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 $\displaystyle \displaystyle x$ means it's true no matter what the value of $\displaystyle \displaystyle x$ is Just perform the binomial expansion.

ii) Expand $\displaystyle \displaystyle (2+x)^4$ which you should have done for part i) and then set it equal to $\displaystyle \displaystyle 16+16x+x^4$ 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)
$\displaystyle (2+x)^4 = x^4+8 x^3+24 x^2+32 x+16$

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

$\displaystyle x^4$ and $\displaystyle 16$ cancel and when you take 16x from both sides you get

$\displaystyle 8x^3+24x^2+16x = 0$

You can factor out $\displaystyle 8x$ from this equation to give $\displaystyle 8x(x^2+3x+2) = 8x(x+2)(x+1) = 0$

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

For example using 0

$\displaystyle (2+0)^4 = 16+16(0)+(0)^4 \implies 16=16$ and so $\displaystyle x=0$ is a solution - Jan 7th 2011, 07:22 AMArchie Meade
$\displaystyle (2+x)^4=x^4+8x^3+24x^2+32x+16$

$\displaystyle (2+x)^4=x^4+16x+16\Rightarrow\ \left(x^4+8x^3+24x^2+32x+16\right)-\left(x^4+16x+16\right)=0$

$\displaystyle \Rightarrow\ 8x^3+24x^2+16x=0\Rightarrow\ x^3+3x^2+2x=0$

You can only verify that $\displaystyle x=0,\;-1,\;-2$ cause the equation $\displaystyle x^3+3x^2+2x$ to be zero. - Jan 7th 2011, 08:09 AMangypangy
Ah you sub in val of x on right too. Yes understand now. Thanks.