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

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

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Solution

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

$\begin{matrix} x^{2} - 1 x^{3} - 3 x + 1 & x^{5} - 4 x^{3} + x^{2} + 3 x + 1 x^{5} - 3 x^{3} + x^{2} - + - ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - x^{3} + 3 x + 1 - x^{3} + 3 x - 1 - ----------------- - 2 \end{matrix}$
Since the remainder $\neq$ 0,

$x^{3} - 3 x + 1$ , couldn't exactly divide $x^{5} - 4 x^{3} + x^{2} + 3 x + 1$ as it leaves 2 as remainder. Hence, it not a factor of $x^{5} - 4 x^{3} + x^{2} + 3 x + 1.$

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Similar questions

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

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

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