# Find the Values of the constants in the following indentities

1. Aug 30, 2005

### ASMATHSHELPME

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.

Probably all find this easy

2. Aug 30, 2005

### TD

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

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: Aug 30, 2005
3. Aug 30, 2005

### iNCREDiBLE

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.

4. Aug 30, 2005

### iNCREDiBLE

TD: It's $+2x$, not $-2x$ as in your post.

5. Aug 30, 2005

### TD

Right, I'll correct :)

6. Aug 30, 2005

### ASMATHSHELPME

Bingo, gee thats easy ... but i always think that, then ill try a new one and get confused again.

These books im 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.

7. Aug 30, 2005

### TD

Seems correct

8. Aug 30, 2005

### ASMATHSHELPME

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????

9. Aug 30, 2005

### iNCREDiBLE

$(a-b)(a-b) = (a-b)^2 = a^2 - 2ab + b^2$

So $-2x^2\frac{A}{x^2}=-2A[/tex] not [itex]2A/X^8$

10. Aug 30, 2005

### ASMATHSHELPME

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 eachother? (The bottom bits) leaving 2A/ Nothing.

11. Aug 30, 2005

### iNCREDiBLE

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. Aug 30, 2005

### ASMATHSHELPME

Wow thats 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 eachother out? Or have i done this basic step wrong in the first place.

13. Aug 30, 2005

### iNCREDiBLE

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

: Missed one +-sign.

Last edited: Aug 30, 2005
14. Aug 30, 2005

### ASMATHSHELPME

So if X^2 is below something(Demoninator i think you call it?) and its multiplyed by itself, It cancels eachother 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. Aug 30, 2005

### iNCREDiBLE

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. Aug 30, 2005

### TD

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

17. Aug 30, 2005

### HallsofIvy

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