If x^2+x-6 is a factor of 2x^3+x^2+ax+b,then | vedclass

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(C) Given that $x^2+x-6$ is a factor of $P(x) = 2x^3+x^2+ax+b$. Factorizing the divisor: $x^2+x-6 = (x+3)(x-2)$. Since $(x+3)$ and $(x-2)$ are factors

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(C) Given that $x^2+x-6$ is a factor of $P(x) = 2x^3+x^2+ax+b$. Factorizing the divisor: $x^2+x-6 = (x+3)(x-2)$. Since $(x+3)$ and $(x-2)$ are factors,$P(-3) = 0$ and $P(2) = 0$. For $x = -3$: $2(-3)^3 + (-3)^2 + a(-3) + b = 0$ $\Rightarrow -54 + 9 - 3a + b = 0$ $\Rightarrow -3a + b = 45$ ... $(i)$. For $x = 2$: $2(2)^3 + (2)^2 + a(2) + b = 0$ $\Rightarrow 16 + 4 + 2a + b = 0$ $\Rightarrow 2a + b = -20$ ... $(ii)$. Subtracting $(ii)$ from $(i)$: $(-3a - 2a) + (b - b) = 45 - (-20)$ $\Rightarrow -5a = 65$ $\Rightarrow a = -13$. Substituting $a = -13$ into $(ii)$: $2(-13) + b = -20$ $\Rightarrow -26 + b = -20$ $\Rightarrow b = 6$. Finally,$6a + 13b = 6(-13) + 13(6) = -78 + 78 = 0$.

If $x^2+x-6$ is a factor of $2x^3+x^2+ax+b$,then $6a+13b=$

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