Question

The expression x3-ax2+bx-6 leaves remainder -8 when divide by (x-1).If (x-2) is a factor of given expression,find the value of a and b.​

Answer 1

Answer:

Let p(x) = x³ + ax²+bx+6

(x-2) is a factor of the polynomial x³ + ax² + bx+6 p(2) = 0

p(2) = 23+a.2²+b.2 +6 =8+4a+2b+6 =14+

4a+ 2b = 0

7+2a+b=0

b = -7-2a-(i)

+ 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 bare -3 and -1 respectively.