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Zeroes of a Polynomial

On dividing ...

Question

On dividing $x^{4} - x^{3} - 3 x^{2} + 3 x + 2$ by polynomial $g (x)$ , the quotient and the remainder are $x^{2} - x - 2$ and $2 x$ respectively. Find $g (x) = x^{2} - m$ .Find $m$

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Solution

We are given the following:

Dividend $= x^{4} - x^{3} - 3 x^{2} + 3 x + 2$
Divisor $= x^{2} - m$
Quotient $= x^{2} - x - 2$
Remainder $= 2 x$

We know that Dividend=Divisor $\times$ Quotient $+$ Remainder
Now, substituting the respective values, we get:

$x^{4} - x^{3} - 3 x^{2} + 3 x + 2 = (x^{2} - m) (x^{2} - x - 2) + 2 x \Rightarrow x^{4} - x^{3} - 3 x^{2} + 3 x + 2 = [x^{2} (x^{2} - x - 2) - m (x^{2} - x - 2)] + 2 x \Rightarrow x^{4} - x^{3} - 3 x^{2} + 3 x + 2 = x^{4} - x^{3} - 2 x^{2} - m x^{2} + m x + 2 m + 2 x \Rightarrow x^{4} - x^{3} - 3 x^{2} + 3 x + 2 = x^{4} - x^{3} - (2 + m) x^{2} + (2 + m) x + 2 m$

Now, comparing the coefficients, we get

$2 + m = 3$ or $2 m = 2$

$⟹ m = 1$

Hence, $m = 1$ .

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