Answer:
[tex](36 {s}^{2} - \frac{1}{64 {t}^{2} } ) \\ \\ = ({6s})^{2} - \frac{1}{ ({8t})^{2} } \\ \\ (6s - \frac{1}{8t} )(6s + \frac{1}{8t} )[/tex]
The factorization of [tex]36s^2-\frac{1}{64t^2}[/tex] will be [tex](6s + \frac{1}{8t} )( 6s - \frac{1}{8t} )[/tex].
Given:
[tex]36s^2-\frac{1}{64t^2}[/tex]
To Find:
The factorization of [tex]36s^2-\frac{1}{64t^2}[/tex] =?
Solution:
The algebraic identity that we are going to use is
a² - b² = (a + b)(a - b)
It is given that [tex]36s^2-\frac{1}{64t^2}[/tex].
It can be rewritten as [tex](6s)^2-\frac{1}{(8t)^2}[/tex].
= [tex](6s)^2-\frac{1}{(8t)^2}[/tex]
= [tex](6s + \frac{1}{8t} )( 6s - \frac{1}{8t} )[/tex]
Hence, the factorization of [tex]36s^2-\frac{1}{64t^2}[/tex] will be [tex](6s + \frac{1}{8t} )( 6s - \frac{1}{8t} )[/tex].
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Verified answer
Answer:
[tex](36 {s}^{2} - \frac{1}{64 {t}^{2} } ) \\ \\ = ({6s})^{2} - \frac{1}{ ({8t})^{2} } \\ \\ (6s - \frac{1}{8t} )(6s + \frac{1}{8t} )[/tex]
The factorization of [tex]36s^2-\frac{1}{64t^2}[/tex] will be [tex](6s + \frac{1}{8t} )( 6s - \frac{1}{8t} )[/tex].
Given:
[tex]36s^2-\frac{1}{64t^2}[/tex]
To Find:
The factorization of [tex]36s^2-\frac{1}{64t^2}[/tex] =?
Solution:
The algebraic identity that we are going to use is
a² - b² = (a + b)(a - b)
It is given that [tex]36s^2-\frac{1}{64t^2}[/tex].
It can be rewritten as [tex](6s)^2-\frac{1}{(8t)^2}[/tex].
[tex]36s^2-\frac{1}{64t^2}[/tex]
= [tex](6s)^2-\frac{1}{(8t)^2}[/tex]
= [tex](6s + \frac{1}{8t} )( 6s - \frac{1}{8t} )[/tex]
Hence, the factorization of [tex]36s^2-\frac{1}{64t^2}[/tex] will be [tex](6s + \frac{1}{8t} )( 6s - \frac{1}{8t} )[/tex].
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