Question

If (x+1)and (x-2)are factors of x^3+(a+1)x^2-(b-2)x-6,find the values of a and b.And then factorise the given expression completely ​

Answer 1

Answer:

a = 1,  b = 7

f(x) = x³ + 2x² - 5x - 6

    = ( x + 1 ) ( x - 2 ) ( x + 3 )

Step-by-step explanation:

Put f(x) = x³ + (a+1) x² - (b-2) x - 6.

(x+1) is a factor of f(x)

=> f(-1) = 0

=> -1 + a+1 + b-2 - 6 = 0

=> a + b = 8    ... (1)

(x-2) is a factor of f(x)

=> f(2) = 0

=> 8 + 4(a+1) - 2(b-2) - 6 = 0

=> 8 + 4a + 4 - 2b + 4 - 6 = 0

=> 4a - 2b = -10

=> 2a - b = -5

=> b = 2a + 5    ... (2)

Substituting (2) into (1):

a + 2a + 5 = 8  =>  3a = 3   =>   a = 1

Then from (2)

b = 2+5 = 7

So

f(x) = x³ + 2x² - 5x - 6

There are three roots:  -1, 2 and call the third one u.

The product of the roots is 6 ( - constant term divided by coefficient of x³), so

-1 × 2 × u = 6  =>  u = -3

Therefore the factorization of f(x) is

f(x) = ( x + 1 ) ( x - 2 ) ( x + 3 )