Chapter 4.6, Problem 16WE

To determine

To Find: Factor of polynomial $5v^2+4v-1$

Answer to Problem 16WE

Factor of polynomial $5v^2+4v-1=\left(v+1\right)\left(5v-1\right)$

Explanation of Solution

Given information: polynomial $5v^2+4v-1$

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 $5v^2+4v-1$

  $pr=5$

The factorization begins

  $\left(v\right)\left(5v\right)$

  $qs=-1$

the coefficient of $v$ is positive.

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

There are some possible factorizations to check

  $\begin{array}{l}TrialFactorization Lineartermofproduct \\ \left(v+1\right)\left(5v-1\right)4v\\ \left(v-1\right)\left(5v+1\right)-4v\end{array}$

Since the required linear term is $v$ , the first factorization is the correct one

  $\therefore 5v^2+4v-1=\left(v+1\right)\left(5v-1\right)$