Chapter 4.6, Problem 8WE

To determine

To Find: Factor of polynomial $h^2-10h+24$

Answer to Problem 8WE

Factor of polynomial $h^2-10h+24=\left(h-4\right)\left(h-6\right)$

Explanation of Solution

Given information: polynomial $h^2-10h+24$

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 $h^2-10h+24$

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

So, $pr=1$

Therefore, let both pand r equal to $1$

So, that the factorization begins

  $\left(h\right)\left(h\right)$

Since the constant term is $24$

So, $qs=24$

There are six possibilities

  $\left(1\right)\left(24\right),\left(2\right)\left(8\right),\left(6\right)\left(4\right),\left(-1\right)\left(-24\right),\left(-2\right)\left(-8\right),\left(-6\right)\left(-4\right)$

Since the coefficient of $h$ is $-10$

So, $ps+qr=-10$ , but p and r are both $1$

So, $\begin{array}{l}ps+qr=-10\\ s+q=-10\end{array}$

Therefore, Choose the pair of factors having sum $-10:-6\text{ and -4}$

  $\therefore h^2-10h+24=\left(h-4\right)\left(h-6\right)$