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disa [49]
3 years ago
13

How do you factor this polynomial expression?

Mathematics
2 answers:
mrs_skeptik [129]3 years ago
6 0

Here's a rule that I learned from my algebra teacher almost 60 years ago. 
It's so handy, and I use it so often, that it's still fresh in my mind, and even
though it's so old, it still works !

In fact, it's so useful that it would be a great item for you to memorize
and keep in your math tool-box.

==> To factor the difference of two squares, write

         <em>(the sum of their square roots) </em>times<em> (the difference of their square roots)</em> .

That's exactly what you need to solve this problem.
I'll show you how it works:

             <u>9x² - 25</u>

You look at this for a few seconds, and you realize that
9x²  is the square of  3x , and  25  is the square of  5 .
So this expression is the difference of two squares,
and you can use the shiny new tool I just handed you.

The square roots are  3x  and  5 .

So the factored form of the polynomial is    <em>(3x + 5) (3x - 5)</em> .

That's all there is to it.  If you FOIL these factors out, you'll see
that you wind up with the original polynomial in the question.


alexdok [17]3 years ago
3 0
9{ x }^{ 2 }-25\\ \\ ={ 3 }^{ 2 }{ x }^{ 2 }-{ 5 }^{ 2 }\\ \\ ={ \left( 3x \right)  }^{ 2 }-{ 5 }^{ 2 }\\ \\ =\left( 3x+5 \right) \left( 3x-5 \right)
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{\tan(\theta)}={\dfrac{\sin(\theta)}{\cos(\theta)}    and   {\sec(\theta)}={\dfrac{1}{\cos(\theta)}

Recall the following reciprocal identity:

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So, the original expression can be written in terms of only sines and cosines:

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\sin(\beta) * \dfrac{\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} {\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}

\sin(\beta) *\dfrac{1 }{\cos(\beta) }

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Working toward one of the answers provided, this is the tangent function.


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Answer:Step-by-step explanation:

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