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

The solution to (x^2+4)^2(x-2i)^2 is (x^6+4x^4-16x^2-64)+(-4x^5-32x^3-64x)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

simplify

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

(x^{6} + 4 x^{4} - 16 x^{2} - 64) + (- 4 x^{5} - 32 x^{3} - 64 x) i

Solution steps

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

Expand

(x^{2} + 4)^{2} : x^{4} + 8 x^{2} + 16

= (x^{4} + 8 x^{2} + 16) (x - 2 i)^{2}

= (x - 2 i)^{2} (x^{4} + 8 x^{2} + 16)

Expand

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

= (x^{2} - 4 x i - 4) (x^{4} + 8 x^{2} + 16)

Distribute parentheses

= x^{2} x^{4} + x^{2} \cdot 8 x^{2} + x^{2} \cdot 16 - 4 x i x^{4} - 4 x i 8 x^{2} - 4 x i 16 - 4 x^{4} - 4 \cdot 8 x^{2} - 4 \cdot 16

Simplify

x^{2} x^{4} + x^{2} \cdot 8 x^{2} + x^{2} \cdot 16 - 4 x i x^{4} - 4 x i 8 x^{2} - 4 x i 16 - 4 x^{4} - 4 \cdot 8 x^{2} - 4 \cdot 16 : x^{6} + 4 x^{4} - 16 x^{2} - 4 x^{5} i - 32 x^{3} i - 64 x i - 64

= x^{6} + 4 x^{4} - 16 x^{2} - 4 x^{5} i - 32 x^{3} i - 64 x i - 64

Rewrite in standard complex form:

(x^{6} + 4 x^{4} - 16 x^{2} - 64) + (- 4 x^{5} - 32 x^{3} - 64 x) i

= (x^{6} + 4 x^{4} - 16 x^{2} - 64) + (- 4 x^{5} - 32 x^{3} - 64 x) i

(x^{4} - 2 x^{3} - 7 x^{2} + 8 x + 12) (x - 11)

(x^{\frac{5}{2}} + y^{\frac{5}{2}}) (x^{\frac{5}{2}} - y^{\frac{5}{2}})

\frac{1}{( x ^{2} - 1 ) ( x - 1 )}

(x cos (a) + y cos (a + b) + z cos (a + b + c))^{2}

\frac{n ^{2} + 2}{n ^{2} ( n ^{2} + 4 )}

What is (x^2+4)^2(x-2i)^2 ?
The solution to (x^2+4)^2(x-2i)^2 is (x^6+4x^4-16x^2-64)+(-4x^5-32x^3-64x)i