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3 years ago

Factor the expression completely. 625x4 - 256y4

Mathematics

1 answer:

scoundrel [369]3 years ago

5 0

Answer:

(625 • (x4)) - 28y4

54x4 - 28y4 3.1 Factoring: 625x4-256y4

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =

A2 - AB + BA - B2 =

A2 - AB + AB - B2 =

A2 - B2Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : 625 is the square of 25

Check : 256 is the square of 16

Check : x4 is the square of x2

Check : y4 is the square of y2

Factorization is : (25x2 + 16y2) • (25x2 - 16y2) 3.2 Factoring: 25x2 - 16y2

Check : 25 is the square of 5

Check : 16 is the square of 4

Check : x2 is the square of x1

Check : y2 is the square of y1

Factorization is : (5x + 4y) • (5x - 4y)this is the answer: (25x2 + 16y2) •(5x + 4y) • (5x - 4y)

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3 years ago

Please Help!! Use Euler’s formula to write in exponential form.

LekaFEV [45]

Answer:

C, $4e^{i(7\pi/4)}$

Step-by-step explanation:

To remind you, Euler's formula gives a link between trigonometric and exponential functions in a very profound way:

$e^{ix}=\cos{x}+i\sin{x}$

Given the complex number $2\sqrt{2}-2i\sqrt{2}$ , we want to try to get it in the same form as the right side of Euler's formula. As things are, though, we're unable to, and the reason for that has to do with the fact that both the sine and cosine functions are bound between the values 1 and -1, and 2√2 and -2√2 both lie outside that range.

One thing we could try would be to factor out a 2 to reduce both of those terms, giving us the expression $2(\sqrt{2}-i\sqrt{2})$

Still no good. √2 and -√2 are still greater than 1 and less than -1 respectively, so we'll have to reduce them a little more. With some clever thinking, you could factor out another 2, giving us the expression $4\left(\frac{\sqrt{2}}{2} -i\frac{\sqrt{2}}{2}\right)$ , and now we have something to work with.

Looking back at Euler's formula $e^{ix}=\cos{x}+i\sin{x}$ , we can map our expression inside the parentheses to the one on the right side of the formula, giving us $\cos{x}=\frac{\sqrt2}{2}$ and $\sin{x}=-\frac{\sqrt2}{2}$ , or equivalently:

$\cos^{-1}{\frac{\sqrt2}{2} }=\sin^{-1}-\frac{\sqrt2}{2} =x$

At this point, we can look at the unit circle (attached) to see the angle satisfying these two values for sine and cosine is 7π/4, so $x=\frac{7\pi}{4}$ , and we can finally replace our expression in parentheses with its exponential equivalent: