Question

Please help! I don't have any idea what to do! Will fan and medal!


Suppose f(x) is a polynomial of degree 4 or greater such that f(1)=2, f(2)=3, and f(3)=5.

Find the remainder when f(x) is divided by (x-1)(x-2)(x-3).

Answer 1

Does this page help at all?

http://mathforum.org/library/drmath/view/60310.html

look at problem (2)

Answer 2

It helps a little but I am still unable to solve the question. Could you explain further?

Answer 3

what is the degree of (x-1)(x-2)(x-3) ? Any idea?

hint: expand out the polynomial

Answer 4

degree 3

Answer 5

so the degree of the remainder is going to be less than 3

http://www.sosmath.com/algebra/factor/fac01/fac01.html

since the degree of the remainder is always less than the degree of the divisor

Answer 6

so the remainder after dividing by (x-1)(x-2)(x-3) will either be

a single number
OR
something in the form Ax+B
OR
something in the form
Ax^2 + Bx + C

hopefully you agree so far?

Answer 7

yeah

Answer 8

Let's assume it's in the form Ax^2 + Bx + C

I'm going to use the same basic approach as the professor is showing on the first link I posted

If that's the case, then

f(x)/[(x-1)(x-2)(x-3)] = Q(x) + R(x)/[(x-1)(x-2)(x-3)]

which turns into

f(x) = (x-1)(x-2)(x-3)*Q(x) + R(x)

f(x) = (x-1)(x-2)(x-3)*Q(x) + Ax^2 + Bx + C

Answer 9

Now we plug in x = 1

f(x) = (x-1)(x-2)(x-3)*Q(x) + Ax^2 + Bx + C

f(1) = (1-1)(1-2)(1-3)*Q(1) + A*(1)^2 + B*(1) + C
f(1) = A+B+C
2 = A+B+C
A+B+C = 2

Answer 10

next plug in x = 2


f(x) = (x-1)(x-2)(x-3)*Q(x) + Ax^2 + Bx + C

f(2) = (2-1)(2-2)(2-3)*Q(2) + A(2)^2 + B(2) + C

3 = 4A + 2B + C

4A + 2B + C = 3

Answer 11

and finally x = 3


f(x) = (x-1)(x-2)(x-3)*Q(x) + Ax^2 + Bx + C

f(3) = (3-1)(3-2)(3-3)*Q(3) + A(3)^2 + B(3) + C

5 = 9A + 3B + C

9A + 3B + C = 5

Answer 12

so I should just solve the system of equations?

Answer 13

so we have this system of equations

A+B+C = 2
4A + 2B + C = 3
9A + 3B + C = 5

are you able to solve for A,B,C ?

Answer 14

Yes

Answer 15

So I can add the first 2 equations to get 5A+3B+2C=5 and then subtract that from the bottom equation to get: 4A-C=0 or 4A=C.

Answer 16

so because C = 4A, we can replace every copy of C with 4A to go from this system

A+B+C = 2
4A + 2B + C = 3
9A + 3B + C = 5

to this system

A+B+4A = 2
4A + 2B + 4A = 3
9A + 3B + 4A = 5

pick any two equations you want and solve for A,B

Answer 17

So the second equation becomes 2C+2B=3 or C+B=1.5

So A=.5, B=-.5, and C=2.

Answer 18

those are the correct values for A,B,C

Answer 19

so IF the remainder is of degree 2, then the remainder is the function

R(x) = 0.5x^2 - 0.5x + 2

Answer 20

Okay. Thanks for all of your help. I'll just check to make sure that is the correct answer.

Answer 21

no problem