Equivalent Expressions

PART II · ADVANCED MATH

Advanced Math

Introduction

Advanced Math is close in size to Algebra and centers on nonlinear expressions, equations, and functions. Unlike a linear relationship, a nonlinear relationship does not have a constant rate of change. Quadratic, exponential, polynomial, rational, radical, and absolute-value structures make up the core of this domain.

The skill that carried you through Algebra carries you through here without modification: choose the form that reveals what you need. A quadratic can be written three different ways, and each one places a different feature in plain sight, where the curve crosses zero, where it turns, where it meets the vertical axis. An expression can be factored or expanded, and the two forms answer different questions. A function can be read forward or backward. None of that habit is new; only the relationships it acts on have started to bend.

FIGURE 1 The parabola, the recurring object of this domain: roots, vertex, intercept, and the symmetry that ties them together.

Quadratics deserve particular attention because roots, vertex, y-intercept, and symmetry can all be read from different representations. Learn to read it for its landmarks; its roots where it crosses the horizontal axis, its vertex where it turns, its intercept on the vertical axis, and the axis of symmetry that mirrors one side onto the other, and a large share of the domain's questions reduce to pointing at the feature you were asked for. The same equation holds all of these at once. As in Algebra, the efficient route is to choose the representation that exposes the requested feature.

The questions to keep in mind shift only slightly from the last domain. What does this form reveal, and what does it keep hidden? What stays genuinely the same when I rewrite an expression, and what merely looks different on the page? And is the question asking for a value, a root, a turning point, or a behavior as the input grows large? Naming which of these is on the table keeps the algebra aimed at the right target instead of producing something true but unasked.

Here is the shape of what is ahead.

In this domain

Equivalent expressions, rewriting to reveal: factoring, expanding, and choosing the form that answers the question rather than fighting the one you were handed.

Nonlinear functions: notation, transformations, composition, and inverses: notation, inputs and outputs, domain, and reading a function in both directions, in symbols and in context.

Nonlinear functions: quadratic forms and features: the three forms and the feature each surfaces, the discriminant, and the graph read as evidence.

Nonlinear equations in one variable: radicals and rational expressions, and the extra solutions that rewriting can introduce and that must be checked.

Exponential growth, where change compounds instead of adding, and how it overtakes any linear rate given enough room.

If Algebra taught you to read a straight relationship no matter which form it takes, this domain asks the same of a bending one. The objects are richer and the graphs more interesting, but the move is the one you already trust: find the form that makes the answer visible, then read it off. The first skill is rewriting an expression into an equivalent form that makes the requested feature easier to identify.

ADVANCED MATH · 1

Equivalent expressions

The same expression can be written many ways without changing its value. The goal is to choose the form that makes the requested feature easiest to read.

Two expressions are equivalent when they produce the same value for every input; they are the same quantity written two different ways. Rewriting changes the form without changing the value. Equivalent rewriting is a central Advanced Math skill. A question that asks “which of the following is equivalent to…” is really asking whether you can move fluently among those forms; a harder question hands you one form and is answerable only after you change it into another. So the skill to build here is two-directional fluency (expanding a product into a sum, and factoring a sum back into a product) together with an eye for which direction the question rewards.

▸ Combine polynomials by adding and subtracting like terms

Before any expanding or factoring, two polynomials can simply be added or subtracted, term matched to term. Like terms, ones with the same variable raised to the same power, combine; unlike terms sit side by side unchanged. A subtraction is the one place a sign slip hides: distribute the minus across every term of the polynomial being subtracted before combining anything.

Add and subtract (3x² + 5x - 2) and (x² - 4x + 7):

Sum: 4x² + x + 5

Difference: 2x² + 9x - 9

The sum matches x², x, and constant terms straight across. The difference first flips every sign in the second polynomial (-x² + 4x - 7), then combines: the most common error is subtracting only the first term and leaving the rest added.

▸ Expand polynomial products

Expanding means carrying out every multiplication until nothing is left in parentheses. Each term in the first factor multiplies each term in the second, and you collect like terms at the end. For two binomials this is the familiar four-product pattern.

WORKED Expand (2x + 3)(x - 4)

Multiply every term in the first factor by every term in the second:

(2x)(x) + (2x)(-4) + (3)(x) + (3)(-4)

2x² - 8x + 3x - 12

2x² - 5x - 12

The two middle products combined; nothing was added or removed, only multiplied out.

There is a picture underneath this that makes the bookkeeping visible. A product of two lengths is an area, so a product like (a + b)(a + b) is the area of a square whose side is a + b, and splitting that square into pieces shows exactly which terms appear.

FIGURE 1 Why (a + b)² = a² + 2ab + b²: the square of side a+b splits into one a², one b², and two ab rectangles.

▸ Recognize difference-of-squares and perfect-square products

A few products appear so often that recognizing them instantly, in either direction, saves real time. They are not separate facts to memorize so much as three shapes to recognize.

Three products to recognize in both directions

Square of a sum: (a + b)² = a² + 2ab + b².

Square of a difference: (a - b)² = a² - 2ab + b².

Sum times difference: (a + b)(a - b) = a² - b², the middle terms cancel.

The third is worth dwelling on, because its middle terms cancel and leave a clean difference of two squares. Read backward, it is one of the most useful factoring patterns there is.

WORKED Two quick expansions

(x + 5)² = x² + 2(5)x + 25 = x² + 10x + 25.

(3x - 2)² = 9x² - 2(3x)(2) + 4 = 9x² - 12x + 4.

(x + 4)(x - 4) = x² - 16, no middle term survives.

▸ Factor polynomial expressions

Factoring reverses expansion: it takes a sum and writes it as a product. This is the more demanding direction, since it requires searching for the pieces rather than multiplying them together. A factored expression reveals features that a sum’s expanded form does not show directly. The first move is to pull out any common factor shared by every term.

WORKED Pull out the common factor

6x³ - 9x². Both terms share 3x².

6x³ - 9x² = 3x²(2x - 3)

Always check for a common factor first; it simplifies everything that follows, and a question that asks to factor completely will demand it.

▸ Use difference-of-squares and perfect-square factoring patterns

Beyond the common factor, a small set of patterns covers the great majority of what the test asks. Learning to recognize the shape of an expression tells you which one to reach for.

A difference of squares, two perfect squares with a minus between them, factors into a sum times a difference, the third product above read backward.

WORKED Difference of squares

9x² - 25 is (3x)² - (5)², so

9x² - 25 = (3x + 5)(3x - 5).

FIGURE 2 A difference of squares is a rectangle: removing a b² corner from an a² square leaves an (a + b) by (a − b) region.

A perfect-square trinomial is the expansion of a squared binomial, recognizable because its first and last terms are squares and its middle term is twice the product of their roots.

WORKED Perfect-square trinomial

x² + 10x + 25 has x² and 25 = 5², and the middle 10x = 2(5)x, the signature of a square.

x² + 10x + 25 = (x + 5)².

For a general trinomial x² + bx + c, you are looking for two numbers that multiply to c and add to b. The area picture shows why: the two numbers are the side lengths added onto x, so their product fills the constant corner and their sum builds the middle term.

FIGURE 3 Factoring x² + bx + c: the two numbers fill the corner (their product is c) and the side strips (their sum is b).

WORKED A trinomial with leading coefficient 1

x² + 7x + 12: find two numbers multiplying to 12 and adding to 7, namely 3 and 4.

x² + 7x + 12 = (x + 3)(x + 4).

When the leading coefficient is not 1, the same idea runs through a short detour. To factor ax² + bx + c, find two numbers multiplying to ac and adding to b, split the middle term using them, and factor by grouping.

WORKED A trinomial with a leading coefficient

2x² + 7x + 3: here ac = (2)(3) = 6, and two numbers multiplying to 6 and adding to 7 are 6 and 1.

Split the middle term and group:

2x² + 6x + 1x + 3

2x(x + 3) + 1(x + 3)

(2x + 1)(x + 3)

That last step, grouping, is a pattern in its own right, the tool for expressions with four terms. Group the terms in pairs, factor each pair, and if a common binomial appears, factor it out.

WORKED Factoring by grouping

x³ + 4x² + 2x + 8. Group and factor each pair:

x²(x + 4) + 2(x + 4)

(x² + 2)(x + 4)

The shared (x + 4) is the signal that grouping was the right move.

▸ Use factored form to identify zeros

Here is the payoff that makes factoring worth the effort. A product equals zero only when one of its factors equals zero, the zero-product idea. So the moment an expression is factored, the values that make it zero are sitting in plain view.

WORKED Reading the zeros

x² - 5x + 6 = 0 is hard to read as written, but factored it opens up:

(x - 2)(x - 3) = 0

A product is zero only if a factor is, so x = 2 or x = 3.

The factored form did not just simplify the expression, it handed over the answer. This is exactly why the quadratics ahead lean on factoring so heavily.

▸ Simplify rational expressions and preserve excluded values

A rational expression is a ratio of polynomials, and the way to simplify one is almost always the same: factor the top and bottom, then cancel what they share. A factor can be cancelled only when it appears in both, and one piece of care travels with it: the values that would make the original denominator zero are excluded from the start, even if cancelling appears to remove them.

WORKED Simplify a rational expression

(x² - 9) / (x² + x - 6). Factor both:

((x + 3)(x - 3)) / ((x + 3)(x - 2))

The shared (x + 3) cancels, leaving

(x - 3) / (x - 2), valid for x ≠ -3 and x ≠ 2.

The restriction x ≠ -3 stays attached even though the factor is gone, because the original expression was never defined there.

On the graph, the two excluded values behave differently, and the difference is the cancelled factor. At x = -3, the factor cancelled, so the function is simply undefined at one point: a hole, a single missing dot on an otherwise ordinary curve. At x = 2, the factor never cancelled; the denominator still reaches zero there while the numerator does not, so the function shoots toward positive or negative infinity on either side: a vertical asymptote, not a hole. A hole comes from a factor shared by top and bottom; an asymptote comes from a factor the denominator has alone.

Multiplying and dividing rational expressions follows the same factor-first habit: factor everything, then cancel across the product (and remember that dividing means multiplying by the reciprocal). Adding and subtracting them, as with ordinary fractions, requires a common denominator first.

WORKED Combine over a common denominator

1/x + 3/(2x). The common denominator is 2x; rewrite the first term:

2/(2x) + 3/(2x) = 5/(2x).

Once the denominators match, only the numerators are added.

▸ Rewrite an expression into the form that exposes the requested feature

All of this fluency serves one judgment: deciding which form a given question wants. If a question asks where an expression equals zero, the factored form puts the answer on the surface. If it asks for the value when the input is zero, the expanded form shows the constant term directly. If it asks for the largest or smallest value, a third form (the completed square, coming in the quadratics section) is the one to reach for. The expressions are all equal; the skill is reading the question, naming the feature it wants, and turning the expression to the form that displays that feature without a fight.

▸ Practice: equivalent expressions

Question 1

Which expression is equal to (x + 6)(x - 2)?

1. AA. x² - 4x - 12
2. BB. x² + 4x - 12
3. CC. x² + 8x - 12
4. DD. x² + 4x + 12

Question 2

Which expression is equal to (2x - 5)²?

1. AA. 4x² + 25
2. BB. 4x² - 10x + 25
3. CC. 4x² - 20x + 25
4. DD. 2x² - 20x + 25

Question 3

Factor x² - 49.

1. AA. (x + 7)(x - 7)
2. BB. (x - 7)²
3. CC. (x + 7)²
4. DD. (x + 49)(x - 1)

Question 4

Factor x² + 9x + 20.

1. AA. (x + 2)(x + 10)
2. BB. (x + 1)(x + 20)
3. CC. (x + 4)(x - 5)
4. DD. (x + 4)(x + 5)

Question 5

Factor completely: 3x² - 12.

1. AA. 3(x² - 4)
2. BB. 3(x + 2)(x - 2)
3. CC. (3x + 6)(x - 2)
4. DD. 3(x - 2)²

Question 6

Factor 2x² + 5x - 3.

1. AA. (2x + 1)(x - 3)
2. BB. (2x - 3)(x + 1)
3. CC. (2x - 1)(x + 3)
4. DD. (x - 1)(2x + 3)

Question 7

For x ≠ 4, the expression (x² - 16)/(x - 4) is equivalent to:

1. AA. x + 4
2. BB. x - 4
3. CC. x² - 4
4. DD. x + 16

Question 8

Which expression is equal to x² - 10x + 25?

1. AA. (x + 5)²
2. BB. (x - 25)(x - 1)
3. CC. (x - 5)(x + 5)
4. DD. (x - 5)²

9. The solutions of (x - 3)(x + 7) = 0 are two numbers. What is their product? (grid-in)

Question 10

For x ≠ 0, simplify (6x³)/(2x).

1. AA. 3x
2. BB. 3x²
3. CC. 12x²
4. DD. 3x³

Question 11

Factor by grouping: x³ + 4x² + 2x + 8.

1. AA. (x² + 2)(x + 4)
2. BB. (x² + 4)(x + 2)
3. CC. (x + 2)(x² + 4)
4. DD. (x³ + 2)(x + 4)

12. If (x + a)(x + b) = x² + 7x + 10 for integers a and b, what is a + b? (grid-in)

Question 13

For x ≠ 1 and x ≠ -3, simplify (x² + 5x + 6)/(x² + 2x - 3).

1. AA. (x + 3)/(x - 1)
2. BB. (x + 2)/(x + 3)
3. CC. (x + 2)/(x - 1)
4. DD. (x + 3)/(x + 2)

Question 14

For x ≠ 0, which expression is equal to 1/x + 3/(2x)?

1. AA. 4/(3x)
2. BB. 4/(2x)
3. CC. 5/x
4. DD. 5/(2x)

15. If (x + 4)(x + k) = x² + 11x + 28 for all x, what is the value of k? (grid-in)

16. If x + 1/x = 5, what is the value of x² + 1/x²? (grid-in)

▸ Solutions: equivalent expressions

Show the answers and explanations

1: B. Expand: (x)(x) + (x)(-2) + (6)(x) + (6)(-2) = x² + 4x - 12. Choice A flips the sign of the middle term, the error of mishandling 6x - 2x.

2: C. (2x - 5)² = (2x)² - 2(2x)(5) + 25 = 4x² - 20x + 25. Choice A drops the middle term entirely: the most common mistake with a squared binomial, which is not just the sum of the squared pieces.

3: A. This is a difference of squares, x² - 7², so it factors as (x + 7)(x - 7). The squared choices would each leave a middle term that does not belong.

4: D. Two numbers multiplying to 20 and adding to 9 are 4 and 5, giving (x + 4)(x + 5). Choice A multiplies to 20 but adds to 12; choice C has the wrong sign.

5: B. Factor out the common 3 first: 3(x² - 4), then recognize the difference of squares inside: 3(x + 2)(x - 2). Choice A stops too early and is not factored completely.

6: C. With ac = -6, the numbers 6 and -1 split the middle term: 2x² + 6x - x - 3 = 2x(x + 3) - (x + 3) = (2x - 1)(x + 3) after checking signs against the -3. Expanding choice C returns the original.

7: A. Factor the numerator as a difference of squares: (x + 4)(x - 4) over (x - 4) cancels to x + 4, valid for x ≠ 4. The cancellation is legal precisely because the restriction was noted.

8: D. First and last terms are x² and 5², and the middle -10x = -2(5)x: a perfect-square trinomial, (x - 5)². Choice C, (x - 5)(x + 5), would give x² - 25, with no middle term.

9: -21. By the zero-product idea, the solutions are x = 3 and x = -7. Their product is (3)(-7) = -21. (Their sum, -4, is the value a hurried reader might give instead.)

Show the answers and explanations

10: B. Divide: 6x³ / 2x = 3x², since 6/2 = 3 and x³ / x = x². Choice A divides the powers as if subtracting to the first power.

11: A. Group: x²(x + 4) + 2(x + 4), and factor the shared binomial to get (x² + 2)(x + 4). Choices B and C pair the terms incorrectly, producing (x + 2) where the cubic does not allow it.

12: 7. The numbers a and b satisfy a + b = 7 (the middle coefficient) and ab = 10 (the constant). The question asks only for the sum, which is 7; read directly off the coefficient of x.

Show the answers and explanations

13: C. Factor both: ((x + 2)(x + 3)) / ((x + 3)(x - 1)). The shared (x + 3) cancels, leaving (x + 2)/(x - 1), valid for x ≠ 1 and x ≠ -3.

14: D. Use the common denominator 2x: 1/x = 2/(2x), so 2/(2x) + 3/(2x) = 5/(2x). Choice B forgets to convert the first fraction before adding the numerators.

15: 7. Expand the left side: (x + 4)(x + k) = x² + (4 + k)x + 4k. For this to match x² + 11x + 28 for every x, the constant terms must agree: 4k = 28, so k = 7. (Checking the middle term confirms it: 4 + 7 = 11.)

16: 23. Square the given relation directly rather than solving for x: (x + 1/x)² = x² + 2 + 1/x². Since the left side is 5² = 25, x² + 1/x² = 25 - 2 = 23. The middle term, 2 · x · (1/x) = 2, is the piece that connects the known structure to the target.

▸ Equivalent expressions: chapter summary

Equivalent expressions are one quantity written many ways, and fluency means moving between those forms at will: expanding a product into a sum, factoring a sum back into a product, and simplifying a ratio by factoring and cancelling. The patterns (common factor, difference of squares, perfect square, the two-number trinomial, and grouping) are a short list you come to recognize on sight. And the reason any of it matters is that the right form shows the answer: factored form reveals the zeros, expanded form reveals the constant, and the requested feature usually signals which one to reach for.

ADVANCED MATH · 2