Given that the polynomial f(x) divided by x2+3x+2 has a remainder of 3bx+a−2 and divided by x2−2x−3 has a remainder of ax−2b. If f(3)+f(−2)=6, then a+b=....
Explanation
Factor the first divisor
The divisor is x2+3x+2=(x+1)(x+2)→x=−1∨x=−2.
The remainder is s(x)=3bx+a−2.
For x=−1
f(−1)=s(−1)
f(−1)=3b(−1)+a−2
f(−1)=−3b+a−2...(1)
For x=−2
f(−2)=s(−2)
f(−2)=3b(−2)+a−2
f(−2)=−6b+a−2...(2)
Factor the second divisor
For the second polynomial, the divisor is x2−2x−3=(x+1)(x−3)→x=−1∨x=3.
The remainder is s(x)=ax−2b.
For x=−1
f(−1)=s(−1)
f(−1)=a(−1)−2b
f(−1)=−a−2b...(3)
For x=3
f(3)=s(3)
f(3)=a(3)−2b
f(3)=3a−2b...(4)
Solve the system of equations
From the first and third equations, we get
f(−1)=f(−1)
−3b+a−2=−a−2b
b=2a−2...(5)
Then combine equations two, four, and five into
f(3)+f(−2)=6
3a−2b+(−6b+a−2)=6
4a−8b=8
a−2b=2
a−2(2a−2)=2
−3a=−2
a=32
Then the value of b=2a−2=2(32)−2=−32.
Therefore, a+b=32+(−32)=0.