On dividing $x^3 - 3x^2 + x + 2$ by a polynomial $g(x)$, the quotient and remainder were $x - 2$ and $-2x + 4$, respectively. Find $g(x)$.
Given:
$x^3 - 3x^2 + x + 2$ is divided by $g(x)$.
The quotient and remainder were $x - 2$ and $-2x + 4$, respectively.
To do:
We have to find $g(x)$.
Solution:
Let $p(x) = x^3 – 3x^2 + x + 2$
Quotient $= x – 2$
Remainder $= -2x + 4$
On dividing $p(x)$ by $g(x)$, we have,
$p(x) = g(x) \times$ quotient $+$ remainder
$x^3– 3x^2 + x + 2 = g(x) (x – 2) + (-2x + 4)$
$x^3 – 3x^2 + x + 2 + 2 x- 4 = g(x) \times (x-2)$
$x^3 – 3x^2 + 3x – 2 = g(x) (x – 2)$
Therefore,
$g(x)=\frac{x^3 – 3x^2 + 3x – 2}{(x – 2)}$
$x-2$)$x^3-3x^2+3x-2$($x^2-x+1$
$x^3-2x^2$
------------------------
$-x^2+3x-2$
$-x^2+2x$
-------------------
$x-2$
$x-2$
------------
$0$
Therefore, $g(x)=x^2-x+1$
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