What is (4x-3)(x-2+i)(x+2-i) ?

The solution to (4x-3)(x-2+i)(x+2-i) is (4x-3)(x^2-3)+(16x-12)i

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simplify (4x-3)(x-2+i)(x+2-i)

simplify

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

(4 x - 3) (x^{2} - 3) + (16 x - 12) i

Solution steps

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

Rewrite

x + 2 - i

in standard complex form:

(x + 2) - i

= (4 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

= (4 x - 3) (((x - 2) (x + 2) - 1 \cdot (- 1)) + ((x - 2) (- 1) + 1 \cdot (x + 2)) i)

Simplify

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

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

Distribute parentheses

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

Expand

(4 x - 3) \cdot 4 : 16 x - 12

= (4 x - 3) (x^{2} - 3) + (16 x - 12) i

20 + 2 (x - 3 = 8)

\frac{( t + 1 ) ^{2}}{3 ( 2 t ^{2} - 1 ) ^{3}}

\frac{( ( x + h ) ^{2} - 5 ) - ( x ^{2} - 5 )}{h}

\frac{8 n}{( n + 4 ) ( n + 5 )}

\frac{- 5 ^{3} \cdot 2 ^{8}}{1 0 ^{6}}

What is (4x-3)(x-2+i)(x+2-i) ?
The solution to (4x-3)(x-2+i)(x+2-i) is (4x-3)(x^2-3)+(16x-12)i