How do you divide $(3x ^ { 3} + 12x ^ { 2} - 21x - 23) \div ( x + 5)$ ?

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Answer 1 · 2017-07-05T20:13:27

$3x^2-3x-6$ $r$ $7$

Firstly, you divide $3x^3$ by $x$ to get $3x^2$ . Then you do $3x^2(x+5)=3x^2+15x^2$ .

$3x^3+12x^2-3x^3-15x^2=-3x^2$ , the first value of the quptient is $3x^2$ .

Now you do $(-3x^2)/x=-3x$ . Then you do $-3x(x+5)=-3x^2-15x$ .
$(3x^2-21x)-(-3x^2-15x)=-21x+15x=-6x$ . $-3x$ is the second part of the quotient.

Now you do $(-6x)/x=-6$ . Then you do $-6(x+5)=-6x-30$ .
$(-6x-23)-(-6x-30)=-23+30=7$ . $-6$ is the last part of the quotient. $7$ is the remainder.

Visual guide:

$" "3x^2-3x-6$ $r$ $7$
$(x+5)"/"(3x^3+12x^2-21x-23)$
$" "-(3x^3+15x^2)$
$" "(0-3x^2)$
$" "-(-3x^2-15x)$
$" "(0-6x)$
$" "-(-6x-30)$
$" "(0+7)$

How do you divide (3x ^ { 3} + 12x ^ { 2} - 21x - 23) \div ( x + 5)(3x ^ { 3} + 12x ^ { 2} - 21x - 23) \div ( x + 5)?