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

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Answer 1 · 2016-12-04T15:04:54

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

Let's do the long division

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

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

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

$a a a a a a a a$ $color(white)(aaaaaaaa)$ $- x^{2} + 2 x$ $-x^2+2x$

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

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

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

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

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

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

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

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

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

Then, $f (2) = 2 \cdot 2^{3} - 5 \cdot 2^{2} + 4 \cdot 2 - 4$ $f(2)=2*2^3-5*2^2+4*2-4$

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

The remainder is $= 0$ $=0$

How do you long divide ( (2x^3) - (5x^2) + (4x) - (4) ) div(x - 2) ((2x3)−(5x2)+(4x)−(4))÷(x−2)( (2x^3) - (5x^2) + (4x) - (4) ) div(x - 2) ?