Find the remainder on dividing the polynomial $(x^{3} - 3 x^{2} + 5 x + 7)$ by $(x - 3)$ . Is $(x - 3)$ a factor of this polynomial? Why?

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Solution

$p (x) = x^{3} - 3 x^{2} + 5 x + 7$
Remainder $= p (3) = (3)^{3} - 3 (3)^{2} + 5 (3) + 7 = 27 - 27 + 15 + 7 = 22$
For $(x - 3)$ to be a factor, p(3) must be equal to zero.
Here, $p (2) \neq 0$ .
Hence, $(x - 3)$ is not a factor of this polynomial.

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Similar questions

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

g (x) = x^{2} - 2

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p (x) = x^{3} - 3 x^{2} + 5 x - 3

g (x) = x^{2} - 2

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

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x - 2

- 2 x + 4

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x^{3} - 3 x^{2} + x + 1

x - 1

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