What is the remainder when $(x^{3} - 2 x^{2} + 5 x - 6) \div (x - 3)$ $(x^3 - 2x^2 + 5x - 6) div(x - 3)$ ?

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Answer 1 · 2018-06-12T06:39:15

The remainder is $= 18$ $=18$

Apply the remainder theorem :

When the polynomial $f (x)$ $f(x)$ is divided by $(x - c)$ $(x-c)$ , then

$f (x) = (x - c) q (x) + r (x)$ $f(x)=(x-c)q(x)+r(x)$

And when $x = c$ $x=c$

$f (c) = 0 \cdot q (x) + r = r$ $f(c)=0*q(x)+r=r$

where $r$ $r$ is the remainder

Here,

$f (x) = x^{3} - 2 x^{2} + 5 x - 6$ $f(x)=x^3-2x^2+5x-6$

and

$c = 3$ $c=3$

Therefore,

$f (3) = 27 - 18 + 15 - 6 = 18$ $f(3)=27-18+15-6=18$

The remainder is $= 18$ $=18$

What is the remainder when (x^3 - 2x^2 + 5x - 6) div(x - 3)(x3−2x2+5x−6)÷(x−3)(x^3 - 2x^2 + 5x - 6) div(x - 3)?