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

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

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

simplify

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

- 4 - 8 x i

Solution steps

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

(2 x - i)^{2} = 4 x^{2} - 4 x i - 1

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

= 4 x^{2} - 4 x i - 1 - ((4 x^{2} + 3) + 4 x i)

Simplify

(4 x^{2} + 3) + 4 x i : 4 x^{2} + 3 + 4 x i

= 4 x^{2} - 4 x i - 1 - (4 x^{2} + 3 + 4 x i)

Expand

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

= 4 x^{2} - 4 x i - 1 - 4 x^{2} - 3 - 4 x i

Group like terms

= 4 x^{2} - 4 x^{2} - 4 x i - 4 x i - 1 - 3

Add similar elements:

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

= - 4 x i - 4 x i - 1 - 3

Add similar elements:

- 4 x i - 4 x i = - 8 x i

= - 8 x i - 1 - 3

Subtract the numbers:

- 1 - 3 = - 4

= - 8 x i - 4

Rewrite in standard complex form:

- 4 - 8 x i

= - 4 - 8 x i

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

x (x - 1) (x - 2) (x - 3) (x - 4)

(4 p^{2})^{3}

3 \cdot \frac{3}{2}

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

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