Question
If x3 + ax2 + 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.
Solution
Leftf(x)=`x^3+ax^2+bx+6`
`x-2=0`⇒ x=2
Since, x-2 is a factor, remainder=0
∴ f(2)=0
`(2)3+a(2)^2+b(2)+6=0`
`8+4a+2b+6=0`
`2a+b+7=0` .........(1)
On dividing f(x)by x-3, it leavvves a remainder 3.
∴ f(3)=3
`(3)^3+a(3)^2+b(3)+6=3`
`27+9a+3b+6=3`
`3a+b+10` ...........(2)
Subtracting (1) from (2), we get,
a+3=0
a=-3
Substituting the value of a in (1), we get,
-6+b+7=0
b=-1
Is there an error in this question or solution?
Solution If X3 + Ax2 + 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. Concept: Remainder Theorem.
