Q 6.48TI. - Intermediate Algebra

Question

Factors:$16z^2-72z+81$.

Step-by-Step Solution

Answer

Factors of $16z^2-72z+81$ is ${\left(4z-9\right)}^2$, that is $\left(4z-9\right)\left(4z-9\right)$.

Step 1. Given information.

Given an expression $16z^2-72z+81$.

Step 2. Find if first and last term are perfect squares or not.

First term is $16z^2$ which can be expressed as ${\left(4z\right)}^2$.

Last term is 81 which can be expressed as $9^2$.

So, both first and last terms are perfect squares.

Consider $a=4z$ and $b=9$

Step 3. Check if the middle term is of the form ± 2 a b .

Middle term is $-72z$ which can be expressed as $-2\times 4z\times 9$.

It follows that middle term is of the form $-2ab$.

Step 4. Write the square of the binomial.

So, $16z^2-72z+81$is a perfect square trinomial of the form $a^2-2ab+b^2$.

Note that ${\left(a-b\right)}^2=a^2-2ab+b^2$.

Therefore, $16z^2-72z+81={\left(4z-9\right)}^2$.

Step 5. Check by multiplying.

Verification can be obtained as follows.

$\begin{array}{rcl}{\left(4z-9\right)}^2& =& {\left(4z\right)}^2-2\times 4z\times 9+9^2\\ & =& 16z^2-72z+81\end{array}$