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How do you multiply $(2 x^{2} - 5 x - 1) (x^{2} + x - 3)$ $(2x^2-5x-1)(x^2+x-3)$ ?

1 Answer

Aug 5, 2015

$(2 x^{2} - 5 x - 1) (x^{2} + x - 3) = 2 x^{4} - 8 x^{3} - 12 x^{2} + 14 x + 3$ $(2x^2-5x-1)(x^2+x-3)=2x^4-8x^3-12x^2+14x+3$

Explanation:

$(p) \cdot (x^{2} + x - 3) = (p x^{2} + p x - 3 p)$ $(p)*(x^2+x-3) = (px^2+px-3p)$

Replacing $(p)$ $(p)$ with $(2 x^{2} - 5 x - 1)$ $(2x^2-5x-1)$

$(2 x^{2} - 5 x - 1) (x^{2} + x - 3)$ $(2x^2-5x-1)(x^2+x-3)$
$XXXX$ $color(white)("XXXX")$ $= (2 x^{2} - 5 x - 1) x^{2} + (2 x^{2} - 5 x - 1) x - 3 (2 x^{2} - 5 x - 1)$ $=(2x^2-5x-1)x^2 +(2x^2-5x-1)x - 3(2x^2-5x-1)$

$XXXX$ $color(white)("XXXX")$ $= (2 x^{4} - 10 x^{3} - x^{2}) + (2 x^{3} - 5 x^{2} - x) - (6 x^{2} - 15 x - 3)$ $=(2x^4-10x^3-x^2)+(2x^3-5x^2-x)-(6x^2-15x-3)$

$XXXX$ $color(white)("XXXX")$ $= (2 x^{4}) + (- 10 x^{3} + 2 x^{3}) + (- x^{2} - 5 x^{2} - 6 x^{2}) + (- x + 15 x) + (3)$ $=(2x^4) + (-10x^3+2x^3) + (-x^2-5x^2-6x^2) + (-x+15x) + (3)$

$XXXX$ $color(white)("XXXX")$ $= 2 x^{4} - 8 x^{3} - 12 x^{2} + 14 x + 3$ $=2x^4-8x^3-12x^2+14x+3$

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