# Page 1-Math: Passport to Advanced Math Study Guide for the SAT Exam

## General Information

Unlike questions about the “Heart of Algebra”, these 16 questions require you to use and manipulate expressions and equations as well as to interpret and build functions. Additionally, the equations with which you work on this type of question tend to be more complex. Some questions will be presented in the calculator-allowed section and some questions will not.

Look through these topics and be sure you know all the ins and outs regarding them. If you find things with which you are not totally comfortable, be sure to seek extra practice in those areas.

## Function Notation

We often wish to formulate a relationship between two quantities, such as how far a vehicle travels over how much time, or how much profit a restaurant will have based on its online ratings. Using these relationships, we can use time to calculate distance traveled or restaurant rating to calculate profit. Because we’re free to plug in whatever value we want for time of rating, we call these quantities variables. Furthermore, we use the term function to describe relationships like these – that is, ones that associate one value to each value the input variable could have. As you will see, the input doesn’t have to be just a variable but can be an expression, so we more generally refer to a function’s input as its argument

If the function is $$f$$ and its input is $$x$$, we write the function as $$f(x)$$ to make it clear that the value of $$f$$ depends on the value of $$x$$. For example, if we wanted $$f$$ to be a function that doubles and adds one, we would write:

$f (x) = 2x + 1$

We could just as well write:

$f (a) = 2a + 1$

The variable in the argument is just a placeholder. In fact, we could even insert another expression in its place, such as::

$f(b - 3) = 2(b - 3) + 1$

The notation just means, “When you replace the placeholder variable in the parentheses with something, replace it with the same something on the other side of the equation.”

When you first learned multiplication, you were probably told to think of it as repeated addition, for example:

$a \times 3 = a + a + a$

Exponents, also known as powers, can first be thought of as repeated multiplication:

$a^3 = a \times a \times a$

We’ll build off this idea to include non-integer powers as well. As for terminology, in $$a^b$$, we call $$a$$ the base and $$b$$ the exponent, or power.

### Rules for Exponents

For any numbers a, b and c, these rules apply:

1) $$a^b \times a^c\;= \;a^{b+c}$$

We define it this way because this is what we’d expect for positive integers:

$a^2 \times a^3 = (a \times a) \times (a \times a \times a) = (a \times a \times a \times a \times a) = a^5$

Now what should a negative exponent mean? Using the first rule:

$a = a^{2 - 1} = (a \times a) \times a^{-1}$

Divide both sides by $$a$$ and we have:

$1 = a \times a^{-1}$

Divide both sides by $$a$$ again to find:

$a^{-1} = \frac{1}{a}$

We can raise this to any power, $$b$$, and use the first rule again to find the second rule:

2) $$a^{-b} =\frac{1}{a^b}$$

That also shows us what we get when we raise to the power of zero:

$a^0 = a^{1-1} = a \times a^{-1} = \frac{a}{a} = 1$

3) $$a^0 = 1$$

Now we know how exponents work for any integer. To go further, we need to think about integers again. Let’s raise a number to two powers:

$(a^3)^2 = (a \times a \times a) \times (a \times a \times a) = a^{3 + 3} = a^{3 x 2}$

We, therefore, write the next rule:

4) $$(a^b)^c = a^{b \, \times \, c}$$

What about fractional exponents? We use the rule we just found:

$a = a^1 = a^{\frac{b}{b}} =(a^b)^\frac{1}{b}$

So raising to a fractional power “undoes” raising to a whole number power. For example:

$$a^\frac{1}{2}$$ is the square root of $$a$$.

Be careful, though, squaring is not invertible:

$a^2 = (-1)^2 \times a^2 = (-a)^2$

So, in general, square roots (and any other even-number roots) can be either positive or negative. Keeping that in mind:

5) $$a^{\frac{b}{c}} = \sqrt[c]{a^b}$$

c√ ab means ”the cth root of ab”, and we call the “√” symbol for the cth root a radical.

One more rule, thinking of positive numbers one more time:

$(a \times b)^2 = (a \times b) \times (a \times b)=(a \times a)(b \times b) = (a^ 2)(b^2)$

Making this a general rule, we have

6) $$(a \times b)^c = (a^c)(b^c)$$

### Exponential Functions

Anytime a rate of some quantity’s change is proportional to the quantity itself, an exponential function describes the situation. The general formula is:

(current amount)=(initial amount)(proportion of change)time

For example, if I earn $$5\%$$ interest on an investment every year and I start out by investing $$\200$$, the formula for how much the investment will have grown to in $$t$$ years is:

$200 \times (1 + 0.05)^t$

Note that the proportion of the change is $$1.05$$, not $$0.05$$. The $$1$$ accounts for the amount already in the investment and the $$0.05$$ accounts for the interest.

We can have a different variable in the place of time in the formula above. As another example, suppose you earn twice as many points in a bean bag toss game for every ten feet further back you are when throwing, starting with $$10$$ points at the closest allowed distance. Then the number of points you could earn with a throw $$x$$ feet behind the closest line is:

$10 \times 2^\frac{x}{10}$

If we’re faced with an equation involving a radical, we need to get rid of it by raising each side to the power corresponding to the root in the radical. For example, if we have a square root in an equation, we need to square that square root to get out whatever is inside it. Let’s solve for $$x$$ in this equation:

$3 = 1 + \sqrt{(x-5)}$

If we square it right now, when we multiply out everything on the right side we’ll have:

$$1 + 2\sqrt{(x-5)}+(x-5)$$ and are still stuck with the radical. We need to isolate the radical first. Subtract $$1$$ from each side and we get:

$2 = \sqrt{(x-5)}$

Now we can get rid of the root by squaring both sides:

$$4 = x - 5$$
$$x = 9$$

We can plug that back in and check that the equation holds. It’s a good idea to check because, as we mentioned earlier, we can’t just “undo” squaring. There are both positive and negative roots.

## Polynomials

Polynomials are functions that can be written as sums of powers of variables. For example, these are all polynomials:

$$x^2 + x - 2$$
$$x^5 + x^4 - 7x^ 3 + x - 4$$
$$x + 1$$
$$2 = 2x^0$$

The powers in polynomials could be any fixed numbers, but not variables, and polynomials can have any number of terms. In general, we can write a polynomial like this:

$a_{n} x^{n} + a_{n-1}x^{n-1}+…a_{2} x^{2} + a_{1}x+a_{0}$

where $$n$$ is any positive integer and each of $$a_1$$, $$a_2$$ , and all the rest up to $$a_n$$ is any constant number (possibly zero). This isn’t the only way to write polynomials, though, as we will soon see.

### Operations

To prepare for exploring other ways of writing polynomials, we should review the order of operations and how it applies to polynomials. To illustrate the steps, let’s go through an example.

$(x^2 + 3)(x^2 - x -1)$

Remember the acronym PEMDAS? Parentheses come first. To deal with them, we distribute:

$$(x^2 - x - 1)$$ through $$(x^2 + 3)$$

to find

$(x^2 + 3)(x^2 -x-1) = x^2 (x^2 - x - 1) + 3(x^2 - x - 1)$

We could also have distributed in the other order, that is:

$$(x^2 + 3)$$ through $$(x^2 - x - 1)$$

and found the same result. Then we distribute again, using the rules of exponents to simplify:

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

That takes care of all the parentheses; next comes exponentiation. Since exponentiation comes before the rest (“MDAS”), we must keep different powers of $$x$$ separate; however, we can combine like powers. In this case, there are two $$x^2$$ terms we can combine.

$x^4 - x^3 - x^2 + 3x^2 - 3x - 3 =x^4 - x^3 + 2x^2 - 3x - 3$

This is the expanded form of the polynomial, just like we saw in the introduction.

### Analysis and Rewriting

Polynomials can also be written as a product of their factors. The factors of polynomials are simply smaller polynomials. For example, let’s look at this:

$x^3 - x^2 + 2x - 2$

This can be factored:

$x^3 - x^2 + 2x - 2 = (x^3 + 2x) - (x^2 + 2) = x(x^2 + 2) - (x^2 + 2) = (x^2 + 2)(x - 1)$

It’s not always easy to see what a polynomial’s factors are, especially since we often will have like terms that combine when the polynomial is expanded.

If you’re given one form of a polynomial and need to find the other, you can always use unknown numbers in the unknown form, compare it to the known form, and then deduce what the numbers must have been. For example, suppose we are given:

$x^3 + ax^2 + 2x - 2=(x^2 + b)(x - 1)$

where $$a$$ and $$b$$ are constant numbers, and we are asked to find them. We can expand the right side:

$x^3 +ax^2 + 2x - 2 = (x^2 + b)(x - 1) = x^3 -x^2 + bx - b$

Now we can compare coefficients, i.e. match the numbers multiplying like powers of $$x$$. One side has a multiplying $$x^2$$ and the other has $$-1$$, so $$a = -1$$. Comparing either of the last two coefficients similarly shows that $$b = 2$$.