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jeka57 [31]
3 years ago
15

What is the factored form of the expression over the complex numbers? 4x2+25y2?

Mathematics
2 answers:
Helen [10]3 years ago
4 0
The factored form is 25y^2 + 8 if the 2 was supposed to be an exponent.
Rzqust [24]3 years ago
3 0

Answer:

(2x+5iy)(2x-5iy)

Step-by-step explanation:

If a and b are two positive numbers, then

a^2+b^2=(a+ib)(a-ib)                .... (1)

The given expression is

4x^2+25y^2

It can be rewritten as

2^2x^2+5^2y^2

Using distributive property of exponent we get

(2x)^2+(5y)^2                 [\because a^mb^m=(ab)^m]

Using formula (1) we get

(2x+5iy)(2x-5iy)                 [\because a^mb^m=(ab)^m]

Therefore, the factored form of the given expression is (2x+5iy)(2x-5iy) .

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Sin5A/SinA-cos5A/cosA=4cos2A​
irina1246 [14]

Answer:

See Explanation

Step-by-step explanation:

\frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}  = 4\cos2A \\  \\ LHS = \frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}   \\  \\  =  \frac{ \sin5A \:\cos A -  \cos5A \:  \sin A}{\sin A \:\cos A }  \\  \\  =  \frac{ \sin(5A -A )}{\sin A \:\cos A}  \\  \\ =  \frac{ \sin 4A}{\sin A \:\cos A}  \\  \\ =  \frac{ 2\sin 2A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =  \frac{ 2 \times 2\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =  \frac{ 4\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =4\cos 2A \\  \\  = RHS \\  \\ thus \\  \\  \frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}  = 4\cos2A \\  \\ hence \: proved

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3 years ago
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