Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.
$x^{3} - 3 x + 1, x^{5} - 4 x^{3} + x^{2} + 3 x + 1$

Yes,

x^{3} - 3 x + 1

is factor.

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No,

x^{3} - 3 x + 1

is not a factor.

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Ambiguous

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Data insufficient

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Solution

The correct option is D No, $x^{3} - 3 x + 1$ is not a factor.
Let's use division method,

$x^{3} - 3 x + 1) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ x^{5} - 4 x^{3} + x^{2} + 3 x + 1$ ( $x^{2} - 1$
${x -}^{5} - + 3 x^{3} + - x^{2} - ------------ -$
$- x^{3} + 3 x + 1$
$- + x^{3} + - 3 x - + 1 - ------------ -$
$2$

Since, the remainder is non-zero, therefore $x^{3} - 3 x + 1$ is not a factor of $x^{5} - 4 x^{3} + x^{2} + 3 x + 1$ .

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Similar questions

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

x^{3} - 3 x + 1, x^{5} - 4 x^{3} + x^{2} + 3 x + 1

Question 2 (iii)
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

$x^{3} - 3 x + 1, x^{5} - 4 x^{3} + x^{2} + 3 x + 1$

g (t) = t^{2} - 3, f (t) = 2 t^{4} + 3 t^{3} - 2 t^{2} - 9 t - 12

g (x) = x^{3} - 3 x + 1, f (x) = x^{5} - 4 x^{3} + x^{2} + 3 x + 1

g (x) = 2 x^{2} - x + 3, f (x) = 6 x^{5} - x^{4} + 4 x^{3} - 5 x^{2} - x - 15

x - 3, x^{3} + 4 x^{2} - 3 x + 5

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