1. ## Factorising this polynomial?

How to divide this polynomial, $f(x) = x^3 - 2x^2 + x - 2$ with x -1 by using long division? I got stuck at $-x^2 + x$ because after that, the remainder is 0 but there is still the -2 remaining. However, the solution is $x^2 + 1$. How?

2. Originally Posted by mark1950
How to divide this polynomial, $f(x) = x^3 - 2x^2 + x - 2$ with x -1 by using long division? I got stuck at $-x^2 + x$ because after that, the remainder is 0 but there is still the -2 remaining. However, the solution is $x^2 + 1$. How?
That's because the solution is incorrect. Dividing $x^3 - 2x^2 + x - 2$ by $x - 1$ gives you a quotient of $x^2 - x$ with a remainder of -2.

Maybe you meant that you want to divide $x^3 - 2x^2 + x - 2$ by $x - {\color{red}2}$? Because if you do that, you do get a quotient of $x^2 + 1$. You sure you copied the problem down correctly?

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3. Originally Posted by mark1950
How to divide this polynomial, $f(x) = x^3 - 2x^2 + x - 2$ with x -1 by using long division? I got stuck at $-x^2 + x$ because after that, the remainder is 0 but there is still the -2 remaining. However, the solution is $x^2 + 1$. How?
Hello : This the
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4. Originally Posted by mark1950
How to divide this polynomial, $f(x) = x^3 - 2x^2 + x - 2$ with x -1 by using long division? I got stuck at $-x^2 + x$ because after that, the remainder is 0 but there is still the -2 remaining. However, the solution is $x^2 + 1$. How?

Use the rational zeros theorem to determine possible roots of the function.
When you have them, make use of the fact that if x=c is a root of f(x), then x-c is a factor of the polynomial function f(x). when you have found all factors x=c, then divide.

Possible zeros. 1,-1,2,-2.