Chapter 4.6, Problem 7OE

To determine

To Find: Factor of polynomial $z^2+z-12$

Answer to Problem 7OE

Factor of polynomial $z^2+z-12=\left(z-3\right)\left(z+4\right)$

Explanation of Solution

Given information: polynomial $z^2+z-12$

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

  $a{x}^2+bx+c=pr{x}^2+\left(ps+qr\right)x+qs$

Setting corresponding coefficients equal gives

  $a=pr,b=ps+qr,c=qs$

Calculation:

polynomial $z^2+z-12$

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

So, $pr=1$

Therefore, let both pand r equal to $1$

So, that the factorization begins

  $\left(\text{z }\right)\left(\text{z }\right)$

Since the constant term is $-12$

So, $qs=-12$

Therefore, q and s must have opposite signs. There are four possibilities

  $\left(-1\right)\left(12\right),\left(1\right)\left(-12\right),\left(-3\right)\left(4\right),\left(3\right)\left(-4\right)$

Since the coefficient of y is $1$

So, $ps+qr=1$ , but p and r are both $1$

So, $\begin{array}{l}ps+qr=1\\ s+q=1\end{array}$

Therefore, Choose the pair of factors having sum $1:-3\text{ and 4}$

  $\therefore z^2+z-12=\left(z-3\right)\left(z+4\right)$