1. ## polynomials

$f(x) = 3x^2 - x^3 + h$. For which values of h will f have three distinct zeros?

How do I do this too.

I guess they need a height for the graph where you get three roots. If I differentiate the function, I see that y would have to be above the x axis on x = 2, and below the x axis on x = 0.

2. Originally Posted by LeoBloom.
$f(x) = 3x^2 - x^3 + h$. For which values of h will f have three distinct zeros?

How do I do this too.
If h doesnt have to be a constant then

$-x^3+3x^2+5x\Rightarrow{-x(x^2+3x+5)}$

Setting it equal to zero we get

$x=0,x=3\pm\frac{\sqrt{29}}{2}$

3. Sorry, forgot to mention h has to be a constant.

The answers in the book are

(a) all h > 4
(b) 0 < h < 4
(c) all h < 0
(d) -4 < h < 0
(e) all h < -4

I think that I could rule out a,c,e because then at some point in the graph, there will only be one root instead of three.

4. And if h is a constant, try negative values such as -1, -2, -3....etc.

5. Hello,

Originally Posted by Mathstud28
If h doesnt have to be a constant then

$-x^3+3x^2+5x\Rightarrow{-x(x^2+3x+5)}$

Setting it equal to zero we get

$x=0,x=3\pm\frac{\sqrt{29}}{2}$
I don't think they're asking for ONE solution, they more likely want the domain of h, such that the condition is OK.

Plus, in géneral with this notation, h will be a constant

6. Originally Posted by LeoBloom.
$f(x) = 3x^2 - x^3 + h$. For which values of h will f have three distinct zeros?

How do I do this too.

I guess they need a height for the graph where you get three roots. If I differentiate the function, I see that y would have to be above the x axis on x = 2, and below the x axis on x = 0.
youve got it spot on from that you get 2 inequalities
$f(0)<0$ ie $3(0^2)+0^3+h<0$

and $f(2)>0$ ie $3(2^2)+2^3+h>0$

solving these two gets you the answer
$-4