SOLUTION: When a polynomial f(x) is divided by x-3 the remainder is -9 and when divided by 2x-1 the remainder is -6.Find the remainder when f(x) is divided by(x-3)(2x-1)

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: When a polynomial f(x) is divided by x-3 the remainder is -9 and when divided by 2x-1 the remainder is -6.Find the remainder when f(x) is divided by(x-3)(2x-1)      Log On


   



Question 1135670: When a polynomial f(x) is divided by x-3 the remainder is -9 and when divided by 2x-1 the remainder is -6.Find the remainder when f(x) is divided by(x-3)(2x-1)
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

When a polynomial f%28x%29 is divided by x-3 the remainder is -9+
x-3=0=>x=3
=>f%283%29=-9
and when divided by 2x-1 the remainder is -6
%282x-1%29=0=>x=1%2F2
=>f%281%2F2%29=-6

We want the remainder when f%28x%29 is divided by %28x-3%29%282x-1%29.
So let f%28x%29+=+q%28x%29%28x-3%29%282x-1%29+%2B+r%28x%29, where q is the quotient and r is the remainder.
Being the remainder on division by a quadratic, r must be a linear polynomial, say
r+=+ax%2Bb.
So f%28x%29+=+q%28x%29%28x-3%29%282x%2B1%29+%2B+ax+%2B+b
f%283%29+=+-9 means 3a%2Bb+=+-+9.....eq.1
f%281%2F2%29+=+-6+means %281%2F2%29a+%2B+b+=+-6, or a+%2B+2b+=+-12.....eq.2
Solve the system
3a%2Bb+=+-+9.....eq.1
a+%2B+2b+=+-12.....eq.2
----------------------------
3a%2Bb+=+-+9.....eq.1.....solve for b
b+=+-+9-3a......substitute that into eq.2
a+%2B+2%28-+9-3a%29+=+-12.....eq.2
a+%2B+-+18-6a+=+-12
-+18-5a+=+-12
-+18%2B12=+5a

-+6=+5a+
a=-6%2F5
b+=+-+9-3a......substitute a
b+=-+9-3%28-6%2F5%29
b+=-+9%2B18%2F5
b+=-+45%2F5%2B18%2F5
b+=-+27%2F5
Conclusion: the remainder is
r+=+ax%2Bb....substitute a and b
r+=+-%286%2F5%29x-+27%2F5


Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

For three linear binomials and three remainders see the lesson

    - Solved problems on the Remainder theorem, Problem 7.

in this site . . .