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

You are not logged in.

- Topics: Active | Unanswered

**Kurtz****Member**- Registered: 2007-10-01
- Posts: 27

Is it possible to solve this problem with the grouping method? It ask to use the grouping method on this to simplify but I think it is not possible.

a²x + bx - a²z - bx

I am a mathemagician. You ask why? I point up

Offline

**Kurtz****Member**- Registered: 2007-10-01
- Posts: 27

Or does it become

(x - z) (a^4 - b²) ?

I am a mathemagician. You ask why? I point up

Offline

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

a²x + bx - a²z - bx

a²x - a²z

a²(x-z)

Why did the vector cross the road?

It wanted to be normal.

Offline

**soroban****Member**- Registered: 2007-03-09
- Posts: 452

Hello, Kurtz!

Is it possible to solve this problem with the grouping method?

. .

. . . . . . . . . .

. . . . . . . . . .

.

Offline

**Kurtz****Member**- Registered: 2007-10-01
- Posts: 27

oops thx that makes a lot more sense because I mess up on writing the problem

it was supposed to be a²x + bx - a²z - bz

thx for overlooking that mistake. and thx so much for helping me.

I am a mathemagician. You ask why? I point up

Offline

**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 1,184

How did soroban know that the last term was supposed to be -bz?

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

If you assume that the person setting the question did make a 'do-able' problem then a sign error making it impossible must be a typo. Over the years I've seen a few.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline