Chapter 4.6, Problem 9ST

To determine

To Find: Factor of polynomial $6t^2+4t-2$

Answer to Problem 9ST

Factor of polynomial $6t^2+4t-2=\left(t+1\right)\left(6t-2\right)$

Explanation of Solution

Given information: polynomial $6t^2+4t-2$

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 $6t^2+4t-2$

  $pr=6$

The positive factors of $6$ are $1,2,3,6$

So, the factorization begins in one of these ways

  $\left(\text{t }\right)\left(\text{6t }\right)\text{ or }\left(\text{2t }\right)\left(\text{3t }\right)$

  $qs=-2$

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

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

There are some possible factorizations to check

  $\begin{array}{l}TrialFactorization Lineartermofproduct \\ \left(t+1\right)\left(6t-2\right)4t\\ \left(t-2\right)\left(6t+1\right)-11t\\ \left(2t-2\right)\left(3t+1\right)-4t\\ \left(2t+1\right)\left(3t-2\right)-t\end{array}$

Since the required linear term is $4t$ , the first factorization is the correct one

  $\therefore 6t^2+4t-2=\left(t+1\right)\left(6t-2\right)$