}^2+bx+c$ can be factored into the product $\left(px+q\right)\left(tx+l\right)$ where $p,q,t,l$ are integers, then

  $a{x}^2+bx+c=pt{x}^2+\left(pl+qt\right)x+ql$

Setting corresponding coefficients equal gives

  $a=pt,b=pl+qt,c=ql$

Calculation:

polynomial $s^2-6s-27$

Since the coefficient of $s^2$ is $1$

So, $pt=1$

Therefore, let both pand t equal to $1$

So, that the factorization begins

  $\left(s\right)\left(s\right)$

Since the constant term is $-27$

So, $ql=-27$

There are four possibilities

  $\left(-1\right)\left(27\right),\left(-27\right)\left(1\right),\left(-9\right)\left(3\right),\left(-3\right)\left(9\right)$

Since the coefficient of $s$ is $-6$

So, $pl+qt=-6$ , but p and tare both $1$

So, $\begin{array}{l}pl+qt=-6\\ l+q=-6\end{array}$

Therefore, Choose the pair of factors having sum $-6:-9\text{ and 3}$

  $\therefore s^2-6s-27=\left(s-9\right)\left(s+3\right)$