Problem 1

We have a sum of squares in the form a^2 + b^2 where

a = x

b = sqrt(20)

It turns out that

a^2 + b^2 = (a+bi)*(a-bi)

where i is the square root of negative 1

So this means,

Answer:

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Problem 2

Similar to problem 1, we use the formula

a^2 + b^2 = (a+bi)*(a-bi)

In this case,

a = x

b = 6

So,

a^2 + b^2 = (a+bi)*(a-bi)

x^2 + 6^2 = (x+6i)*(x-6i)

x^2 + 36 = (x+6i)*(x-6i)

Answer: (x+6i)*(x-6i)

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Problem 3

Use the difference of squares rule twice to get

x^4 - 81 = (x^2)^2 - (9)^2

x^4 - 81 = (x^2-9)(x^2+9)

x^4 - 81 = (x^2-3^2)(x^2+9)

x^4 - 81 = (x-3)(x+3)(x^2+9)

x^4 - 81 = (x-3)(x+3)(x^2+3^2)

x^4 - 81 = (x-3)(x+3)(x-3i)(x+3i)

Answer: (x-3)(x+3)(x-3i)(x+3i)

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Problem 4

This is in the form a^2 + 2*a*b + b^2 where

a = y^2

b = 7

So,

Answer:

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Problem 5

I'm assuming the expression should be y^3+2y^2+16y+32

Use factoring by grouping

y^3+2y^2+16y+32

(y^3+2y^2)+(16y+32)

y^2(y+2)+(16y+32)

y^2(y+2)+16(y+2)

(y^2+16)(y+2)

(y^2+4^2)(y+2)

(y-4i)(y+4i)(y+2)

Answer: (y-4i)(y+4i)(y+2)