What is (2+i)^3-3(2+i)^2+4(2+i)-8 ?

The solution to (2+i)^3-3(2+i)^2+4(2+i)-8 is -7+3i

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simplify (2+i)^3-3(2+i)^2+4(2+i)-8

simplify

(2 + i)^{3} - 3 (2 + i)^{2} + 4 (2 + i) - 8

- 7 + 3 i

Solution steps

(2 + i)^{3} - 3 (2 + i)^{2} + 4 (2 + i) - 8

(2 + i)^{3} = 11 i + 2

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

= 11 i + 2 - 3 (4 i + 3) + 4 (2 + i) - 8

Expand

- 3 (4 i + 3) : - 12 i - 9

= 11 i + 2 - 12 i - 9 + 4 (2 + i) - 8

Expand

4 (2 + i) : 8 + 4 i

= 11 i + 2 - 12 i - 9 + 8 + 4 i - 8

Group like terms

= 11 i - 12 i + 4 i + 2 - 9 + 8 - 8

Add similar elements:

11 i - 12 i + 4 i = 3 i

= 3 i + 2 - 9 + 8 - 8

Add the numbers:

2 - 9 + 8 - 8 = - 7

= 3 i - 7

Rewrite in standard complex form:

- 7 + 3 i

= - 7 + 3 i

∣ x ∣ = - 15

14 x^{2} - 5 x - 6

(y + 5) (y - 1) = 27

x^{2} - 5 x + 20 = 0

- 4 x^{2} + 56 x + 0 = 0

What is (2+i)^3-3(2+i)^2+4(2+i)-8 ?
The solution to (2+i)^3-3(2+i)^2+4(2+i)-8 is -7+3i