How do you use the remainder theorem to find the remainder for each division $(x^{2} + 2 x - 15) \div (x - 3)$ $(x^2+2x-15)div(x-3)$ ?

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Answer 1 · 2016-11-26T18:57:26

The remainder $= 0$ $=0$

When we divide a polynomial $f (x)$ $f(x)$ by $(x - a)$ $(x-a)$

we get, $f (x) = (x - a) q (x) + r$ $f(x)=(x-a)q(x)+r$

If $x = a$ $x=a$

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

So, $f (a) = r$ $f(a)=r$

Here, $f (x) = x^{2} + 2 x - 15$ $f(x)=x^2+2x-15$ is divided by $(x - 3)$ $(x-3)$

$f (3) = 3^{2} + 2 \cdot 3 - 15 = 9 + 6 - 15 = 0$ $f(3)=3^2+2*3-15=9+6-15=0$

Remainder $= 0$ $=0$

$f (x)$ $f(x)$ is divisible by $(x - 3)$ $(x-3)$

$(x^2+2x-15)/(x-3)=(cancel(x-3)(x+5))/cancel(x-3)$

How do you use the remainder theorem to find the remainder for each division (x^2+2x-15)div(x-3)(x2+2x−15)÷(x−3)(x^2+2x-15)div(x-3)?