Homework Help: 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.