Factor Theorem: When (x+k) Is A Factor Of F(x)

Hey math whizzes! Let's dive into a super cool concept in algebra: the Factor Theorem. Ever wonder what it really means when we say $(x+k)$ is a factor of a polynomial $f(x)$? It sounds fancy, but it's actually a pretty straightforward idea that unlocks a lot of problem-solving power. We're going to break down exactly what this means and why it's so important, especially when you see questions like, "If $(x+k)$ is a factor of $f(x)$, which of the following must be true?" Get ready to totally nail these types of problems, guys!

Understanding Factors and Roots

Before we get into the nitty-gritty of the Factor Theorem, let's quickly recap what 'factors' and 'roots' are in the context of polynomials. Think of a polynomial, $f(x)$, like a mathematical expression made up of variables (like $x$) and constants, combined using addition, subtraction, and multiplication. For example, $f(x) = x^2 - 4$ is a polynomial. Now, when we talk about factors of a polynomial, we're talking about smaller polynomials that divide evenly into the original polynomial. So, for $f(x) = x^2 - 4$, we know we can factor it as $(x-2)(x+2)$. In this case, $(x-2)$ and $(x+2)$ are factors of $f(x)$. Pretty neat, right?

Now, what about roots? The roots of a polynomial $f(x)$ are the values of $x$ that make the polynomial equal to zero. So, if $f(x) = x^2 - 4$, we can find the roots by setting $f(x) = 0$: $x^2 - 4 = 0$. Solving this gives us $x^2 = 4$, so $x = 2$ and $x = -2$. These are the roots of the polynomial. Notice something interesting? The roots are closely related to the factors. If $(x-a)$ is a factor of $f(x)$, then $x=a$ is a root of $f(x)$. Conversely, if $x=a$ is a root of $f(x)$, then $(x-a)$ is a factor of $f(x)$. This connection is key to understanding the Factor Theorem.

The Factor Theorem Explained

Alright, let's get to the main event: the Factor Theorem. This theorem is a direct consequence of the Remainder Theorem, which states that when a polynomial $f(x)$ is divided by $(x-c)$, the remainder is $f(c)$. The Factor Theorem takes this one step further. It says that $(x-c)$ is a factor of $f(x)$ if and only if $f(c) = 0$. In simpler terms, $(x-c)$ divides $f(x)$ evenly (meaning the remainder is zero) if and only if $c$ is a root of $f(x)$.

Now, let's apply this to our specific question: "If $(x+k)$ is a factor of $f(x)$, which of the following must be true?" Our theorem talks about $(x-c)$ being a factor, and our problem gives us $(x+k)$. We need to make these match up. We can rewrite $(x+k)$ as $(x - (-k))$. So, in the context of the Factor Theorem, our 'c' value is actually $-k$.

According to the Factor Theorem, if $(x - (-k))$ (which is the same as $(x+k)$) is a factor of $f(x)$, then the value of the polynomial when $x$ is equal to $-k$ must be zero. In other words, $f(-k) = 0$. This is a fundamental rule, guys, and it's super powerful for solving polynomial equations.

Let's look at the options given in the question:

• A. $f(k)=0$: This would mean that $x=k$ is a root. But our factor is $(x+k)$, which corresponds to a root of $x=-k$, not $x=k$. So, this is generally not true.
• B. $f(-k)=0$: This aligns perfectly with the Factor Theorem. If $(x+k)$ is a factor, then substituting $x=-k$ into the polynomial $f(x)$ should result in 0. This must be true.
• C. A root of $f(x)$ is $x=k$: Similar to option A, this suggests $k$ is a root. However, as we've established, the factor $(x+k)$ implies a root of $x=-k$. So, this isn't necessarily true. It could be true if $k=0$, but the theorem applies generally.
• D. A $y$-intercept of $f(x)$ is $x=-k$: The $y$-intercept of a function $f(x)$ is the value of $y$ when $x=0$. So, the $y$-intercept is $f(0)$. The statement is saying that $f(0) = -k$. This has no direct relationship with $(x+k)$ being a factor of $f(x)$ and $f(-k)=0$. The $y$-intercept occurs at $x=0$, while our factor implies a root at $x=-k$. These are different points on the graph.

Therefore, the statement that must be true is $f(-k)=0$. This is the direct application of the Factor Theorem. It's such a crucial concept that it's worth drilling into your heads!

Why Does This Work? The Remainder Theorem Connection

The Factor Theorem is essentially a special case of the Remainder Theorem. The Remainder Theorem states that if you divide a polynomial $f(x)$ by a linear expression $(x-c)$, the remainder of that division will be equal to $f(c)$. You can write this relationship as:

$f(x) = q(x)(x-c) + R$

where $q(x)$ is the quotient and $R$ is the remainder. Now, if we substitute $x=c$ into this equation, we get:

$f(c) = q(c)(c-c) + R$ $f(c) = q(c)(0) + R$ $f(c) = R$

This shows us that the value of the polynomial at $x=c$ is indeed the remainder when $f(x)$ is divided by $(x-c)$.

Now, for the Factor Theorem, we're interested in the case where $(x-c)$ is a factor of $f(x)$. If something is a factor, it means it divides the polynomial evenly, with no remainder. In other words, the remainder $R$ must be 0.

So, if $(x-c)$ is a factor of $f(x)$, then $R=0$. From the Remainder Theorem, we know that $f(c) = R$. If $R=0$, then it must be that $f(c) = 0$. This is exactly what the Factor Theorem states!

Let's re-apply this logic to our problem where the factor is $(x+k)$. We need to express $(x+k)$ in the form $(x-c)$. We can do this by writing $(x+k)$ as $(x - (-k))$. So, in this case, $c = -k$.

According to the Factor Theorem, $(x - (-k))$ is a factor of $f(x)$ if and only if $f(-k) = 0$. This means that when you substitute $-k$ for $x$ in the polynomial $f(x)$, the result will be 0. This is why option B, $f(-k)=0$, is the correct answer. It’s a direct application of the theorem derived from the Remainder Theorem, and it’s a fundamental property connecting factors and roots of polynomials.

Visualizing with Graphs

Let's think about what $f(-k)=0$ means graphically. Remember, the roots of a polynomial are the $x$-values where the graph of the function $f(x)$ crosses or touches the $x$-axis. These points are also called the $x$-intercepts.

If $f(-k)=0$, it means that when $x$ takes on the value $-k$, the output of the function, $f(x)$, is 0. On a graph, this translates to the point $(-k, 0)$ being on the $x$-axis. Therefore, $x = -k$ is an $x$-intercept, which also means $x = -k$ is a root of the polynomial $f(x)$.

Since the Factor Theorem states that $(x-c)$ is a factor if and only if $f(c)=0$, and we have the factor $(x+k)$, which we rewrote as $(x-(-k))$, it means that $c = -k$. Thus, $f(-k)=0$. This tells us that $-k$ is a root, and graphically, the function's graph will intersect the $x$-axis at $x=-k$. This graphical interpretation reinforces the algebraic concept.

Consider the options again in light of the graph:

• A. $f(k)=0$: This would mean the graph intersects the $x$-axis at $x=k$. This is not necessarily true just because $(x+k)$ is a factor.
• B. $f(-k)=0$: This means the graph intersects the $x$-axis at $x=-k$. This is true because $(x+k)$ is a factor.
• C. A root of $f(x)$ is $x=k$: Same as A, it means the graph intersects the $x$-axis at $x=k$. Not necessarily true.
• D. A $y$-intercept of $f(x)$ is $x=-k$: The $y$-intercept is where the graph crosses the $y$-axis, which happens when $x=0$. So the $y$-intercept is $f(0)$. The statement suggests $f(0) = -k$. This is unrelated to the factor $(x+k)$ and the root $x=-k$, unless $k=0$ and $f(0)=0$, which is a specific case.

So, the graphical perspective confirms that $f(-k)=0$ is the only statement that must be true when $(x+k)$ is a factor of $f(x)$.

Putting It All Together: Example Time!

Let's solidify our understanding with a concrete example. Suppose we have the polynomial $f(x) = x^2 + 5x + 6$. We want to know what must be true if $(x+2)$ is a factor of $f(x)$.

Here, our factor is $(x+2)$. We can write this in the form $(x-c)$ as $(x - (-2))$. So, $c = -2$.

According to the Factor Theorem, if $(x+2)$ is a factor of $f(x)$, then $f(-2)$ must be equal to 0.

Let's check: $f(-2) = (-2)^2 + 5(-2) + 6$ $f(-2) = 4 - 10 + 6$ $f(-2) = 0$

It works! So, $f(-2)=0$ is true.

Now let's consider the other possibilities:

• Is $f(2)=0$? $f(2) = (2)^2 + 5(2) + 6 = 4 + 10 + 6 = 20$. So, $f(2) eq 0$. Option A is false.
• Is $x=2$ a root? No, because $f(2) eq 0$. Option C is false.
• Is the $y$-intercept $x=-2$? The $y$-intercept is $f(0) = (0)^2 + 5(0) + 6 = 6$. So, the $y$-intercept is 6 (at $x=0$), not $-2$. Option D is false.

This example clearly demonstrates that when $(x+k)$ is a factor of $f(x)$, the only statement that must be true is $f(-k)=0$. It's a direct consequence of the Factor Theorem and its connection to the roots of the polynomial. So, the next time you see a question like this, just remember to match your factor to the $(x-c)$ form, find your $c$, and then apply $f(c)=0$. You guys got this!