Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫  π  -¹ ² ³ °

You are not logged in.

## #1 2005-11-28 05:18:02

RickyOswaldIOW
Member
Registered: 2005-11-18
Posts: 212

### More factorising revision

I have y = 2 - 3x - 2x^2.

Normally I would take out the coefficient of x^2 and thus have;

y = -2[x^2 + 1(1/2) - 1]

But I can't seem to factorise this?...

Aloha Nui means Goodbye.

Offline

## #2 2005-11-28 05:21:53

RickyOswaldIOW
Member
Registered: 2005-11-18
Posts: 212

### Re: More factorising revision

Okay, maybe I should not take the 2 from the x^2 and instead have;

-[2x^2 + 3x - 2].

I think this will work but I have to go out now, thanks anyway

Aloha Nui means Goodbye.

Offline

## #3 2005-11-28 06:08:15

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: More factorising revision

That would work. -[2x² + 3x - 2] factorises to give -(x+2)(2x-1).

Why did the vector cross the road?
It wanted to be normal.

Offline