Roots of a polynomial having arbitrary constant term

Question

Let $f_r(x)=x(x-1)(x-2)(x-3)+r$, where $r$ is a real number. Then

$(A)$ $f_r(x)$ has a real root only for finitely many values of $r$;

$(B)$ $f_r(x)$ has a real root for every value of $r$;

$(C)$ $f_r(x)$ can never have a repeated root;

$(D)$ $f_r(x)$ can have a repeated root only for finitely many values of $r$.

My attempt: Considering a function $\phi(x)=f_r(x)-r$, then the all the roots of $\phi(x)$ are real and distinct and so $\phi'(x)$ is also so. Also $\phi'(x)={f_r'}(x)$. So the roots of ${f_r'}(x)$ are all real and distinct, but we don't say that roots of $f_r(x)$ are all real and distinct (since converse may not not true). I am stuck here.

Another way is by choosing different value of $r$ solve $f_r(x)$ and discard one by one option, but this is too broad.

The problem appears in ISI B-Math 2006 admission test. Thank's for your valuable time.

Answer 1

Hint: consider the shape when $r=0$: it's a quartic with four roots, at $0, 1, 2, 3$, with positive leading coefficient. What, then, does that quartic look like?

Now translate that shape vertically. How does the number of roots change as we do this?

Answer 2

Note that this is a polynomial with four real roots. So you can visualize what the graph looks like.

As x approaches negative infinity y will approach positive infinity.

As x approaches positive inifinity y will approach positive infinity as well.

The shape of the graph is like a "W"

As we vary r the shape will not change, only the graph will shift up or down.

As we increase r the graph will move up and as we decrease r the graph will move down.

But notice that since r never changes the relative placement of the roots, and the function has no repeated root for $r=0$, this proves that the function can never have any repeated roots.

Answer 3

Consider the equation without the constant. It has a plot that looks like this.

What happens when you add the constant $r$. It moves the graph up, i.e values of $f(x)$ at all $x$ are increased by $r$. Now how much can you increase or decrease the $r$. It should be obvious from looking at the graph that you can move the graph down as much as you want and you will always have two roots. But of you move the graph up by more than $1$ then it all the roots start to disappear. So $r$ can't be larger than $1$.

If you reduce the value of $r$ the function comes down, it is possible to chose a value for $r$ such that the maxima point just touches the x axis. This means that $x=1.5$ can be a repeated root of $f(x)$ for one value of $r$. Hence the option $D$ is the correct one.

Answer 4

A) false, $f_r(0)=r$ and $f_r(4)=24+r$, so for all $r\in(-24,0)$ there is at least a change of sign.

B) false, $x(x-1)(x-2)(x-3)>-\dfrac{81}{16}^*$ and there is no real root for $r=6$.

C) false, the polynomial is bounded below so it has at least a local minimum (zero of the derivative), and setting $r$ to be the opposite of this minimum yields a multiple root (simultaneous zero of the function and derivative).

D) true, there are at most three local extrema (three real roots of the derivative).

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$^*$ The minimum of $(x-a)(x-b)$ occurs at $x=\dfrac{a+b}2$ and has the value $-\dfrac{(b-a)^2}4$. Hence

$$x(x-3)\ge -\frac94\text{ and }(x-1)(x-2)\ge-\frac94$$ and the claim follows because the two minima coincide.