Question

1. factor the binomial $144x^{2}y^{4}-9x^{8}y^{6}$ completely.

$\bigcirc$ $(12xy^{2}+3x^{4}y^{3})(12xy^{2}-3x^{4}y^{3})$

$\bigcirc$ $(16xy^{2}+3x^{4}y^{3})(9xy^{2}-3x^{4}y^{3})$

$\bigcirc$ $(36xy^{2}+3x^{4}y^{3})(4xy^{2}-3x^{4}y^{3})$

$\bigcirc$ $(72xy^{2}+3x^{4}y^{3})(2xy^{2}-3x^{4}y^{3})$

Explanation:

Step1: Identify the GCF

First, find the greatest common factor (GCF) of \(144x^{2}y^{4}\) and \(9x^{8}y^{6}\). The GCF of \(144\) and \(9\) is \(9\), the GCF of \(x^{2}\) and \(x^{8}\) is \(x^{2}\), and the GCF of \(y^{4}\) and \(y^{6}\) is \(y^{4}\). So the GCF is \(9x^{2}y^{4}\). Factor out the GCF:

$$
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Wait, actually, we can also recognize this as a difference of squares directly. Recall that \(a^{2}-b^{2}=(a + b)(a - b)\). Let's rewrite the given binomial:
\(144x^{2}y^{4}-9x^{8}y^{6}=(12xy^{2})^{2}-(3x^{4}y^{3})^{2}\)

Step2: Apply the difference of squares formula

Using the formula \(a^{2}-b^{2}=(a + b)(a - b)\) where \(a = 12xy^{2}\) and \(b=3x^{4}y^{3}\), we get:
\((12xy^{2}+3x^{4}y^{3})(12xy^{2}-3x^{4}y^{3})\)
We can check the other options:

• For the second option, \((16xy^{2}+3x^{4}y^{3})(9xy^{2}-3x^{4}y^{3})\), multiplying the first terms \(16xy^{2}\times9xy^{2}=144x^{2}y^{4}\), but the second terms multiplication \(3x^{4}y^{3}\times(- 3x^{4}y^{3})=-9x^{8}y^{6}\), however, when we expand \((16xy^{2}+3x^{4}y^{3})(9xy^{2}-3x^{4}y^{3})\), we get \(16xy^{2}\times9xy^{2}-16xy^{2}\times3x^{4}y^{3}+3x^{4}y^{3}\times9xy^{2}-3x^{4}y^{3}\times3x^{4}y^{3}=144x^{2}y^{4}-48x^{5}y^{5}+27x^{5}y^{5}-9x^{8}y^{6}=144x^{2}y^{4}-21x^{5}y^{5}-9x^{8}y^{6}

eq144x^{2}y^{4}-9x^{8}y^{6}\)

• For the third option, \((36xy^{2}+3x^{4}y^{3})(4xy^{2}-3x^{4}y^{3})\), multiplying the first terms \(36xy^{2}\times4xy^{2}=144x^{2}y^{4}\), but expanding the whole product gives extra terms, so it's incorrect.
• For the fourth option, \((72xy^{2}+3x^{4}y^{3})(2xy^{2}-3x^{4}y^{3})\), multiplying the first terms \(72xy^{2}\times2xy^{2}=144x^{2}y^{4}\), but expanding gives extra terms, so it's incorrect.

Answer:

\((12xy^{2}+3x^{4}y^{3})(12xy^{2}-3x^{4}y^{3})\)