Question

find 'a' and 'β' if (x+1) and (x+2) are factors of x³ + 3x²-2ax + 2β

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Answer 1

Here I am writing Alpha as a, and Beta as b.

Given f(x) = x^3 + 3x^2 - 2ax + 2b.

Given that (x + 1) is a factor of f(x).

By factor theorem, 

f(-1) = 0.

plug x = -1, we get

= > (-1)^3 + 3(-1)^2 - 2a(-1) + 2b = 0

= > -1 + 3 + 2a + 2b = 0

= > 2 + 2a + 2b = 0

= > 2a + 2b = -2 ------ (1)



Given that (x + 2) is a factor of f(x). we get

f(-2) = 0

plug x = -2 in f(x), we get

= > (-2)^3 + 3(-2)^2 - 2(-2)a + 2b = 0

= > -8 + 12 + 4a + 2b = 0

= > 4 + 4a + 2b = 0

= > 4a + 2b = -4  ----- (2)


On solving (1) & (2), we get

2a + 2b = -2

4a + 2b = -4

-----------------------

-2a = 2

a = -1.


Substitute a = -1 in (1), we get

2a + 2b = -2

2(-1)  + 2b = -2

-2 + 2b = -2

2b = -2 + 2

2b = 0

b = 0


Therefore a = -1 and b = 0.


Hope this helps!

Answer 2

Beta is written as B,
××××××××××××××××××××××××



Given that (x + 1) and (x + 2) are the Factors

Then, x = -1 or - 2

Taking x = -1

x³ + 3x² - 2ax + 2B = 0

(-1)³ + 3(-1)² - 2a(-1) + 2B =0

-1 + 3 + 2a + 2B = 0

2 + 2a + 2B = 0

2(a + B) = -2

a + B = -1 -----1equation

×××××××××××××××××

Taking x = -2

x³ + 3x² - 2ax + 2B = 0

(-2)³ + 3(-2)² - 2(-2)a + 2B = 0

-8 + 12 + 4a + 2B = 0

4 + 4a + 2B =0

2a + B= -2 -----2equation

×××××××××××××××××××

Subtract eq(2) from (1)

a + B = -1
2a + B = -2
(-) (-) ___+
- a = 1
_______

a = -1

Putting the value of a in 1equation,

a + B = -1
-1 + B = -1
B = -1 + 1
B = 0



I hope this will help you

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