Cubic Function Solving Problem I have question from my son:
A cubic function $f$ has the rule $f(x) =(x-3) (x+4) (x-5)$
Find the values of $h$ such that $f(x-h) = 0$ has only one positive solution.
Could anyone explain about this please, since in my opinion the answer will be a range of numbers like $a<h<b$, but my son said it must be one real number for $h$.
 A: Notice that to solve $0=f(x)= (x-3)(x+4)(x-5)$, because $f$ is a product, one of the terms in the product must be $0$. So either $x-3=0$ or $x+4=0$ or $x-5=0$. Then the solutions are $x=3, -4,5$, so there are two positive solutions. We can write out $f(x-h)$,
$$
f(x-h)= (x-h-3)(x-h+4)(x-h-5)
$$
So for this to be zero, one of the terms in the product must be zero. So $x-h-3=0$ or $x-h+4=0$ or $x-h-5=0$. Then $x=h+3, h-4, h+5$ are the solutions. Can you start plugging in $h$ values on a number line and see which $h$'s will give you only one positive solution, i.e. so that only one of those $3$ values will be positive?
For instance, if $h=0$, then the solutions are $x=3,-4,5$, but that is two positive answers so $h=0$ does not work. However, if $h= -4$, then the solutions are $x= -4,-8,1$ and that does work! Play around with a few values on the number line and you should come to a prediction about which ones work and which do not. Staring at the $+3,-4,+5$ in the $x$ solutions and you should start to see why your guess is correct, finally bringing you to the final interval(s) which give valid values for $h$.
As for your sons comment, this is probably just a misunderstanding that asking to 'solve for' a variable must result in only one answer. But this will not always be the case. For instance, if I want to 'solve' for $x$ and $y$ in $xy=0$, notice once one of $x,y$ is zero, the product is $0$. So the other variable can be anything! So there are infinitely many possible values for $x,y$ that make this equation hold. One the other hand, an equation like $x^2=-1$ has no (real) solutions. So when you solve for a variable, you can have none, one, a few, or even infinitely many solutions. 
