If the expression x^3+3x^2-9x+lambda is of th | vedclass

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(D) Given $x^3+3x^2-9x+\lambda = (x-\alpha)^2(x-\beta) = (x^2-2\alpha x+\alpha^2)(x-\beta) = x^3 - (2\alpha+\beta)x^2 + (2\alpha\beta+\alpha^2)x - \al

Vedclass · Accepted Answer

$-27, 5$

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(D) Given $x^3+3x^2-9x+\lambda = (x-\alpha)^2(x-\beta) = (x^2-2\alpha x+\alpha^2)(x-\beta) = x^3 - (2\alpha+\beta)x^2 + (2\alpha\beta+\alpha^2)x - \alpha^2\beta$. Comparing coefficients: $2\alpha+\beta = -3$ $(i)$ $2\alpha\beta+\alpha^2 = -9$ (ii) $-\alpha^2\beta = \lambda$ (iii) From $(i)$,$\beta = -3-2\alpha$. Substitute into (ii): $2\alpha(-3-2\alpha) + \alpha^2 = -9$ $\Rightarrow -6\alpha - 4\alpha^2 + \alpha^2 = -9$ $\Rightarrow 3\alpha^2 + 6\alpha - 9 = 0$ $\Rightarrow \alpha^2 + 2\alpha - 3 = 0$. Solving for $\alpha$: $(\alpha+3)(\alpha-1) = 0$,so $\alpha = 1$ or $\alpha = -3$. If $\alpha = 1$,then $\beta = -3-2(1) = -5$. From (iii),$\lambda = -\alpha^2\beta = -(1)^2(-5) = 5$. If $\alpha = -3$,then $\beta = -3-2(-3) = 3$. From (iii),$\lambda = -\alpha^2\beta = -(-3)^2(3) = -27$. Thus,the values of $\lambda$ are $-27$ and $5$.

If the expression $x^3+3x^2-9x+\lambda$ is of the form $(x-\alpha)^2(x-\beta)$,then the values of $\lambda$ are

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