Question

if x^3
+ ax^2
+ bx + 6 has (x – 2) as a factor and leaves a remainder 3 when divided by (x– 3),
find the values of a and b.
*pls explain*​

Answer 1

Step-by-step explanation:

f(x)= x³+ax²+bx+6

(x-2) is a factor

Therefore , zero of the polynomial=2

f(2) = 2³+a.2²+b.2+6=14+4a+2b

(x-3) leaves a remainder 3

f(3)=3³+a.3²+b.3+6= 33+9a+3b

by remainder theorem,

14+4a+2b=0

⇒2a+b=-7...............(i)

33+9a+3b=3

⇒3a+b = 1-11=-10..........(ii)

(ii)-(i)

⇒a= -10+7=-3

⇒2.(-3)+b=-7[ on putting the value of a in eq (i) ]

⇒b=-1

a=-3,b=-1