How do you use the remainder theorem to determine the remainder when the polynomial $(x^{3} + 2 x^{2} - 3 x + 5) \div (x - 3)$ $(x^3+2x^2-3x+5) div(x-3)$ ?

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Answer 1 · 2017-03-03T12:48:17

The remainder is $= 41$ $=41$

When we divide the polynomial $f (x)$ $f(x)$ by $(x - c)$ $(x-c)$

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

and

$f (c) = 0 + r = r$ $f(c)=0+r=r$

Here,

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

and

$c = 3$ $c=3$

So,

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

$= 27 + 18 - 9 + 5$ $=27+18-9+5$

$= 41$ $=41$

How do you use the remainder theorem to determine the remainder when the polynomial (x^3+2x^2-3x+5) div(x-3)(x3+2x2−3x+5)÷(x−3)(x^3+2x^2-3x+5) div(x-3)?