How do you multiply $(2 x - 3) (x^{4} - 2 x^{2} + 3)$ $(2x-3)(x^4-2x^2+3)$ ?

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Answer 1 · 2018-05-13T07:06:16

$2 x^{5} - 3 x^{4} - 4 x^{3} + 6 x^{2} + 6 x - 9$ $2x^5-3x^4-4x^3+6x^2+6x-9$

Multiply each term in the second bracket by $2 x$ $2x$ and then multiply each term by -3.

$(2 x - 3) (x^{4} - 2 x^{2} + 3)$ $(2x-3)(x^4-2x^2+3)$

$2 x^{5} - 4 x^{3} + 6 x - 3 x^{4} + 6 x^{2} - 9$ $2x^5-4x^3+6x-3x^4+6x^2-9$

$2 x^{5} - 3 x^{4} - 4 x^{3} + 6 x^{2} + 6 x - 9$ $2x^5-3x^4-4x^3+6x^2+6x-9$

How do you multiply (2x-3)(x^4-2x^2+3)(2x−3)(x4−2x2+3)(2x-3)(x^4-2x^2+3)?