Find the Values of the constants in the following indentities

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Discussion Overview

The thread discusses the process of finding the values of constants in algebraic identities, specifically focusing on polynomial equations and their simplifications. Participants explore the steps involved in expanding and rearranging expressions, comparing coefficients, and addressing errors in calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant presents an identity and attempts to find constants A, B, and C by comparing coefficients after expanding both sides of the equation.
  • Another participant suggests rearranging the equation to group terms by the power of x before comparing coefficients.
  • Several participants point out errors in simplification, specifically regarding the combination of like terms and the correct coefficients.
  • There is a discussion about the simplification of terms involving fractions, with confusion arising over the cancellation of variables in the numerator and denominator.
  • One participant expresses uncertainty about the correctness of their simplifications and seeks clarification on the cancellation process.
  • Another participant provides a detailed explanation of how to properly simplify the expression and corrects misconceptions about the operations involved.

Areas of Agreement / Disagreement

Participants generally agree on the need to simplify expressions correctly and compare coefficients, but there are multiple competing views regarding specific simplifications and the handling of fractions. The discussion remains unresolved in some areas, particularly concerning the understanding of cancellation in algebraic expressions.

Contextual Notes

Some participants express confusion about the terminology and processes involved in algebraic manipulation, indicating a potential gap in foundational understanding. There are also unresolved questions about specific steps in the simplification process.

Who May Find This Useful

This discussion may be useful for students learning algebra, particularly those struggling with polynomial identities, simplification techniques, and the manipulation of fractions in mathematical expressions.

ASMATHSHELPME
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Find the Values of the constants in the following indentities.

A(X^2 -1) + B(X - 1) + C = (3x -1)(x +1)

AX^2 - A + Bx - B + C = 3x^2 + 3x -x -1

Ok, Is this correct so far?

AX^2 = 3x^2 :. A = 3
Is correct ?

Bx = 3x = b = 3 ?
C = -1 ?

I'm sure this is wrong, I just need help explaining why and correcting my errors. It seems i need to simplify it more before comparing the co-efficients.

Thanks in advanced guys,
Probably all find this easy :blushing:
 
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So, you start with

[tex]A\left( {x^2 - 1} \right) + B\left( {x - 1} \right) + C = \left( {3x - 1} \right)\left( {x + 1} \right)[/tex]

You worked out both sides

[tex]Ax^2 - A + Bx - B + C = 3x^2 + 3x - x - 1[/tex]

Now, before you continue, rearrange both sides so you group per power of x

[tex]Ax^2 + Bx - A - B + C = 3x^2 + 2x - 1[/tex]

Now compare the coefficients of the equal powers of x of both sides. Remember that the all the constants are the coefficients of x^0.
 
Last edited:
Just a small error. You forgot to simplify [itex]3x-x[/itex] to [itex]2x[/itex].

[tex]A(x^2-1) + B(x-1) + C = (3x-1)(x+1)[/tex]
<=>
[tex]Ax^2-A+Bx-B+C = 3x^2+3x-x-1 = 3x^2+2x-1[/tex]
<=>
[tex]Ax^2+Bx-A-B+C = 3x^2+2x-1[/tex]

Now try again.
 
TD: It's [itex]+2x[/itex], not [itex]-2x[/itex] as in your post.
 
Right, I'll correct :)
 
Bingo, gee that's easy ... but i always think that, then ill try a new one and get confused again.

These books I am learning from arn't to clear.

Thanks guys,
Ax^2 + Bx - A - B +C = 3x^2 + 2x - 1
A = 3
B = 2
-A -B + C = -1
-3 -2 + C = - 1
C = 4
-5 + 4 = - 1

Excellent.
 
Seems correct :smile:
 
One more question, a new identity but basically expanding with fractions.

(X^2 - A/X^2) (X^2 - A/X^2)

= X^4 - A/X^4 - A/X^4 + A^2/X^4

X^4 - 2A/x^8 + A^2/X^4 ?

But is the 2A/X^8 Correct? Sure its wrong?
 
[itex](a-b)(a-b) = (a-b)^2 = a^2 - 2ab + b^2[/itex]

So [itex]-2x^2\frac{A}{x^2}=-2A[/tex] not [itex]2A/X^8[/itex][/itex]
 
  • #10
How do you make it look like that?

My layouits are unclear so i think you've given the wrong explanation(My fault).

Wondering how its 2A/Nothing

Rather than
2A/x^8

From -A/x^4 -A/X^4

SOmehow they cancel each other? (The bottom bits) leaving 2A/ Nothing.
 
  • #11
I assume you know that [itex](a-b)(a-b) = (a-b)^2 = a^2 - 2ab + b^2[/itex] (*)

So:
[tex](x^2 - \frac{A}{x^2}) (x^2 - \frac{A}{x^2}) = (x^2 - \frac{A}{x^2})^2 = (*) = (x^2)^2 -2x^2\frac{A}{x^2} + (\frac{A}{x^2})^2 = x^4 - 2A + \frac{A^2}{x^4}[/tex]

What I'm saying is that:
[tex]-2x^2\frac{A}{x^2} = \frac{-2Ax^2}{x^2} = -2A[/tex]
since the [itex]x^2[/itex] in the numerator and denominator cancels out.
 
  • #12
Wow that's confusing!

So can you quickly type out[itex](X^2 - A/x^2)^2[/itex] Please

Because i get:

[itex]+x^4 - A/x^4 - A/x^4 + A^2/x^4[/itex]
Just want to know how X^4 cancel each other out? Or have i done this basic step wrong in the first place.
 
  • #13
Yes, you seem to believe that
[tex]x^2\frac{A}{x^2} = \frac{A}{x^4}[/tex]
which is not true.

In fact [itex]x^2\frac{A}{x^2} = A[/itex]

So
[tex](x^2-A/x^2)(x^2-A/x^2) = x^4 - A - A + A^2/X^4[/tex]


[Edit]: Missed one +-sign.
 
Last edited:
  • #14
So if X^2 is below something(Demoninator i think you call it?) and its multiplyed by itself, It cancels each other out?

Just trying to get my head around this, Quite confused lol.

Basically [itex]X^2 [Multiplied by ] A/X^2[/itex] Cancels X^2 leaving A/ Nothing.
 
  • #15
That's right. Just take a look at this trivial example:

[tex]1 = \frac{1}{2} + \frac{1}{2} = (definition) = 2*\frac{1}{2} = 1[/tex]
 
  • #16
ASMATHSHELPME said:
So if X^2 is below something(Demoninator i think you call it?) and its multiplyed by itself, It cancels each other out?

Just trying to get my head around this, Quite confused lol.

Basically [itex]X^2 [Multiplied by ] A/X^2[/itex] Cancels X^2 leaving A/ Nothing.
A little bit of terminology:

- we call [itex]\frac{A}{B}[/itex] a fraction
- A is the nominator
- B is the denominator

Perhaps that makes it easier to follow :smile:
 
  • #17
ASMATHSHELPME said:
So if X^2 is below something(Demoninator i think you call it?) and its multiplyed by itself, It cancels each other out?

Just trying to get my head around this, Quite confused lol.

Basically [itex]X^2 [Multiplied by ] A/X^2[/itex] Cancels X^2 leaving A/ Nothing.

Well, not A over Nothing! That would be A/0 which makes no sense.
The cancelling leaves 1. x2(A/x2= A/1= A.
 

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