Find the Values of the constants in the following indentities

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The discussion focuses on finding the values of constants A, B, and C in polynomial identities. Participants clarify that A equals 3, B equals 2, and C equals 4 after correctly simplifying and comparing coefficients. There is also confusion regarding the expansion of the expression (X^2 - A/X^2)², particularly about how terms cancel and the correct interpretation of fractions. The conversation emphasizes the importance of careful simplification and understanding of algebraic identities. Overall, the participants work through their misunderstandings to arrive at correct conclusions about the constants and polynomial expansions.
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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

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

You worked out both sides

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

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

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

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 3x-x to 2x.

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

Now try again.
 
TD: It's +2x, not -2x 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?
 
(a-b)(a-b) = (a-b)^2 = a^2 - 2ab + b^2

So -2x^2\frac{A}{x^2}=-2A[/tex] not 2A/X^8
 
  • #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 (a-b)(a-b) = (a-b)^2 = a^2 - 2ab + b^2 (*)

So:
(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}

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

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

Because i get:

+x^4 - A/x^4 - A/x^4 + A^2/x^4
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
x^2\frac{A}{x^2} = \frac{A}{x^4}
which is not true.

In fact x^2\frac{A}{x^2} = A

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


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

1 = \frac{1}{2} + \frac{1}{2} = (definition) = 2*\frac{1}{2} = 1
 
  • #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 X^2 [Multiplied by ] A/X^2 Cancels X^2 leaving A/ Nothing.
A little bit of terminology:

- we call \frac{A}{B} 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 X^2 [Multiplied by ] A/X^2 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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