How do you use the remainder theorem to find the remainder for each division $(x^{2} - 2) \div (x - 1)$ $(x^2-2)div(x-1)$ ?

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Answer 1 · 2017-10-11T04:43:54

Quotient $(x + 1)$ $(x+1)$ & Remainder $3$ $3$

          x          1
      ………………………………………………………

x-1 | x^2 0 2
| x^2 -x
|__
0 x 2
x -1
_
0 3 (Remainder)

Answer 2 · 2017-10-11T07:03:06

The remainder is $- 1$ $-1$ . See explanation.

The remainder theorem says that:

For a polynomial $P (x)$ $P(x)$ the remainder of division $P (x) \div (x - a)$ $P(x)-:(x-a)$ is equal to $P (a)$ $P(a)$ .

So instead of using the long division to calculate the remainder only you can substitute $a$ $a$ to $P (x)$ $P(x)$ . Here we have:

$a = 1$ $a=1$ , $P (x) = x^{2} - 2$ $P(x)=x^2-2$ , so:

$P (1) = 1^{2} - 2 = 1 - 2 = - 1$ $P(1)=1^2-2=1-2=-1$

The remainder of dividision $(x^{2} - 2) \div (x - 1)$ $(x^2-2)-:(x-1)$ is $- 1$ $-1$

How do you use the remainder theorem to find the remainder for each division (x2−2)÷(x−1)(x^2-2)div(x-1)?