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# Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial. (i) $$t^2 - 3 , 2t^4 + 3t^3 -2t^2 - 9t -12$$ (ii) $$x^2 + 3x + 1 , 3x^4 + 5x^3 - 7x^2 + 2x + 2$$ (iii) $$x^2 - 3x + 1 , x^5 - 4x^3 + x^2 + 3x + 1$$

Answer :

(i) Solving it using long division:
$$\begin{array}{rrrr|ll} 3t^3 + 2t^4 & - 2t^2 & - 9t & -12 & t^2 - 3 \\-2t^4 & +6t^2 & & & 2t^2 + 3t + 4 \\ \hline 3t^3 & + 4t^2 & -9t & -12\\ \phantom{-}-3t^3 & & + 9t & & & \\ \hline & 4t^2 & & -12 \\ & -4t^2 & & +12 \\ \hline & & & 0\\ \hline \end{array}$$
Here, Remainder = 0 so the first polynomial $$t^2 - 3$$ is a factor of second polynomial $$2t^4 + 3t^3 -2t^2 - 9t -12$$.

(ii) Solving it using long division:
$$\begin{array}{rrrr|ll} 3x^4 + 5x^3 & - 7x^2 & + 2x & +2 & x^2 + 3x + 1 \\-3x^4 - 9x^3 & -3x^2 & & & 3x^2 - 4x + 2 \\ \hline -4x^3 & - 10x^2 & +2x & +2\\ \phantom{-}+4x^3 & +12x^2 & + 4x & & & \\ \hline & 2x^2 & +6x & +2 \\ & -2x^2 & -6x & -2 \\ \hline & & & 0\\ \hline \end{array}$$
Here, Remainder = 0 so the first polynomial $$x^2 + 3x + 1$$ is a factor of second polynomial $$3x^4 + 5x^3 - 7x^2 + 2x + 2$$.

(iii) Solving it using long division:
$$\begin{array}{rrrr|ll} x^5 - 4x^3 & + x^2 & + 3x & +1 & x^3 - 3x + 1\\ -x^5 + 3x^3 & -x^2 & & & x^2 - 1 \\ \hline & -x^3 & +3x & + 1\\ & \phantom{-} +x^3 & -3x & + 1 & & & & \\ \hline & & & 2 \\ \hline \end{array}$$
Here, Remainder = 2 so the first polynomial $$x^2 - 3x + 1$$ is not a factor of second polynomial $$x^5 - 4x^3 + x^2 + 3x + 1$$.