Question

If (x+1) and (x-2) are factors of x³+ax²-bx-6, then find the values of 'a' and 'b' respectively

(1)2,3
(2)3,5
(3)5,3
(4)2,5

Help me doing this sum!!!​

Answer 1

Answer:

if that is the case then -1 and 2 are zeroes so substitute them in the equation

-1+a+b-6=0

a+b=7

8+4a-2b-6=0

b-2a =1

now subtract them u will get a is 2 and b is 5 so option 4 mate

Answer 2

ANSWER

A = 2 AND B = 5

SOLUTION

GIVEN

(x + 1 ) and ( x - 2 ) are factor of

x^3 + ax^2 - bx - 6

TO FIND VALUE OF A AND B

( x + 1 ) = 0

x = -1

put x = -1 in above equation we get,

(-1)^3 + a (-1) ^2 - b(-1) - 6 = 0

-1 + a + b - 6 = 0

a + b = 7

b = 7 - a .......(1)

(x - 2 ) = 0

x = 2

put x = 2 in above equation we get,

(2) ^3 + a (2) ^2 - b (2) - 6 = 0

8 + 4a - 2b - 6 = 0

4a - 2b +2 = 0

2a - b + 1 = 0 ....... (2)

put equation (1) on equation (2)

we get,

2a - (7- a) +1 = 0

2a - 7 + a + 1 = 0

3a - 6 = 0

a = 2

put a = 2 in equation (1)

b = 7 - a

b = 7 - 2 = 5

Hence, a = 2 and b = 5 = ANSWER