(A) Let $p(x) = x^{3}-3x^{2}+ax+24$. Since $(x-2)$ is a factor of $p(x)$,by the Factor Theorem,$p(2) = 0$. Substituting $x = 2$ into the polynomial: $

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(A) Let $p(x) = x^{3}-3x^{2}+ax+24$. Since $(x-2)$ is a factor of $p(x)$,by the Factor Theorem,$p(2) = 0$. Substituting $x = 2$ into the polynomial: $p(2) = (2)^{3} - 3(2)^{2} + a(2) + 24 = 0$. $8 - 3(4) + 2a + 24 = 0$. $8 - 12 + 2a + 24 = 0$. $20 + 2a = 0$. $2a = -20$. $a = -10$.

Class 9 Mathematics Polynomials Factorisation of Polynomials

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If $x-2$ is a factor of $x^{3}-3x^{2}+ax+24$,then $a = \ldots$

A
$-10$
B
$10$
C
$5$
D
$-5$

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If $x-2$ is a factor of $x^{3}-3x^{2}+ax+24$,then $a = \ldots$

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Find the quotient and the remainder when $2x^2 - 7x - 15$ is divided by $2x + 3$.

Dividing $p(x) = x^{2} + 12x + 36$ by $(x + 5)$,the remainder is.......

Factorise the following quadratic polynomial by splitting the middle term:$6x^2 + 19x + 10$

If $(x-1)$ is a factor of $x^{3}+7x^{2}+ax-3$,then find the value of $a$.

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Factorise the following quadratic polynomial by splitting the middle term:
$6x^2 + 19x + 10$