Problem 1
[tex]x^2+20 = (x)^2 + \left(\sqrt{20}\right)^2[/tex]
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
[tex]i=\sqrt{-1}[/tex]
So this means,
[tex]x^2+20 = (x)^2 + \left(\sqrt{20}\right)^2[/tex]
[tex]x^2+20 = \left(x+\sqrt{20}*i\right)\left(x-\sqrt{20}*i\right)[/tex]
Answer: [tex]\left(x+\sqrt{20}*i\right)\left(x-\sqrt{20}*i\right)[/tex]
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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,
[tex]a^2 + 2*a*b + b^2 = (a+b)^2[/tex]
[tex](y^2)^2 + 2*(y^2)*7 + 7^2 = (y^2+7)^2[/tex]
[tex]y^4 + 14y^2 + 49 = (y^2+7)^2[/tex]
[tex]y^4 + 14y^2 + 49 = (y^2+(\sqrt{7})^2)^2[/tex]
[tex]y^4 + 14y^2 + 49 = ((y-\sqrt{7})(y+\sqrt{7}))^2[/tex]
Answer: [tex]y^4 + 14y^2 + 49 = ((y-\sqrt{7})(y+\sqrt{7}))^2[/tex]
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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)
