Irritating math qn?

Question

The expression f(x)=ax^3-(a+3b)x^2+2bx+c is exactly divisible by x^2-2x. When f(x) is divided by x-1, the remainder is 8 more than when it is divided by x+1. Factorise f(x) completely. Find the range of values of x for which f(x)<0.

How do you get rid of c? It doesn't allow me to do simultaneous eqn. if you factorise x^2-2x=x(x-2), does it mean if you let x=0, then f(0)=c, c=0"?

Thx, to those who have replied...
chin_hung, you 're a genius..hate that 4hr thingy...

Answer 1

f(x) = ax^3 - (a + 3b)x^2 + 2bx + c

Given: f(x) is divisble by x^2 - 2x
which means that f(x) is divisible by x(x - 2)

Since f(x) is divisible by x,
f(0) = 0
a(0)^3 - (a + 3b)(0)^2 + 2b(0) + c = 0
c = 0 --- (1)

Since f(x) is divisble by (x - 2),
f(2) = 0
a(2)^3 - (a + 3b)(2)^2 + 2b(2) + c = 0
8a - 4a - 12b + 4b + c = 0
4a - 8b + c = 0 --- (2)

Sub (1) into (2)
4a - 8b = 0
a = 2b --- (3)

Also given: when f(x) is divided by (x - 1), remainder is 8 more than when it is divided by (x + 1)
f(1) = 8 + f(-1)
a(1)^3 - (a + 3b)(1)^2 + 2b(1) + c
= 8 + a(-1)^3 - (a + 3b)(-1)^2 + 2b(-1) + c
a - a - 3b + 2b + c = 8 - a - a - 3b - 2b + c
c - b = 8 - 2a - 5b + c
2a + 4b = 8
a + 2b = 4 --- (4)

Sub (3) into (4)
2b + 2b = 4
b = 1
when b = 1, a = 2(1)
a = 2

With the values of a, b and c, rewriting the expression f(x)......
f(x) = 2x^3 - [2 + 3(1)]x^2 + 2(1)x + 0
f(x) = 2x^3 - 5x^2 + 2x

Factorising f(x)......
f(x) = x(x - 2)(2x - 1)

when f(x) < 0,
x(x - 2)(2x - 1) < 0
x < 0, 0.5 < x < 2

Therefore, the range of x for which f(x) < 0 is
x < 0 and 0.5 < x < 2.

Answer 2

I don't know if you really get rid of c or not.
If you carry out the long division by x^2-2x, to get an exactly divisible quotient, the quotient must be
ax+(a-3b). The last term being divided is
(a-3b)x^2+2bx. The partial dividend is
(a-3b)x^2-2(a-3b)x. Subtracting the partial dividend from the term being divided and setting the result to 0 means that b =(a-3b) or a=4b

You can now convert all the a's to b's to get an equation of the form 4bx^3- 7bx^2+2bx+c.
I haven't done the rest, but you will have to divide the above by x-1 and x+1. You will probably wind up with 2 equations in b and c, reflecting the difference of 8 in the remainder. You should be able to solve for b and c, and thus wind up only x's and numbers. Since you have already shown that x^2-2x is a root, you have to go back to the quotient of the first division, which is 4bx+b