Question

If x+1, x+2 are two factors of xcube + 3xsquare - 2alpha x+ beta ,then the value of (alpha + beta) is​

Answer 1

The required value of α = -2 and β = 2.

Here in the solution I used the remainder theorem.

Refer to the above attachment.

Answer 2

SOLUTION:-

Given:

Two factors are, x+1 & x+2.

Equation: x³ + 3x² - 2alphax + beta.

So,

=) x+1=0 or x+2=0

=) x= -1 or x= -2

$= > {x}^{3} + 3 {x}^{2} - 2 \alpha x + \beta \\ \\ = > f( - 1) = 0 \\ = > ( - 1) {}^{3} + 3( { - 1)}^{2} - 2 \alpha ( - 1) + \beta = 0 \\ \\ = > - 1 + 3 + 2 \alpha + \beta = 0 \\ \\ = > 2 + 2 \alpha + \beta = 0 \\ \\ = > 2 \alpha + \beta = - 2..............(1)$

&

f(-2)=0

$= > {x}^{3} + 3 {x}^{2} - 2 \alpha x + \beta \\ \\ = > ( - 2) {}^{3} + 3( - 2) {}^{2} - 2 \alpha ( - 2) + \beta = 0\\ \\ = > - 8 + 12 + 4 \alpha + \beta = 0 \\ \\ = > 4 + 4 \alpha + \beta = 0 \\ \\ = > 4 \alpha + \beta = - 4..............(2)$

Now,

Subtract equation (1) & (2), we get;

$2 \alpha + \beta = - 2 \\ 4 \alpha + \beta = - 4 \\ - \: \: \: \: \: - \: = \: \: \: + \\ = > - 2 \alpha = 2 \\ \\ = > \alpha = \frac{2}{ - 2} \\ \\ = > \alpha = - 1$

So,

Putting the value of alpha in equation (1), we get;

$= > 2 \alpha + \beta = - 2 \\ \\ = > 2 ( - 1) + \beta = - 2 \\ \\ = > - 2 + \beta = - 2 \\ \\ = > \beta = - 2 + 2 \\ \\ = > \beta = 0$

Thus,

The value of alpha is -1.

The value of beta is 0.

Hope it helps ☺️