# Factors of a polynomial

1. Oct 16, 2004

### footprints

$$3x^2 + 4x + C \equiv A(x + 1)^2 + B(x + 1) + 7$$
Find all values of A, B and C.
Could someone teach how to do this?

2. Oct 16, 2004

### arildno

Rewrite
$$x^{2}=((x+1)-1)^{2}$$
and do the necessary operations.

3. Oct 16, 2004

### footprints

Huh? Sorry I don't understand what u mean.

4. Oct 16, 2004

### arildno

$$((x+1)-1)^{2}=(x+1)^{2}-2(x+1)+1$$

5. Oct 16, 2004

### footprints

How is $$((x+1)-1)^{2}=(x+1)^{2}-2(x+1)+1$$
related to $$3x^2 + 4x + C \equiv A(x + 1)^2 + B(x + 1) + 7 ?$$
Sorry but i don't get u.

6. Oct 16, 2004

### arildno

It's equal to $$x^{2}$$!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Make a similar rewriting of x:
$$x=((x+1)-1)$$
Now, substitute these expressions for $$x,x^{2}$$ into your LEFT-HAND SIDE.
Reorganize the terms you get, and derive conditions so that your new expression equals your ORIGINAL RIGHT-HAND SIDE.
This will determine A,B,C.

7. Oct 16, 2004

### Hurkyl

Staff Emeritus
Alternatively, you could expand the right hand side, and equate coefficients to solve for A, B, and C.
Or you could plug in a few values for x to generate equations.

8. Oct 16, 2004

### footprints

I probably wasn't clear. Firstly, how does $$x^{2}=((x+1)-1)^{2}.$$ Secondly, which ones do i substitute in for x? Is it $$((x+1)-1)^{2} = 3 or (x + 1)^2 ?$$

9. Oct 16, 2004

### arildno

You know that 1-1=0, right?
So (x+1)-1=x+1-1=x.
To be nice, I'll do this for once:
We have:
$$3x^{2}=3((x+1)-1)^{2}=3(x+1)^{2}-6(x+1)+3$$
$$4x=4((x+1)-1)=4(x+1)-4$$
$$C=C=C$$
Now, add these equations together, downwards. The outermost terms then turn into:
$$3x^{2}+4x+C=3(x+1)^{2}-2(x+1)+(C-1)$$
Do you understand this?

10. Oct 16, 2004

### Hurkyl

Staff Emeritus
What does (x + 1) - 1 equal? Then what does ((x+1) - 1)^2 equal?

11. Oct 16, 2004

### Sariaht

I'm just curious, has this got something to do with modulus?

12. Oct 16, 2004

### arildno

Not that I know of..
It is simply to substitute "equal for equal"

13. Oct 16, 2004

### MiGUi

<Sarcastic mode ON>
arildno you have to be a little more diplomatic, I think
<Sarcastic mode OFF>

14. Oct 16, 2004

### Derivative86

you guys are making the problem too complicated
since
$$3x^2 + 4x + C = A(x + 1)^2 + B(x + 1) + 7$$
expand $$A(x + 1)^2 + B(x + 1) + 7$$
then you have $$Ax^2 + 2Ax + A + Bx + B + 7$$
and $$3x^2 + 4x + C = Ax^2 + 2Ax + A + Bx + B + 7$$
equate the coeffients of the powers you have
$$3 = A$$ for $$x^2$$
$$2A + B = 4$$ for $$x^1$$
$$A + B + 7 = C$$ for $$x^0$$
solve the system and you got the answer

Last edited: Oct 16, 2004
15. Oct 17, 2004

### footprints

Ahh... Sorry but i was a bit slow on "(x+1)-1" part (sorry if i pissed u arildno ). Thanks for the help guys!

16. Oct 17, 2004

### arildno

I wasn't exactly pissed off; rather, I felt resignation sneak up on me..