How do you long divide $( (2x^3) - (5x^2) + (4x) - (4) ) div(x - 2)$ ?

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Answer 1 · 2016-12-04T15:04:54

The remainder is $=0$ and the quotient is $=2x^2-x+2$

Let's do the long division

$color(white)(aaaa)$ $2x^3-5x^2+4x-4$ $∣$ $x-2$

$color(white)(aaaa)$ $2x^3-4x^2$ $color(white)(aaaaaaaa)$ $∣$ $2x^2-x+2$

$color(white)(aaaaaa)$ $0-x^2+4x$

$color(white)(aaaaaaaa)$ $-x^2+2x$

$color(white)(aaaaaaaaaaa)$ $0+2x-4$

$color(white)(aaaaaaaaaaaaa)$ $+2x-4$

$color(white)(aaaaaaaaaaaaaa)$ $+0-0$

so, the remainder is $0$ and the quotient is $2x^2-x+2$

You can also use the remainder theorem to see that the remainder is $=0$

If $p(x)$ is a polynomial, and $(x-a)$ is a factor of that polynomial

Then, $p(a)=(x-a)q(a)+0$

$q(a)$ is the quotient and $0$ the remainder

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

Then, $f(2)=2*2^3-5*2^2+4*2-4$

$=16-20+8-4=24-24=0$

The remainder is $=0$

How do you long divide ( (2x^3) - (5x^2) + (4x) - (4) ) div(x - 2) ( (2x^3) - (5x^2) + (4x) - (4) ) div(x - 2) ?