How do you divide $(6x^4 + 2x^3 – 6x^2 – 14x – 1 )/( 3x + 1)$ ?

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Answer 1 · 2017-03-02T08:10:05

The remainder is $=3$
The quotient is $=2x^3-2x-4$

We perform a polynomial long division

$color(white)(aaaa)$ $6x^4+2x^3-6x^2-14x-1$ $color(white)(aaaa)$ $|$ $3x+1$

$color(white)(aaaa)$ $6x^4+2x^3$ $color(white)(aaaaaaaaaaaaaaaaaa)$ $|$ $2x^3-2x-4$

$color(white)(aaaaaaa)$ $0+0-6x^2-14x$

$color(white)(aaaaaaaaaaaa)$ $-6x^2-2x$

$color(white)(aaaaaaaaaaaaaaa)$ $0-12x-1$

$color(white)(aaaaaaaaaaaaaaaaa)$ $-12x-4$

$color(white)(aaaaaaaaaaaaaaaaaaa)$ $-0+3$

The remainder is $=3$

The quotient is $=2x^3-2x-4$

How do you divide (6x^4 + 2x^3 – 6x^2 – 14x – 1 )/( 3x + 1) (6x^4 + 2x^3 – 6x^2 – 14x – 1 )/( 3x + 1) ?