What must be added to the polynomial $p (x) = x^{4} + 2 x^{3} - 2 x^{2} + x - 1$ so that the resulting polynomial is exactly divisible by $x^{2} + 2 x - 3$

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Solution

We know, $p (x) = [g (x) \times q (x)] + r (x)$

$∴ p (x) - r (x) = g (x) \times q (x)$

$∴ p (x) + {- r (x)} = g (x) \times q (x)$

It is clear that RHS is divisible by $g (x)$ . $∴ L H S$ is also divisible by $g (x)$

Thus, if we add $- r (x)$ to $p (x)$ , then the resulting polynomial is divisible by $g (x)$ .

Let us divide $p (x) = x^{4} + 2 x^{3} - 2 x^{2} + x - 1$ by $g (x) = x^{2} + 2 x - 3$ to find the remainder $r (x)$ .

$∴ r (x) = - x + 2 \Rightarrow {- r (x)} = x - 2$

Hence, we should add $(x - 2)$ to $p (x) = x^{4} + 2 x^{3} - 2 x^{2} + x - 1$ so that the resulting polynomial is exactly divisible by $x^{2} + 2 x - 3$ .

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What must be added to the polynomial p(x)=x4+2x3−2x2+x−1 so that the resulting polynomial is exactly divisible by x2+2x−3

What must be added to the polynomial $p (x) = x^{4} + 2 x^{3} - 2 x^{2} + x - 1$ so that the resulting polynomial is exactly divisible by $x^{2} + 2 x - 3$