Chapter 4.6, Problem 7WE

To determine

To Find: Factor of polynomial $p^2-8p+9$

Answer to Problem 7WE

polynomial $p^2-8p+9$ is prime

Explanation of Solution

Given information: polynomial $p^2-8p+9$

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 $p^2-8p+9$

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

So, $pr=1$

Therefore, let both pand r equal to $1$

So, that the factorization begins

  $\left(p\right)\left(p\right)$

Since the constant term is $9$

So, $qs=9$

There are six possibilities

  $\left(-1\right)\left(-9\right),\left(1\right)\left(9\right),\left(3\right)\left(3\right),\left(-3\right)\left(-3\right)$

Since the coefficient of $p$ is $-8$

So, $ps+qr=-8$ , but p and r are both $1$

So, $\begin{array}{l}ps+qr=-8\\ s+q=-8\end{array}$

Therefore, Choose the pair of factors having sum $-8$

No two factors have a product of $9$ and sum of $-8$

Therefore $p^2-8p+9$ is prime