Question

If (x + 1) and (x – 2) are factors of x3

+ ax2

– bx – 6, then find the values of a and b respectively.​

Answer 1

Answer:

Let p(x) = x3 + ax2 + bx +6

(x-2) is a factor of the polynomial x3 + ax2 + b x +6

p(2) = 0

p(2) = 23 + a.22 + b.2 +6 =8+4a+2b+6 =14+ 4a+ 2b = 0

7 +2 a +b = 0

b = - 7 -2a -(i)

x3 + ax2 + bx +6 when divided by (x-3) leaves remainder 3.

p(3) = 3

p(3) = 33 + a.32 + b.3 +6= 27+9a +3b +6 =33+9a+3b = 3

11+3a +b =1 => 3a+b =-10 => b= -10-3a -(ii)

Equating the value of b from (ii) and (i) , we have

(- 7 -2a) = (-10 - 3a)

a = -3

Substituting a = -3 in (i), we get

b = - 7 -2(-3) = -7 + 6 = -1

Thus the values of a and b are -3 and -1 respectivel

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