The square of the remainder when $9 x^{3} - 3 x^{2} - 5 x - 1$ is divided by $x + 1$ is:

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Solution

To find the remainder, we will put Divisor $= 0$
$\Rightarrow x + 1 = 0 \Rightarrow x = - 1$

$9 x^{3} - 3 x^{2} - 5 x - 1$

$= 9 (- 1)^{3} - 3 (- 1)^{2} - 5 (- 1) - 1 = - 9 - 3 + 5 - 1 = - 8$

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9 x^{3} - 3 x^{2} + x - 5

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The remainder when x⁴ – 3x² + 5x – 7 is divided by x + 1 is:

Statement1: The polynomial $P (x) = 4 x^{3} - 3 x^{2} + 5 x - 6$ when divided by $x - 1$ gives zero as the remainder.

Statement2: $(x - 1)$ is a factor of the polynomial $P (x) = 4 x^{3} - 3 x^{2} + 5 x - 6$ .

p (x) = - x + 3 x^{2} - 1

x + 1

The remainder of the polynomial P(x) = 3 $x^{2}$ + 5x + 8, when divided by x + 2 is___.

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