In the attached table, which of the following must be a factor of f(x)?

Question

$$\begin{array}{c|c} x & f(x) \\\hline 0 & 3 \\ 2 & 1 \\ 4 & 0 \\ 5 & -2 \end{array}$$

The function $f$ is defined by a polynomial. Some values of $x$ and $f(x)$ are shown in the above. What must be a factor of $f(x)$?

The question is from a SAT practice test.

I'm not entirely sure what the question is asking.

At first I thought that the $x$ table values could be substituted in the equation and checked if they get the respective $f(x)$ table values. For example, in the equation $x-3$, if $0$ is $x$ then $f(x)$ should be $-3$. However, that doesn't work for any of the equations.

Then I thought, that the $x$ table values must be multiplied by the equations to get $f(x)$, for example, in the equation $x-3$ multiplied by $0$ if $x$ is $0$ to get $0(0-3)$, however that clearly doesn't work either.

So basically, I'm not sure where to start. According to the answer key, the answer is $(x-4)$, however I don't understand how to get to that answer, hence can anyone please provide a step by step explanation and solution to the problem.

Answer 1

So when you have some polynomial $g(x)$, for instance, consider: $$g(x)=x^4 -4x^3 +8x$$ another way of writing it is using it's zeros, that is, where the polynomial is equal to zero. In this case, we would have $$g(x)=-x (x - 2) (1 + \sqrt{5} - x) (-1 + \sqrt{5} + x)$$ getting a polynomial into this form sometimes takes quite a bit of hard work -- but an easy way of finding it in some cases is to look at its graph and see where it crosses the $x$-axis (because the $y$-axis is the value of $f(x)$, so $y=0$ if and only if $f(x)=0$). In the above example I gave, if you were to look at it's graph, you'll notice crosses the axis at $0$, $2$, $1+ \sqrt{5}$, and $1- \sqrt{5}$, precisely as indicated by its factored form.

Now using that above information you can see how that table helps you find the factors of $f(x)$. It tells you that at $x=4$ we have $f(4)=0$. That must mean somewhere in our function we have the factor $(x-4)$ -- because when $x$ is $4$, then the factored form contains a $0$ and since it's just one large product -- multiplying by zero causes the who function to be zero -- just as the table says.

Answer 2

The problem definition assumes that the polynomial $f(x)$ can be factored into $$ f(x) = g(x) h(x) $$ where $g(x)$ and $h(x)$ are polynomials. Furthermore, it can be shown that if $x=a$ is a root for the polynomial, then $x-a$ is a factor for it. This makes sense, since substituting, for ex. $h(x) = x-a$ gives us $$ f(x) = g(x) (x-a) $$ which clearly has a root at $x=a$ ($f(a)= 0$). Reading the table, the polynomial has a root at $x=4$, so therefore $x-4$ is a factor of $f(x)$.

Answer 3

The Factor Theorem states that a polynomial has factor $x - k$ if and only if $f(k) = 0$.

We are given $f(4) = 0$. Thus, by the Factor Theorem, $x - 4$ must be a factor.