Hi

I have this question:

(i) show that (2 + x)^4 = 16 + 32x + 24x^2 + 8x^3 + x^4 for all x.

for all x? What does that mean?

(ii) Find the values of x for which (2 + x)^4 = 16 + 16x + x^4

??

Can someone please explain?

Angus

- Jan 7th 2011, 04:17 AMangypangybinomial expansion question - don't know what is being asked?
Hi

I have this question:

(i) show that (2 + x)^4 = 16 + 32x + 24x^2 + 8x^3 + x^4 for all x.

for all x? What does that mean?

(ii) Find the values of x for which (2 + x)^4 = 16 + 16x + x^4

??

Can someone please explain?

Angus - Jan 7th 2011, 04:27 AMProve It
i) For all means it's true no matter what the value of is Just perform the binomial expansion.

ii) Expand which you should have done for part i) and then set it equal to and solve. - Jan 7th 2011, 04:32 AMPlato
- Jan 7th 2011, 04:48 AMgrgrsanjay
i think this is the equation for second one

x(x^2 + 3x +2) = 0

the values of x are 0, -1 ,-2 - Jan 7th 2011, 07:02 AMangypangy
Hi

I am still a little confused with this question. I can see how to get factors 0, -1 and -2. But if I use them in (2 + x)^4 - ie 2+ -2 = 0 - 0^4 = 0 - not 16 + 16x + x^4

Can someone please explain?

Angus - Jan 7th 2011, 07:18 AMe^(i*pi)

x^4+8 x^3+24 x^2+32 x+16 = x^4+16x+16[/tex]

and cancel and when you take 16x from both sides you get

You can factor out from this equation to give

edit: You have to sub in 0, -1 and 2 on the RHS too

For example using 0

and so is a solution - Jan 7th 2011, 07:22 AMArchie Meade
- Jan 7th 2011, 08:09 AMangypangy
Ah you sub in val of x on right too. Yes understand now. Thanks.