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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 6: Q 6.48TI. (page 594)

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

Short Answer

Expert verified

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

Step by step solution

01

Step 1. Given information.

Given an expression $16 z^{2} - 72 z + 81$ .

02

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

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

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

So, both first and last terms are perfect squares.

Consider $a = 4 z$ and $b = 9$

03

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

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

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

04

Step 4. Write the square of the binomial.

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

Note that ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ .

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

05

Step 5. Check by multiplying.

Verification can be obtained as follows.

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

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Most popular questions from this chapter

In the following exercises, factor completely using the sums and differences of cubes pattern, if possible.
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