What is (x-3)(x-4-4i)(x+4+4i) ?

The solution to (x-3)(x-4-4i)(x+4+4i) is (x-3)x^2+(-32x+96)i

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simplify (x-3)(x-4-4i)(x+4+4i)

simplify

(x - 3) (x - 4 - 4 i) (x + 4 + 4 i)

(x - 3) x^{2} + (- 32 x + 96) i

Solution steps

(x - 3) (x - 4 - 4 i) (x + 4 + 4 i)

Rewrite

x - 4 - 4 i

in standard complex form:

(x - 4) - 4 i

Rewrite

x + 4 + 4 i

in standard complex form:

(x + 4) + 4 i

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

Apply complex arithmetic rule:

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

= (x - 3) (((x - 4) (x + 4) - (- 4) \cdot 4) + ((x - 4) \cdot 4 - 4 (x + 4)) i)

Simplify

((x - 4) (x + 4) - (- 4) \cdot 4) + ((x - 4) \cdot 4 - 4 (x + 4)) i : x^{2} - 32 i

= (x - 3) (x^{2} - 32 i)

Distribute parentheses

= (x - 3) x^{2} - (x - 3) \cdot 32 i

Expand

- (x - 3) \cdot 32 : - 32 x + 96

= (x - 3) x^{2} + (- 32 x + 96) i

(e^{x} + e^{- x})^{4}

y^{2} cos (\frac{1}{2}) (x - \frac{π}{2}) + 2

(x^{2} + y^{2}) e^{- x^{2} - 4 y^{2}}

(2 + x) e^{x}

(x + 5 + 2 i) (x + 5 - 2 i) (x + 1 - 3 i) (x + 1 + 3 i)

What is (x-3)(x-4-4i)(x+4+4i) ?
The solution to (x-3)(x-4-4i)(x+4+4i) is (x-3)x^2+(-32x+96)i