What is (x-2)(x-1+i)(x+1-i) ?

The solution to (x-2)(x-1+i)(x+1-i) is (x-2)x^2+(2x-4)i

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simplify (x-2)(x-1+i)(x+1-i)

simplify

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

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

Solution steps

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

Rewrite

x + 1 - i

in standard complex form:

(x + 1) - i

= (x - 2) (x - 1 + i) ((x + 1) - i)

Apply complex arithmetic rule:

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

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

Simplify

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

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

Distribute parentheses

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

Expand

(x - 2) \cdot 2 : 2 x - 4

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

\frac{1}{( 1 + j w ) ( 3 + j w )}

\frac{x + 10}{( x - 2 ) ( x + 2 )}

x^{\frac{1}{4}} (2 x - 2)

(x - a) (x + a)

(x - \frac{a + b}{2})^{2}

What is (x-2)(x-1+i)(x+1-i) ?
The solution to (x-2)(x-1+i)(x+1-i) is (x-2)x^2+(2x-4)i