How do you divide $(4x^3 - 8x^2 + 7x - 9) ÷ (x + 3)$ ?

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Answer 1 · 2016-11-23T07:51:10

The quotient is $=(4x^2-20x+67)$ and the remainder is $=-210$

You can do a long division

$color(white)(aaaa)$ $4x^3-8x^2+7x-9$ $∣$ $x+3$

$color(white)(aaaa)$ $4x^3+12x^2$ $color(white)(aaaaaaa)$ $∣$ $4x^2-20x+67$

$color(white)(aaaaaa)$ $0-20x^2+7x$

$color(white)(aaaaaaaa)$ $-20x^2-60x$

$color(white)(aaaaaaaaaaa)$ $+0+67x-9$

$color(white)(aaaaaaaaaaaaaaa)$ $+67x+201$

$color(white)(aaaaaaaaaaaaaaaaaaa)$ $0-210$

The quotient is $=(4x^2-20x+67)$

and the remainder is $=-210$

You can get the remainder directly with the remainder theorem

Let $f(x)=4x^3-8x^2+7x-9$

Then, $f(-3)=4*(-3)^3-8(-3)^2+7(-3)-9$

$=-108-72-21-9=-210$

This is the remainder.

How do you divide (4x^3 - 8x^2 + 7x - 9) ÷ (x + 3)(4x^3 - 8x^2 + 7x - 9) ÷ (x + 3)?