Find the remainder when x^5 - 3x^3 + x^2 - 3 is divided by a) x - 2 b) x + 1?

Question

I got 3 and -6

Could someone show the workings so I can figure out where i went wrong

Answer 1

There is a theorem related to polynomial divisions called the remainder theorem. Let p(x) be a polynomial of any degree. When it is divided by a linear polynomial of the form x - a, the remainder is p(a).
Answer 1:
Let p(x) = x^5 - 3x^3 + x^2 - 3
Divisor = x - 2
Here, a = 2
p(2) = 2^5 - 3(2)^3 + 2^2 - 3
= 32 - 24 + 4 - 3
= 32 - 24 + 1
= 8 + 1
= 9

Answer 2:
p(x) = x^5 - 3x^3 + x^2 - 3
Divisor = x + 1
a = -1
p(-1) = (-1)^5 -3(-1)^3 + (-1)^2 - 3
= -1 + 3 + 1 - 3
= 0

You can check with actual division if you like. Looks like some of us have it wrong.

Answer 2

Im getting a)9 and b)0.

There is an easier methor to find remainders than actually doing the division:
You assume that the denominator is a root and shove the root in the function. i.e. for the first question, you say that the x=2 is a root for the function (since you are assuming x-2 = 0). Now substitute all the x's in the question with 2. Since 2^5-3(2)^3+(2)^2-3 = 9, 9 is the remainder of the first question. Similarly, the root of the second question in x=-1. Accordingly, the remaider comes out be 0 which means that x=-1 is actually a root of the function.

i rechecked my answers by doing the actual division and my answers match.

Answer 3

3 and -6