---
title: "Advanced Algebra: Simplifying Polynomials, Compound Interest"
---

# Simplifying Polynomials and Compound Interest

## Simplifying Polynomials: Overview

The basic goal that you have, throughout all these problems, is to take a scary-looking, complicated expression and simplify it into a form where someone could look at it and immediately see what the **solutions** are. For example, if you show someone the expression

$$
x^3 - 6 x^2 + 11 x - 6 = 0
$$

they're going to have no idea what values of $x$ make this true... BUT, if you do the work to factor this, it turns out to be just

$$
(x-1)(x-2)(x-3),
$$

and when it's written out this way, you can immediately see that the solutions are $x = 1$, $x = 2$, and $x = 3$.

## Domain Restrictions

The only tricky thing on top of this, that your teacher is adding in, is the idea of **domain restrictions**. But all this really means is, what values are **well-defined** in these equations? Or, put another way, are there cases where you can "break" math? We know that dividing by zero is *undefined*, so that if we see

$$
\frac{x-1}{x-2} = 0,
$$

we can see from the **numerator** that plugging in $x = 1$ makes this true:

$$
\frac{(1) - 1}{(1) - 2} = \frac{0}{-1} = 0 \; ✅
$$

but *also* from the **denominator** that plugging in $x = 2$ "breaks" this equation:

$$
\frac{(2)-1}{(2)-2} = \frac{1}{0} = \; ??? \text{ (undefined)} \; ❌
$$

So, we now have the **two things** we're looking for in all these problems:

1. The $x$ values that make the equation *true*, and
2. The $x$ values that *break* the equation.

And, our **toolbox** for finding both of these is, those three bullet points you had in your notes, which I'm trying to replicate from memory:

* Factoring
* Cancellation
* Domain Restrictions

## Practice Problems

1. Simplify the expression:

$$
\frac{x^3 + x^2 - 20x}{5x}
$$

First, we see if we can factor anything from the numerator or denominator. And, in this case, we see that all of the terms in the numerator have an $x$:

$$
\frac{x^2 + x^2 - 20x}{5x} = \frac{x(x^2 + x - 20)}{5x},
$$

so that now we can cancel the $x$ in the numerator and the $x$ in the denominator:

$$
\frac{x(x^2 + x - 20)}{5x} = \frac{x^2 + x - 20}{5}
$$

Now, you might be tempted to try and cancel the $20$ in the numerator and the $5$ in the denominator, but, remember! that the $20$ is part of an **addition/subtraction** (subtraction in this case), meaning that it **cannot be separated** from the other terms in the numerator.

So, instead, let's see if we can do the "fancier" kind of factoring, where we factor the polynomial $x^2 + x - 20$ into something like $(x - a)(x + b)$. In this case, we see that $20$ can be factored into $5 \cdot 4$, and that $5 - 4 = 1$, the coefficient on $x$, so that we can factor the numerator:

$$
\frac{x^2 + x - 20}{5} = \frac{(x+5)(x-4)}{5}
$$

Now, it might look like we're done, since now someone looking at this could immediately see, "oh, the solutions here will be $x = -5$ and $x = 4$". **But**, remember the final bullet point from above: **domain restrictions**. Notice how, when we cancelled out the $x$ term above, we crossed it out in both the numerator **and** the denominator. So, what this means is, even though it looks like $x = 0$ is perfectly fine to plug in at the end, it's actually "hiding" the fact that $x = 0$ will "break" the expression. We can see this by just taking our original expression and plugging in $0$:

$$
\frac{(0)^3 + (0)^2 - 20(0)}{5(0)} = \frac{0}{0},
$$

which is **undefined** since it has a $0$ in the denominator. So, remember: the nice, factored and simplified form we got at the end was **hiding** this undefined value, so we have to make sure to go back and find all the places where we **removed something from the denominator**, and record these removals by indicating the **domain restrictions** they imply. In this case, this means that we write the simplified from **along with** the domain restrictions, so that our final answer is:

$$
\frac{(x+5)(x-4)}{5}, \; x \neq 0.
$$

2. Simplify the expression

$$
\frac{x^2 + x - 6}{x} \div \frac{x^2}{2x^2 + 2x - 12}
$$

The issue here is that, if we forget to **flip the second term** into its reciprocal to turn the whole thing into a multiplication --- meaning, if we just forget that $\div$ is the division symbol and accidentally see it as a multiplication symbol, we get tricked into thinking a ton of stuff cancels out:

$$
\frac{x^2 + x - 6}{x} \cdot \frac{x^2}{2x^2 + 2x - 12} = \frac{(x+3)(x-2)}{x}\cdot \frac{x^2}{2(x+3)(x-2)} = \frac{x}{2} \; \text{ (WRONG!)}
$$

But, this is a **trap**, don't fall into it! When you see the division symbol $\div$, the first thing you should do is flip the second term to turn the whole expression into a **multiplication** instead of a division:

$$
\frac{x^2 + x - 6}{x} \div \frac{x^2}{2x^2 + 2x - 12} = \frac{x^2 + x - 6}{x}\cdot \frac{2x^2 + 2x - 12}{x^2}.
$$

*This* form will get you the right answer, since you can proceed with this the same way as you've been doing all the other, non-division-based problems:

$$
\frac{x^2 + x - 6}{x}\cdot \frac{2x^2 + 2x - 12}{x^2} = \frac{(x+3)(x-2)}{x}\cdot \frac{2(x+3)(x-2)}{x^2} = \frac{2(x+3)^2(x-2)^2}{x^3},
$$

and unfortunately this can't be simplified any further. In this case, the domain restrictions are pretty easy to see, since at every point the only thing in the denominator is $x$, $x^2$, or $x^3$, and in all those cases this means that only $x = 0$ will "break math". So, the final answer is:

$$
\frac{2(x+3)^2(x-2)^2}{x^3}, \; x \neq 0.
$$

And, in case your teacher asks, the fact that $(x+3)$ appears twice and $(x-2)$ appears twice means that we have solutions $x = -3$ with degree $2$ and $x = 2$ with degree $2$. (If they only appeared once each, we'd have $x = -3$ with degree $1$ and $x = 2$ with degree $1$, but for degree-$1$ solutions we usually just don't mention the degree, since degree-$1$ is assumed).

## More Practice Problems

For these problems, I'm providing solutions (that default to hidden!) but less explanation, since the above two cover the most difficult cases, imo.

3. Simplify the expression

$$
\frac{6x^2 + 5x - 4}{9x^2 - 16} \div \frac{4x^2 - 1}{9x^2 - 24x + 16}
$$

:::spoiler **Click to show/hide solution**

$$
\frac{3x - 4}{2x + 1}, x \neq \pm \frac{4}{3}, x \neq \pm \frac{1}{2}
$$

:::

 

4. A rectangle has a width of $2x$ and a height of $x - 3$. Find the ratio of its **perimeter** to its **area**.

:::spoiler **Cick to show/hide solution**

In this case, we break the problem up into: get the numerator by writing out an equation for the rectangle's perimeter, then get the denominator by writing out an equation for the rectangle's area. The perimeter, we know, is just the sum of the side lengths. So in this case, since we have two sides of length $2x$ and two sides of length $x - 3$, this gives us a perimeter of

$$
2(2x) + 2(x-3) = 4x + 2x - 6 = 6x - 6 = 6(x-1)
$$

Now, we also know that the area of a rectangle is $\text{width} \times \text{length}$, so in this case our area is

$$
(2x)(x-3) = 2x^2 - 6x.
$$

Taking these two together, then, the ratio of perimeter to area is

$$
\frac{6(x - 1)}{2x(x-3)} = \frac{3(x+1)}{x(x-3)}.
$$

Sadly (like most real-world problems), there's not much cancellation, but we did cancel the $\frac{6}{2}$ into just $3$ in the numerator. Then, we just have to figure out our domain restrictions, and we're done. Final answer:

$$
\frac{3(x+1)}{x(x-3)}, \; x \neq 0, x \neq 3.
$$
:::

 

## Compound Interest Practice Problems

For these, the idea is I'm just picking each element of the compound interest formula (I know there are multiple, but they all have generally the same "pieces"), and creating a scenario where it's missing and you need to solve for it.

5. Missing rate $r$: Avery puts $10,000 into a bank account, which compounds continuously for 5 years, at which point they see that they now has $12,000. Find the rate of interest of their bank account.

**Solution:** I'm going to use $M(t)$ to represent the "output" of the continuous compounding equation, meaning that $M(t)$ represents the amount of Money you have at time $t$:

$$
M(t) = Pe^{rt}.
$$

In this case, we're given the $P$ value ($10,000), the $t$ value (5 years), and the $M(t)$ value ($12,000, after 5 years), so that all that's left as a variable is $r$, which we need to solve for:

$$
$12,000 = $10,000 e^{r(5)}.
$$

In this case, the first thing we can do to make our lives a bit easier is to divide both sides by $10,000, to simplify the equation before we worry about solving for $r$:

$$
$12,000 = $10,000 e^{r(5)} \Leftrightarrow \frac{6}{5} = e^{r(5)}
$$

Now, since we have an $r$ we need to solve for, but it's "trapped" up in the exponent above $e$, we need to pull out our $\log$ function, which (if we apply it to both sides) will bring the $r$ down from being an exponent to being a multiplication term. Since we see that our **base** is "Euler's number" $e$, we know that we can use the natural logarithm $\ln(x)$ to ensure that the $e$ will get cancelled out by the log:

$$
\frac{6}{5} = e^{r(5)} \Leftrightarrow \ln\left( \frac{6}{5} \right) = \ln\left( e^{r(5)} \right),
$$

and now we can use the log identities we know to simplify this! First, we know that

$$
\log\left( \frac{m}{n} \right) = \log(m) - \log(n),
$$

which means that the left side turns into

$$
\ln(6) - \ln(5).
$$

Then, we can use another log identity,

$$
\log \left( m^n \right) = n \log(m),
$$

to simplify our right-hand side:

$$
\ln\left( e^{r(5)} \right) = r(5)\ln(e).
$$

Since we know that the natural log $\ln()$ is the **inverse** of the exponential $e$, $\ln(e) = 1$, so that this simplifies to just $5r$. Therefore, putting the two sides together, we now have

$$
\ln(6) - \ln(5) = 5r.
$$

Solving for $r$, we get

$$
r = \frac{\ln(6) - \ln(5)}{5},
$$

which we can solve on our calculators or [type into WolframAlpha](https://www.wolframalpha.com/input?i=%28ln%286%29-ln%285%29%29%2F5) to get

$$
r \approx 0.036,
$$

meaning that the **rate** of continuous compounding in Avery's bank account was about $3.6\%$.

6. Missing principal $P$: Blair puts some amount of money into a bank account with a 5\% interest rate, compounded continuously, and finds that after 11 years it has grown to be $50,000. How much did Blair put into the account initially?

Using the same basic logic as in the previous problem, we see that they've given us values for $r$ (5\%), $t$ (11 years), and $M(t)$ ($50,000). So, we need to solve for the missing piece, the principal $P$ (remember to convert the interest rate $r$ from a **percentage** into a **decimal** when using this equation!):

$$
$50,000 = Pe^{(0.05)(11)}
$$

In this case, we actually have it easier than in the last problem, since we don't need to use logs at all: the variable we want to solve for, $P$, is already a multiplication term, not an exponent. So, to solve for it, we just have to divide the $e$ term over to the other side!

$$
$50,000 = Pe^{(0.05)(11)} \Leftrightarrow P = \frac{$50,000}{e^{(0.05)(11)}} = \frac{$50,000}{e^{0.55}},
$$

so we can calculate this or type it into [WolframAlpha](https://www.wolframalpha.com/input?i=%28%2450000%29%2F%28exp%280.55%29%29) (we can even leave the dollar sign $ in, and Wolfram will make sure the answer stays in the unit of dollars! But, we make sure to write "exp(0.55)" in place of $e^{0.55}$, to make sure that it knows we're talking about the specific constant $e$ not just some unknown variable named $e$) to get

$$
P \approx $28,847.50.
$$

So, at the beginning, they put about $28,847.50$ in the account, which grew to $50,000$ after 11 years of continuously-compounding interest.

## Additional Practice Problems

These I give less explanation for, but I do give answers!

7. Missing $M(t)$: Cedar puts $8,500 into a bank account with an annual interest rate of 8\%. How much will Cedar have in this account 7 years later?

:::spoiler **Click to show/hide solution**

Note that here we do **not** have compound interest, just annual interest, so we need to use the formula

$$
M(t) = P\left(1 + \frac{r}{n} \right)^{nt}.
$$

In this case,

$$
M(t) = $8,500\left( 1 + \frac{0.08}{1} \right)^{1(7)} = $8,500(1.08)^7,
$$

and we can calculate or type into WolframAlpha to get

$$
M(t) = $14,568.00
$$

:::

 

8. Missing $t$: Devyn put $5,000 into a bank account, with interest compounding quarterly at a rate of 3\%, and came back at some later time to withdraw the entire account balance, which was $6,700. How many years passed between the initial deposit and the withdrawal?

:::spoiler **Click to show/hide solution**

Since we have interest compounding **quarterly**, our $n$ value in the non-continuous interest formula is going to be $4$ (since $n$ means "number of times the compounding happens in a year"). Remembering to include $n$ **both inside the parentheses and in the exponent**, we get

$$
$6,700 = $5,000 \left( 1 + \frac{0.03}{4} \right)^{4(t)}
$$

which simplifies to

$$
\frac{67}{50} = 1.0075^{4t} \Leftrightarrow \log(67) - \log(50) = 4t \log(1.0075) \Leftrightarrow t = \frac{\log(67) - \log(50)}{4 \log(1.0075)},
$$

and we calculate or [plug into WolframAlpha](https://www.wolframalpha.com/input?i=%5Cfrac%7B%5Clog%2867%29+-+%5Clog%2850%29%7D%7B4+%5Clog%281.0075%29%7D) to get

$$
t \approx 9.79
$$

And, since it just asks for years, we can just put "9.79 years" as our final answer. But, if they asked for something weirder, like years and days, we could take that $.79$ and convert it into days by [multiplying by 365](https://www.wolframalpha.com/input?i=%280.79%29*%28365%29):

$$
(0.79)(365) \approx 288,
$$

so we could say "9 years and 288 days" in this case.

:::

9. Half-life problem:

[These are a bit weird, because they mean you just *leave* $P$ alone as a variable, since the $P$ on the left side and $P$ on the right side of the equation will cancel out.]

The half-life of a newly-discovered element, Jeffonium, is 3 years. [Remember, this just means that if you start with some amount of Jeffonium $P$, and wait three years, you'll have half of the original amount, $P / 2$] What is this element's rate of decay?

:::spoiler **Click to show/hide solution**

The values they give us allow us to write an equation for this exponential decay (the exponent is *negative* because this is a process of *decay* rather than *growth*):

$$
\frac{P}{2} = Pe^{-r(3)}
$$

And by writing this out, we see why it's ok for us to just leave $P$ as a variable instead of trying to solve for it or plug something in for it: we can just divide both sides by $P$ to get rid of it entirely:

$$
\frac{P}{2} = Pe^{-r(3)} \Leftrightarrow \frac{1}{2} = e^{-3r}.
$$

So, using logs to get the $r$ we want to solve for out of the exponent:

$$
\ln \left( \frac{1}{2} \right) = \ln \left( e^{-3r} \right) \Leftrightarrow \ln(1) - \ln(2) = -3r \Leftrightarrow r = -\frac{\ln(1)-\ln(2)}{3},
$$

and calculator or [WolframAlpha](https://www.wolframalpha.com/input?i=-%28ln%281%29-ln%282%29%29%2F3) give us:

$$
r \approx 0.2310,
$$

meaning that the rate of decay is approximately $23.1\%$.

:::