Given that the polynomial f(x) divided by 2x2−x−1 has a remainder of 4ax−b and divided by 2x2+3x+1 has a remainder of −2bx+a−11. If f(x−2) is divisible by x−3, then a+2b+6=....
Explanation
Factor the first divisor
2x2−x−1=(2x+1)(x−1)=0
x=−21∪x=1
The remainder is s(x)=4ax−b. The roots of the divisor are −21 and 1, so
For x=−21
f(−21)=s(−21)
f(−21)=4a(−21)−b
f(−21)=−2a−b...(1)
For x=1
f(1)=s(1)
f(1)=4a−b...(2)
Factor the second divisor
Given the polynomial f(x) with divisor 2x2+3x+1=(2x+1)(x+1)=0, then x=−21∪x=−1.
The remainder is s(x)=−2bx+a−11.
For x=−21
f(−21)=s(−21)
f(−21)=−2b(−21)+a−11
f(−21)=a+b−11...(3)
Use the divisibility condition
f(x−2) is divisible by x−3, so the divisor x−3=0→x=3.
The remainder is s(x)=0 (because it is divisible).
x=3→f(x−2)=0
f(3−2)=0
f(1)=0...(4)
Solve the system of equations
From the second and fourth equations
4a−b=0
b=4a...(5)
From the first and third equations, with b=4a
f(−21)=f(−21)
a+b−11=−2a−b
3a+2b=11
3a+2(4a)=11
3a+8a=11
11a=11
a=1
Substitute the value a=1 into the fifth equation
b=4a=4(1)=4
Therefore, the value of a+2b+6=1+2(4)+6=15.