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

The solution to (x+4)(x+4+2i)(x-4-2i) is (x+4)(x^2-12)+(-16x-64)i

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

simplify

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

(x + 4) (x^{2} - 12) + (- 16 x - 64) i

Solution steps

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

Rewrite

x + 4 + 2 i

in standard complex form:

(x + 4) + 2 i

Rewrite

x - 4 - 2 i

in standard complex form:

(x - 4) - 2 i

= (x + 4) ((x + 4) + 2 i) ((x - 4) - 2 i)

Apply complex arithmetic rule:

(a + bi) (c + d i) = (a c - b d) + (a d + b c) i

= (x + 4) (((x + 4) (x - 4) - 2 (- 2)) + ((x + 4) (- 2) + 2 (x - 4)) i)

Simplify

((x + 4) (x - 4) - 2 (- 2)) + ((x + 4) (- 2) + 2 (x - 4)) i : (x^{2} - 12) - 16 i

= (x + 4) ((x^{2} - 12) - 16 i)

Distribute parentheses

= (x + 4) (x^{2} - 12) - (x + 4) \cdot 16 i

Expand

- (x + 4) \cdot 16 : - 16 x - 64

= (x + 4) (x^{2} - 12) + (- 16 x - 64) i

\frac{3 ( 2 - x ) ^{2}}{2}

\frac{4 ( a - 1 )}{4 a - 3}

(2 + 7) 3

\frac{( a + 1 ) ^{2}}{( a + 3 ) ( a + 5 )}

\frac{\partial}{\partial t} ((y) (2 y + e^{2 t}))

What is (x+4)(x+4+2i)(x-4-2i) ?
The solution to (x+4)(x+4+2i)(x-4-2i) is (x+4)(x^2-12)+(-16x-64)i