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 nd b.

Answer 1

here is your answer
we know x-2 is a factor
so, x-2=0 which gives x=2
put value of x in the given equation
we get, 2^3 +a2^2+b2 +6=0[sice x=2 is a zero]
so, 8+4a+2b+6=0
4+2a+b+3=0[divided the eq with 2]
b=-2a-7————eq 1
we are also given,when x=3 remainder is 3[x-3]
so, 27 +9a+3b+6=3
9+3a+b+2=1[divided both sides with 3]
b=-3a-10————eq 2
from eq 1 and 2
-2a-7=-3a-10
a=-3 put this value in eq 1
b=6-7=-1
so, a=-3 and b=-1
hope it helps

Answer 2

(x-3)(x-2)+3

=x^2 - 2x - 3x + 6

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