What is (x-3)(x-2-i)(x+2+i) ?

The solution to (x-3)(x-2-i)(x+2+i) is (x-3)(x^2-3)+(-4x+12)i

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simplify (x-3)(x-2-i)(x+2+i)

simplify

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

(x - 3) (x^{2} - 3) + (- 4 x + 12) i

Solution steps

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

Rewrite

x - 2 - i

in standard complex form:

(x - 2) - i

= (x - 3) ((x - 2) - i) (x + 2 + i)

Apply complex arithmetic rule:

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

= (x - 3) (((x - 2) (x + 2) - (- 1) \cdot 1) + ((x - 2) \cdot 1 - 1 \cdot (x + 2)) i)

Simplify

((x - 2) (x + 2) - (- 1) \cdot 1) + ((x - 2) \cdot 1 - 1 \cdot (x + 2)) i : (x^{2} - 3) - 4 i

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

Distribute parentheses

= (x - 3) (x^{2} - 3) - (x - 3) \cdot 4 i

Expand

- (x - 3) \cdot 4 : - 4 x + 12

= (x - 3) (x^{2} - 3) + (- 4 x + 12) i

(ln (x))^{s e c (x)}

(3 - cos (x))^{2}

(x - 3 + 1)^{2}

\frac{( 4 x + 9 ) ^{\frac{3}{2}} - 27}{6}

(cos (t) - 4 sin (t))^{2}

What is (x-3)(x-2-i)(x+2+i) ?
The solution to (x-3)(x-2-i)(x+2+i) is (x-3)(x^2-3)+(-4x+12)i