Find the values of a, b and c

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So we have a = 1, b = 1, and c = 1. In summary, the conversation discusses factoring polynomials and finding the values of a, b, and c in the polynomial x^3-1=(x-1)(ax^2+bx+c). Various methods are suggested, including manual factorisation, long division, identity, and substitution. Ultimately, it is determined that a = 1, b = 1, and c = 1.
  • #1
aisha
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I looked at this question and said WHAT? i don't know what its talking about can someone please help me out, I don't have a clue as to what to do, or where to start. :cry:

For x^3-1=(x-1)(ax^2+bx+c) find the values of a, b, and c
 
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  • #2
Try dividing [tex]x^3 -1 [/tex] by [tex]x - 1[/tex].
Do you know how to do that?
 
  • #3
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
aisha said:
I looked at this question and said WHAT? i don't know what its talking about can someone please help me out, I don't have a clue as to what to do, or where to start. :cry:

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

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
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 learned in Grade K10 I think.
x^3-1=(x-1)(x^2+x+1)
 
  • #6
This is what I got:
[tex]x^3-1=(x-1)(ax^2+bx+c)[/tex]

[tex]ax^2+bx+c=\frac{x^3-1}{x-1}=x^2[/tex]

[tex]ax^2+bx+c=x^2[/tex]

[tex]a=\frac{x^2-bx-c}{x^2}[/tex]

The [tex]x^2[/tex] get canceled out and I got:

[tex]a=bx-c[/tex]


*NOTE: I might be wrong
 
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  • #7
Yeah, Raza you are... wrong that is.
[tex]a=\frac{x^2-bx-c}{x^2}[/tex] does not cancel out to get [tex]a=bx-c[/tex]
Plus, you say that [tex]ax^2+bx+c=x^2[/tex]. 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:
  • #8
I think I've got the answer

Hi, thanks for all ur help I think I've got it tell me if u think I am 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
aisha said:
Hi, thanks for all ur help I think I've got it tell me if u think I am 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

Sounds right to me.
 

1. What is the purpose of finding the values of a, b, and c?

The values of a, b, and c are used to solve mathematical equations, specifically quadratic equations in the form of ax^2 + bx + c = 0. These values help determine the roots or solutions of the equation.

2. How do you find the values of a, b, and c?

The values of a, b, and c can be found by using various methods such as factoring, completing the square, or using the quadratic formula. These methods involve manipulating the equation to isolate the values of a, b, and c.

3. What information do you need to find the values of a, b, and c?

To find the values of a, b, and c, you need to have a quadratic equation in the form of ax^2 + bx + c = 0. This equation should have known coefficients, which are the numbers in front of the variables a, b, and c.

4. Can the values of a, b, and c be imaginary or complex numbers?

Yes, the values of a, b, and c can be imaginary or complex numbers. This can occur when the quadratic equation has no real solutions, which happens when the discriminant (b^2 - 4ac) is negative.

5. What can the values of a, b, and c tell us about the graph of a quadratic equation?

The values of a, b, and c can tell us the shape and position of the graph of a quadratic equation. The value of a determines whether the graph opens upwards or downwards, b affects the position of the vertex, and c is the y-intercept of the graph.

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