Find the Values of the constants in the following indentities

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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.
 
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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.
 
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  • #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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