Chapter 4.6, Problem 6ST

To determine

To Find: Factor of polynomial $9t^2+6t+1$

Answer to Problem 6ST

Factor of polynomial $9t^2+6t+1=\left(3t+1\right)\left(3t+1\right)$

Explanation of Solution

Given information: polynomial $9t^2+6t+1$

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 $9t^2+6t+1$

  $pr=9$

The positive factors of $9$ are $1,3,3,9$

So,the factorization begins in one of these ways

  $\left(\text{t }\right)\left(\text{9t }\right)\text{ or }\left(\text{3t }\right)\left(\text{3t }\right)$

  $qs=1$

Since the product $qs$ is positive, its factors are both positive or both negative. But the coefficient of t is positive. So, choose positive factor of $1$

  $\left(\text{ 1}\right)\left(\text{ 1 }\right)$

There are some possible factorizations to check

Trial Factorization Linear term of product

  $\begin{array}{l}\left(t+1\right)\left(9t+1\right)\\ \left(3t+1\right)\left(3t+1\right)\end{array}$$\begin{array}{l}10t\\ 6t\end{array}$

Since the required linear term is $6t$ , the second factorization is the correct one

  $\therefore 9t^2+6t+1=\left(3t+1\right)\left(3t+1\right)$