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

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To do:

We have to check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial in each case.

Solution:

(i) On applying the division algorithm,

Let dividend $f(t)\ =\ 2t^4\ +\ 3t^3\ –\ 2t^2\ –\ 9t\ –\ 12$

Divisor $g(t)\ =\ t^2\ –\ 3$

If $g(t)$ is a factor of $f(t)$ then the remainder on long division should be $0$.

$t^2-3$)$2t^4+3t^3-2t^2-9t-12$($2t^2+3t+4$

$2t^4 -6t^2$

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$3t^3+4t^2-9t-12$

$3t^3 -9t$

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$4t^2-12$

$4t^2-12$

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$0$

Therefore, $g(t)$ is a factor of $f(t)$.

(ii) On applying the division algorithm,

Let dividend $f(x)=3x^4+5x^3-7x^2+2x + 2$

Divisor $g(x) =x^2 + 3x + 1$

If $g(x)$ is a factor of $f(x)$ then the remainder on long division should be $0$.

$x^2+3x+1$)$3x^4+5x^3-7x^2+2x+2$($3x^2-4x+2$

$3x^4+9x^3+3x^2$

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$-4x^3-10x^2+2x+2$

$-4x^3-12x^2-4x$

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$2x^2+6x+2$

$2x^2+6x+2$

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$0$

Therefore, $g(x)$ is a factor of $f(x)$.

(iii) On applying the division algorithm,

Let dividend$f(x)\ =\ x^5\ –\ 4x^3\ +\ x^2\ +\ 3x\ +\ 1$

Divisor$g(x)\ =\ x^3\ –\ 3x\ +\ 1$

If $g(x)$ is a factor of $f(x)$ then the remainder on long division should be $0$.

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

$x^5-3x^3+x^2$

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$-x^3+3x+1$

$-x^3+3x-1$

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$0$

Therefore, $g(x)$ is a factor of $f(x)$.

Updated on 10-Oct-2022 13:19:37