Chapter 4.6, Problem 10ST

To determine

To Find: Factor of polynomial $4z^2+z-14$

Answer to Problem 10ST

Factor of polynomial $4z^2+z-14=\left(z+2\right)\left(4z-7\right)$

Explanation of Solution

Given information: polynomial $4z^2+z-14$

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 $4z^2+z-14$

  $pr=4$

The positive factors of $4$ are $1,2,2,4$

So, the factorization begins in one of these ways

  $\left(z\right)\left(4z\right)\text{ or }\left(2z\right)\left(2z\right)$

  $qs=-14$

Since the product $qs$ is negative, its factors are both positive or both negative. But the coefficient of $z$ is positive.

  $\begin{array}{l}\left(\text{ -2}\right)\left(\text{ 7}\right)\\ \left(\text{ 2}\right)\left(\text{ -7 }\right)\\ \left(\text{ 1}\right)\left(\text{ -14}\right)\\ \left(\text{ 14}\right)\left(\text{ -1 }\right)\end{array}$

There are some possible factorizations to check

  $\begin{array}{l}TrialFactorization Lineartermofproduct \\ \left(z-2\right)\left(4z+7\right)\text{ -z }\\ \left(2z-2\right)\left(2z+7\right)10z\\ \left(z+2\right)\left(4z-7\right)\text{ z}\\ \left(2z+2\right)\left(2z-7\right)-10z\end{array}$

Since the required linear term is $z$ , the third factorization is the correct one

  $\therefore 4z^2+z-14=\left(z+2\right)\left(4z-7\right)$