Question

Check whether p(x) is a multiple of g(x) or not where p(x)=x3-x+1 g(x)=2-3x

Answer 1

Hola Mate !!

Here's your answer :-

$g(x) = 0 \\ = > 2 - 3x = 0 \\ = > - 3x = - 2 \\ = > x = \frac{ - 2}{ - 3} = \frac{2}{3}$

$p(x) = {x}^{3} - x + 1 \\ = > p( \frac{2}{3} ) = {( \frac{2}{3})}^{3} - \frac{2}{3} + 1 \\ = \frac{8}{27} - \frac{2}{3} + 1 \\ = \frac{8 - 18 + 27}{27} \\ = \frac{35 - 18}{27} \\ = \frac{17}{27}$

$as \: p(x) \: is \: not \: equal \: to \: zero \: \\ so \: p(x) \: is \: not \: a \: multiple \: of \: g(x).$

Hope it helps

Answer 2

$\bf p(x) = {x}^{3} - x + 1$ is not a multiple of $\bf g(x) = 2 - 3x$ or g(x) is not a factor of p(x).

Given:

• A polynomial
• $p(x) = {x}^{3} - x + 1$
• $g(x) = 2 - 3x$

To find:

• p(x) is a multiple of g(x) or not.

Solution:

Concept to be used: Apply long division or factor theorem.

Perform long division.

$\: \: \: \: \: \:- 3x + 2 \: ) \: {x}^{3} - x + 1 \: ( - \frac{ {x}^{2} }{3} - \frac{2x}{9} + \frac{5}{27} \\ \: \: \: \: \: \: {x}^{3} \: \: \: - \frac{2 {x}^{2} }{3} \: \: \: \: \: \: \\ ( - ) \: \: \: ( + ) \\ - - - - - \\ \frac{2 {x}^{2} }{3} - x + 1 \\\\ \frac{2 {x}^{2} }{3} - \frac{4x}{9} \\ ( - ) \: \: ( + ) \\ - - - - - - \\ - \frac{5x}{9} + 1 \\ - \frac{5x}{9} + \frac{10}{27} \\ ( + ) \: \: ( - ) \\ - - - - - \\ \frac{17}{27} \\ - - - - - -$

g(x) cannot divides p(x) completely as remainder is present.

Thus,

p(x) is not a multiple of g(x) or g(x) is not a factor of p(x).

Learn more:

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