What is (x^2+2i)^2(x+4) ?

The solution to (x^2+2i)^2(x+4) is (x^5+4x^4-4x-16)+(4x^3+16x^2)i

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simplify (x^2+2i)^2(x+4)

simplify

(x^{2} + 2 i)^{2} (x + 4)

(x^{5} + 4 x^{4} - 4 x - 16) + (4 x^{3} + 16 x^{2}) i

Solution steps

(x^{2} + 2 i)^{2} (x + 4)

Expand

(x^{2} + 2 i)^{2} : x^{4} + 4 x^{2} i - 4

= (x^{4} + 4 x^{2} i - 4) (x + 4)

Distribute parentheses

= x^{4} x + x^{4} \cdot 4 + 4 x^{2} i x + 4 x^{2} i 4 - 4 x - 4 \cdot 4

Simplify

x^{4} x + x^{4} \cdot 4 + 4 x^{2} i x + 4 x^{2} i 4 - 4 x - 4 \cdot 4 : x^{5} + 4 x^{4} + 4 x^{3} i + 16 x^{2} i - 4 x - 16

= x^{5} + 4 x^{4} + 4 x^{3} i + 16 x^{2} i - 4 x - 16

Rewrite in standard complex form:

(x^{5} + 4 x^{4} - 4 x - 16) + (4 x^{3} + 16 x^{2}) i

= (x^{5} + 4 x^{4} - 4 x - 16) + (4 x^{3} + 16 x^{2}) i

sec (a) (1 - sin (a)) (sec (a) + tan (a))

(x - 2)^{2} (x - (6 i)) (x - (- 6 i))

x^{2 (a - 2)} - y^{6 b}

2 (e^{\frac{π}{2}} - 1)

(x - 1) \times x

What is (x^2+2i)^2(x+4) ?
The solution to (x^2+2i)^2(x+4) is (x^5+4x^4-4x-16)+(4x^3+16x^2)i