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True [87]
3 years ago
7

Factor the following polynomial completely. -x 2 y 2 + x 4 + 4y 2 - 4x 2

Mathematics
2 answers:
zlopas [31]3 years ago
5 0

Answer:

(x + y)(x - y) (x + 2)(x - 2)

Step-by-step explanation:

-x^2 y^2 + x^4 + 4y^2 - 4x^2

Lets rearrange the terms

x^4 -x^2 y^2 - 4x^2 + 4y^2

Then we group first two terms and last two terms

(x^4 -x^2 y^2)+(- 4x^2 + 4y^2)

Factor out each group

x^2(x^2 - y^2) - 4 (x^2 - y^2)

(x^2 - y^2)(x^2 - 4)

Apply formula a^2 - b^2 = (a+b)(a-b)

(x + y)(x - y) (x + 2)(x - 2)

fgiga [73]3 years ago
3 0
<span>-x^2 y^2 + x^4 + 4y^2 - 4x^2
= </span>x^4 -x^2 y^2 - 4x^2 + 4y^2<span>
= x^2 (x^2 - y^2) - 4 (x^2 - y^2)
= (x^2 - y^2)(x^2 - 4)
= (x + y)(x - y) (x + 2)(x - 2) (....using a^2 - b^2 = (a+b)(a-b) )

hope it helps</span>
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ratelena [41]

By the definition of the hyperbolic function tanh x, we have proven that \frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}

<h3>Hyperbolic functions & Proof of identities </h3>

By definition

tanh \ x=\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}}

Then,

\frac{1-tanh \ x}{1 + tanh \ x}=\frac{1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }{1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }

=1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} \div (1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} })

=\frac{e^{x} +e^{-x}-(e^{x} -e^{-x})}{e^{x} +e^{-x}} \div (\frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} })

=\frac{e^{x} +e^{-x}-e^{x} +e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} }

=\frac{e^{-x}+e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{x} }{e^{x} +e^{-x}} }

=\frac{2e^{-x}}{e^{x} +e^{-x}} \div \frac{2e^{x} }{e^{x} +e^{-x}} }

=\frac{2e^{-x}}{e^{x} +e^{-x}} \times \frac{e^{x} +e^{-x}}{2e^{x}}

=\frac{2e^{-x}}{1} \times \frac{1}{2e^{x}}

=\frac{2e^{-x}}{2e^{x}}

=\frac{e^{-x}}{e^{x}}
=e^{-x} \times \frac{1}{e^{x}}

= e^{-x} \times e^{-x}

= e^{-x+-x}

= e^{-x-x}

= e^{-2x}

Hence, we have proven that \frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}

Learn more on Proof of Identities here: brainly.com/question/2561079

#SPJ1

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