Question

Use a Special Factoring Formula to factor the expression

$16z^2-24z+9$

Short answer

The solution$16z^2-24z+9={\left(4z-3\right)}^2$

Step-by-step solution

Step 1. Apply the concept of factoring trinomials.

To factor a trinomial of the form $a{x}^2+bx+c;a\ne 0$, we note that

$a{x}^2+bx+c=\left(px+r\right)\left(qx+s\right)=pq{x}^2+\left(ps+qr\right)x+rs$

So, we need to choose numbers p, q, r and s so that $pq=a$,$ps+qr=b$ and $rs=c$.

Step 2. Analyze the trinomial.

Given$16z^2-24z+9$

This is of the form $a{x}^2+bx+c$where $a=16$$b=-24$and$c=9$

The factorization of the trinomial will be$\left(px+r\right)\left(qx+s\right)=pq{x}^2+\left(ps+qr\right)x+rs$

So, we need to choose numbers r and s so that $pq=a$,$ps+qr=b$ and $rs=c$.

Step 3. Find the integers to factor the trinomial.

We need to choose integers r and s so that $pq=16$, $ps+qr=-24$and $rs=9$.

By trial and error we find that the two integers are $p=4,q=4,r=-3,s=-3$

$pq=4\cdot 4=16$

$ps+qr=-12-12=-24$

$rs=-3\cdot (-3)=9$

Step 4. Factor the trinomials.

The factorization of the trinomial is

$\begin{array}{l}16z^2-24z+9=\left(pz+r\right)\left(qz+s\right)\\ \Rightarrow 16z^2-24z+9=\left(4z-3\right)\left(4z-3\right)\\ \Rightarrow 16z^2-24z+9={\left(4z-3\right)}^2\end{array}$