How do you multiply $(5x^3-x^2+x-4) (x+1)$ ?

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Answer 1 · 2015-07-04T22:45:40

Use distributivity to find:

$(5x^3-x^2+x-4)(x+1) = 5x^4+4x^3-3x-4$

$(5x^3-x^2+x-4)(x+1)$

$=(5x^3-x^2+x-4)*x + (5x^3-x^2+x-4)*1$

$=5x^4-x^3+x^2-4x+5x^3-x^2+x-4$

$=5x^4-x^3+5x^3+x^2-x^2-4x+x-4$

$=5x^4+(-1+5)x^3+(1-1)x^2+(1-4)x-4$

$=5x^4+4x^3-3x-4$

Alternatively, look at each power of $x$ in descending order and total up the relevant products of the coefficients:

$x^4$ : $5*1 = 5$

$x^3$ : $(5*1) + (-1*1) = 4$

$x^2$ : $(-1*1) + (1*1) = 0$

$x$ : $(1*1) + (-4*1) = -3$

$1$ : $-4*1 = -4$

Hence: $5x^4+4x^3-3x-4$

How do you multiply (5x^3-x^2+x-4) (x+1)(5x^3-x^2+x-4) (x+1)?