# Finding the total number of roots of a graph

So I was given this graph: and I was told the function was defined as:

$$f(x) = (x-c_1)(x-c_2)...(x-c_n)$$

Then I was told to find n. So I assumed they meant the total number of real roots. So since from the graph I could tell there were 2 real roots, then it was impossible for it to be 2 reals roots and 1 non-real root. So the answer must be 3 real roots. It was correct, but their explanation was:

The graph crosses the x‑axis at −10, so f must have a positive odd number of factors corresponding to that x‑intercept

The graph crosses the x‑axis at 15, so f must have a positive even number of factors corresponding to that x‑intercept

There are no other x‑intercepts, so this accounts for all the factors. The total number of factors of f is, therefore, an odd number plus an even number, and must be at least 3.

I don't understand their explanation, because they first say f must have a positive odd number of factors, then they say it has positive even number of factors, and then they say there are no other x-intercepts? when there is another third factor? Does anyone know what is meant by that explanation?

• They probably mean roots with multiplicity. The term $(x-15)$ appears at least twice in the factorization (otherwise the sign would change at $15$) – Astyx May 11 '17 at 20:29

## 4 Answers

The third root comes from the fact that one of the zeros is tangent to the x-axis, i.e. it is touching the x axis but not crossing through. Kind of odd, but let me try to explain why.

Imagine the function was shifted down a small amount, then there would clearly be three zeros. Now let's slowly shift the function back to its original position. As you do, the two zeros on the right would get closer and closer to each other until finally when the function finally reaches its original position, both zeros have merged into one single position. That's why it seems like the third zero is kind of "hidden".

This is the graph of $$P(x)=(x+10)(x-15)^2$$ You are forgetting about the multiplicities of the roots of a function. If a function has a zero crosses the x-axis at $x=a$, then $(x-a)$ is a factor of $P(x)$ with odd multiplicity. If it touches the x-axis without crossing at $x=a$, then $(x-a)$ is a factor of $P(x)$ with even multiplicity. For example, here is the graph of $y=(x-2)(x-4)^2(x-6)^3$: Notice how at the root of multiplicity $1$, the function behaves linearly, and at the root of multiplicity $2$, it behaves like a quadratic, and at the root of multiplicity $3$, it behaves like a cubic. This is because as the multiplicity of a factor increases, its "influence" on the value of $P(x)$ increases, so $P(x)$ begins to behave as it would when that factor approaches zero.

There is obviously an error in their explanation (or your quoting of it):

The graph crosses the x‑axis at 15,

is obviously not correct, the graph touches the $x$-axis (or it looks so) but doesn't cross it. Correcting that makes it more reasonable that the conclude different about the multiplicity of that root than they do about the root at $x=-10$, where they also writes that the graphs crosses the $x$-axis.

Hint:

$x=15$ is (at least) a double root because here the graph touch the $x$ axis and also the first (at least) derivative of the function is null.