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

The solution to (x-3)(x+4+i)(x-4-i) is (x-3)(x^2-15)+(-8x+24)i

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

simplify

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

(x - 3) (x^{2} - 15) + (- 8 x + 24) i

Solution steps

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

Rewrite

x - 4 - i

in standard complex form:

(x - 4) - i

= (x - 3) (x + 4 + i) ((x - 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) - 1 \cdot (- 1)) + ((x + 4) (- 1) + 1 \cdot (x - 4)) i)

Simplify

((x + 4) (x - 4) - 1 \cdot (- 1)) + ((x + 4) (- 1) + 1 \cdot (x - 4)) i : (x^{2} - 15) - 8 i

= (x - 3) ((x^{2} - 15) - 8 i)

Distribute parentheses

= (x - 3) (x^{2} - 15) - (x - 3) \cdot 8 i

Expand

- (x - 3) \cdot 8 : - 8 x + 24

= (x - 3) (x^{2} - 15) + (- 8 x + 24) i

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

(3 - 23 x)^{2}

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

(x - 16 + 72) (x - 16 - 72)

cos (t) U (t - \frac{π}{2})

What is (x-3)(x+4+i)(x-4-i) ?
The solution to (x-3)(x+4+i)(x-4-i) is (x-3)(x^2-15)+(-8x+24)i