# Find the values of a, b and c

1. Nov 21, 2004

### aisha

I looked at this question and said WHAT? i dont know what its talking about can someone please help me out, I dont have a clue as to what to do, or where to start.

For x^3-1=(x-1)(ax^2+bx+c) find the values of a, b, and c

2. Nov 21, 2004

### check

Try dividing $$x^3 -1$$ by $$x - 1$$.
Do you know how to do that?

3. Nov 21, 2004

### Hurkyl

Staff Emeritus
Or he could multiply out the RHS and set both sides equal to each other.
Or he could plug in values of x to generate a system of equations.

4. Nov 21, 2004

### TenaliRaman

Many polynomials can be factored into products of smaller polynomials.

E.g -
(x^2 - 1) = (x+1)(x-1)

This is known as factorisation of polynomials.

x^3-1 can be factored in a similar way.
Now,
x^3-1=(x-1)(ax^2+bx+c)

What does this mean?
This means that x^3 - 1 can be factored into two polynomials (x-1) and (ax^2+bx+c).

You are supposed to find the coefficients of the second polynomial.

Ways to do it?
1> Factorise x^3-1 manually and see what u get?
Suppose u get,
x^3-1 = (x-1)(px^2+qx+r)
then a = p,b = q and c = r.

2> Now if x^3-1=(x-1)(ax^2+bx+c)
then (x^3-1)/(x-1) = (ax^2+bx+c)
That means u can get (ax^2+bx+c) by dividing x^3-1 by x-1.

3> Multiply RHS. That is multiply (x-1) with (ax^2+bx+c) . Then compare the coefficients of this with x^3 - 1 so that u can determine a,b and c

4> Note that,
x^3-1=(x-1)(ax^2+bx+c)
is true for all x
So substitute x = 0 and u will note that u can get c
sub in x = -1 and u will get an equation in terms of a and b
sub in x = 2 and u will again get an equation in terms of a and b
solve them simultaneously to find a and b.

-- AI

5. Nov 21, 2004

### primarygun

Ya, this question is rather simple.
Three simple method:
1. Long division. Don't forget to wite down 0x^2 and 0x
2.Identity. Comparing coefficient. Expand the given function.
Compare the term to x^3-1
3.The most common one. It is learnt in Grade K10 I think.
x^3-1=(x-1)(x^2+x+1)

6. Nov 21, 2004

### Raza

This is what I got:
$$x^3-1=(x-1)(ax^2+bx+c)$$

$$ax^2+bx+c=\frac{x^3-1}{x-1}=x^2$$

$$ax^2+bx+c=x^2$$

$$a=\frac{x^2-bx-c}{x^2}$$

The $$x^2$$ get cancelled out and I got:

$$a=bx-c$$

*NOTE: I might be wrong

Last edited: Nov 21, 2004
7. Nov 21, 2004

### check

Yeah, Raza you are... wrong that is.
$$a=\frac{x^2-bx-c}{x^2}$$ does not cancel out to get $$a=bx-c$$
Plus, you say that $$ax^2+bx+c=x^2$$. If this is true for all (since a,b and c are constants) that would mean that either a,b and c are all = 0, (which the aren't) or that a = 1* and b and c = 0 (which, they don't)

Last edited: Nov 21, 2004
8. Nov 21, 2004

### aisha

I think I've got the answer

Hi, thanks for all ur help I think I've got it tell me if u think im wrong.
the question was for X^3-1=(X-1)(ax^2+bx+c)
what are the values of a, b and c

well I brought x-1 to the LHS and divided x^3-1 by x-1 I used long division and i got an answer of x^2+x+1 with no remainder I made this equal to ax^2+bx+c and I think a=1 b=1 and c=1 ? I'm not sure

9. Nov 21, 2004

### check

Sounds right to me.