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
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
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.
$\displaystyle (2+x)^4 = x^4+8 x^3+24 x^2+32 x+16$
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
$\displaystyle (2+x)^4=x^4+8x^3+24x^2+32x+16$
$\displaystyle (2+x)^4=x^4+16x+16\Rightarrow\ \left(x^4+8x^3+24x^2+32x+16\right)-\left(x^4+16x+16\right)=0$
$\displaystyle \Rightarrow\ 8x^3+24x^2+16x=0\Rightarrow\ x^3+3x^2+2x=0$
You can only verify that $\displaystyle x=0,\;-1,\;-2$ cause the equation $\displaystyle x^3+3x^2+2x$ to be zero.