Find $\alpha$ and $\beta$, if $x + 1$ and $x + 2$ are factors of $x^3 + 3x^2 - 2 \alpha x + \beta$.

Given:

Given expression is $x^3 + 3x^2 - 2 \alpha x + \beta$.

$x + 1$ and $x + 2$ are factors of $x^3 + 3x^2 - 2 \alpha x + \beta$.

To do:

We have to find the values of $\alpha$ and $\beta$.

Solution:

We know that,

If $(x-m)$ is a root of $f(x)$ then $f(m)=0$.

$x+1$ and $x+2$ are factors of $x^3 + 3x^2 - 2 \alpha x + \beta$.

Therefore,

$f(-1)=0$

$\Rightarrow (-1)^3 + 3(-1)^2 - 2\alpha(-1) + \beta=0$

$\Rightarrow -1+3+2\alpha+\beta=0$

$\Rightarrow \beta=-2\alpha-2$........(i)

$f(-2)=0$

$\Rightarrow (-2)^3 + 3(-2)^2 - 2\alpha(-2) + \beta=0$

$\Rightarrow 4\alpha + \beta-8+3(4)=0$

$\Rightarrow 4\alpha -2\alpha-2-8+12=0$           [From (i)]

$\Rightarrow 2\alpha-10+12=0$

$\Rightarrow 2\alpha=-2$

$\Rightarrow \alpha=-1$

$\Rightarrow \beta=-2(-1)-2$

$\Rightarrow \beta=2-2=0$

The values of $\alpha$ and $\beta$ are $-1$ and $0$ respectively.       

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Updated on: 10-Oct-2022

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