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Mathematical Architects · Algebra II · Foundations

Get Ready: Module 4 — Extending Beyond Polynomials

Before we build the rooms, we pour the floor. This page rebuilds the six skills you'll lean on the most in Module 4 — slowly, plainly, and with no pressure. Everyone starts somewhere, and starting here is exactly the right move.

Go at your own pace. You don't have to master all six skills today. Pick the one that feels shakiest, read the worked example, try a problem, and reveal the answer when you're ready. There's no clock and no judgment here — just a floor being laid one board at a time.
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Skills to Build First

Six prerequisites for Module 4. Read the "why," follow the worked example, then try one yourself.

1 Fraction Operations & Simplifying

What it is: adding, multiplying, and reducing fractions, and finding a common denominator. Why you need it: a rational function is literally one big fraction with \(x\)'s in it. Every move you make in Module 4 — simplifying \(\dfrac{x^2-1}{x-1}\), combining \(\dfrac{1}{x}+\dfrac{2}{x+3}\) — is fraction work wearing a costume. Get comfortable here and the scary parts stop being scary.

Worked example — simplify \(\dfrac{2}{3}+\dfrac{1}{4}\)

Step 1. Find a common denominator. The smallest number both \(3\) and \(4\) divide into is \(12\).

Step 2. Rewrite each fraction over \(12\): \(\dfrac{2}{3}=\dfrac{8}{12}\) and \(\dfrac{1}{4}=\dfrac{3}{12}\).

Step 3. Now the denominators match, so just add the tops: \(\dfrac{8}{12}+\dfrac{3}{12}=\dfrac{11}{12}\).

Step 4. Can it reduce? \(11\) is prime and doesn't divide \(12\), so \(\dfrac{11}{12}\) is the final answer.

Try it

(a) Simplify \(\dfrac{12}{18}\) to lowest terms.

Show answer
\(\dfrac{2}{3}\). The biggest number that divides both \(12\) and \(18\) is \(6\). Divide top and bottom by \(6\): \(\dfrac{12\div 6}{18\div 6}=\dfrac{2}{3}\).

(b) Compute \(\dfrac{3}{5}\times\dfrac{10}{9}\).

Show answer
\(\dfrac{2}{3}\). Multiply straight across: \(\dfrac{3\times 10}{5\times 9}=\dfrac{30}{45}\). Then reduce by dividing top and bottom by \(15\): \(\dfrac{30}{45}=\dfrac{2}{3}\). (You could also cancel first: \(\dfrac{\cancel{3}}{\cancel{5}}\times\dfrac{\cancel{10}^2}{\cancel{9}^3}\).)

2 Factoring

What it is: rewriting an expression as a product of simpler pieces, like \(x^2-x-6=(x-3)(x+2)\). Why you need it: Module 4 finds asymptotes and holes by factoring the top and bottom of a fraction and seeing what cancels. If you can't factor, you can't tell a hole from an asymptote — and that's the whole point of the rational-functions unit.

Worked example — factor \(x^2-x-6\)

Step 1. It's \(x^2 + bx + c\) with \(b=-1\) and \(c=-6\). We need two numbers that multiply to \(-6\) and add to \(-1\).

Step 2. List pairs that multiply to \(-6\): \((1,-6),(-1,6),(2,-3),(-2,3)\). Which pair adds to \(-1\)? The pair \(2\) and \(-3\): \(2+(-3)=-1\). ✓

Step 3. Drop those numbers into the parentheses: \(x^2-x-6=(x+2)(x-3)\).

Step 4. Check by expanding: \((x+2)(x-3)=x^2-3x+2x-6=x^2-x-6\). ✓

Try it

(a) Factor \(x^2+5x+6\).

Show answer
\((x+2)(x+3)\). Two numbers that multiply to \(6\) and add to \(5\) are \(2\) and \(3\). Check: \((x+2)(x+3)=x^2+5x+6\). ✓

(b) Factor the difference of squares \(x^2-9\).

Show answer
\((x-3)(x+3)\). A difference of squares \(a^2-b^2\) always factors as \((a-b)(a+b)\). Here \(a=x\) and \(b=3\) since \(9=3^2\).

3 Square Roots & Radicals

What it is: the \(\sqrt{\phantom{x}}\) symbol asks "what number, times itself, gives this?" Simplifying radicals means pulling out perfect squares. Why you need it: the radical-functions half of Module 4 graphs things like \(y=\sqrt{x-2}+1\). To understand the shape and where it starts, you need to be at home with square roots and what they will and won't accept.

Worked example — simplify \(\sqrt{48}\)

Step 1. Look for a perfect-square factor of \(48\). Perfect squares are \(4,9,16,25,\dots\) Here \(48=16\times 3\), and \(16\) is a perfect square.

Step 2. Split the radical: \(\sqrt{48}=\sqrt{16\times 3}=\sqrt{16}\cdot\sqrt{3}\).

Step 3. Take the root of the perfect square: \(\sqrt{16}=4\), so \(\sqrt{48}=4\sqrt{3}\).

Step 4. The \(3\) has no perfect-square factors left, so \(4\sqrt{3}\) is fully simplified.

Try it

(a) Find \(\sqrt{81}\).

Show answer
\(9\). Because \(9\times 9 = 81\). (\(81\) is a perfect square.)

(b) Simplify \(\sqrt{50}\).

Show answer
\(5\sqrt{2}\). Since \(50=25\times 2\) and \(25\) is a perfect square: \(\sqrt{50}=\sqrt{25}\cdot\sqrt{2}=5\sqrt{2}\).

4 Exponent Rules

What it is: the shortcuts for combining powers — \(x^a\cdot x^b=x^{a+b}\), \(\dfrac{x^a}{x^b}=x^{a-b}\), and \(x^{-1}=\dfrac{1}{x}\). Why you need it: rational expressions are full of \(x\) terms divided by \(x\) terms, and a square root is secretly a \(\tfrac12\) power (\(\sqrt{x}=x^{1/2}\)). Knowing the rules keeps the algebra clean instead of getting tangled.

Worked example — simplify \(\dfrac{x^5}{x^2}\)

Step 1. Same base \(x\), divided. The rule for dividing like bases is to subtract the exponents: \(\dfrac{x^a}{x^b}=x^{a-b}\).

Step 2. Subtract: \(5-2=3\).

Step 3. So \(\dfrac{x^5}{x^2}=x^{3}\). (Sanity check: \(\dfrac{x\cdot x\cdot x\cdot x\cdot x}{x\cdot x}\) — two \(x\)'s cancel, three remain.)

Try it

(a) Simplify \(x^3\cdot x^4\).

Show answer
\(x^{7}\). Multiplying like bases means adding exponents: \(x^{3}\cdot x^{4}=x^{3+4}=x^{7}\).

(b) Rewrite \(x^{-2}\) without a negative exponent.

Show answer
\(\dfrac{1}{x^{2}}\). A negative exponent means "reciprocal": \(x^{-n}=\dfrac{1}{x^{n}}\), so \(x^{-2}=\dfrac{1}{x^{2}}\).

5 Domain — Which \(x\)-Values Are Allowed

What it is: the domain is the list of \(x\)-values you're allowed to plug in. Two things are forbidden: dividing by zero and taking the square root of a negative. Why you need it: this is Module 4. Rational functions are undefined where the denominator is zero (that's where asymptotes and holes live); radical functions are undefined where the inside goes negative. Finding the domain is finding the function's edges.

Worked example — find the domain of \(f(x)=\dfrac{1}{x-3}\)

Step 1. This is a fraction, so the only danger is the denominator being zero. Set the denominator equal to zero to find the forbidden value: \(x-3=0\).

Step 2. Solve: \(x=3\). So \(x=3\) is not allowed (it would mean dividing by \(0\)).

Step 3. Every other number is fine. Domain: all real numbers except \(x=3\), written \(x\neq 3\).

Try it

(a) What is the domain of \(g(x)=\dfrac{x+1}{x+5}\)?

Show answer
All real numbers except \(x=-5\). Set the denominator to zero: \(x+5=0\Rightarrow x=-5\). That single value is forbidden; everything else is allowed.

(b) What is the domain of \(h(x)=\sqrt{x-2}\)?

Show answer
\(x\ge 2\). The inside of a square root can't be negative, so we need \(x-2\ge 0\). Add \(2\) to both sides: \(x\ge 2\).

6 Solving Equations

What it is: finding the value of \(x\) that makes an equation true, by doing the same thing to both sides until \(x\) is alone. Why you need it: finding where a denominator is zero, where a graph crosses an axis, or where two functions meet all comes down to solving an equation. It's the engine under every other skill on this page.

Worked example — solve \(2x+5=13\)

Step 1. Get the \(x\)-term by itself. Subtract \(5\) from both sides: \(2x+5-5=13-5\), so \(2x=8\).

Step 2. Now \(x\) is multiplied by \(2\); undo that by dividing both sides by \(2\): \(\dfrac{2x}{2}=\dfrac{8}{2}\).

Step 3. So \(x=4\).

Step 4. Check by plugging back in: \(2(4)+5=8+5=13\). ✓ It works.

Try it

(a) Solve \(3x-7=8\).

Show answer
\(x=5\). Add \(7\) to both sides: \(3x=15\). Divide both sides by \(3\): \(x=5\). Check: \(3(5)-7=15-7=8\). ✓

(b) Solve \(x^2-4=0\). (Two answers!)

Show answer
\(x=2\) or \(x=-2\). Add \(4\): \(x^2=4\). Take the square root of both sides, remembering both signs: \(x=\pm 2\). (Or factor: \((x-2)(x+2)=0\), so each factor can be zero.)
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Quick Readiness Check

Five short questions across the six skills. Try them in your head or on scratch paper.

  1. Simplify \(\dfrac{1}{2}+\dfrac{1}{3}\).
  2. Factor \(x^2-x-12\).
  3. Simplify \(\sqrt{72}\).
  4. Simplify \(\dfrac{x^6}{x^2}\), and rewrite \(x^{-3}\) without a negative exponent.
  5. What value of \(x\) is not allowed in \(\dfrac{x}{x-4}\)? And for \(\sqrt{x+1}\), what \(x\)-values are allowed?
Show all answers
1. \(\dfrac{5}{6}\) (common denominator \(6\): \(\tfrac{3}{6}+\tfrac{2}{6}\)).
2. \((x-4)(x+3)\) (multiply to \(-12\), add to \(-1\)).
3. \(6\sqrt{2}\) (since \(72=36\times 2\)).
4. \(\dfrac{x^6}{x^2}=x^{4}\); and \(x^{-3}=\dfrac{1}{x^{3}}\).
5. \(x=4\) is forbidden in \(\dfrac{x}{x-4}\); for \(\sqrt{x+1}\) the allowed values are \(x\ge -1\).

If these feel comfortable, you're ready for the module. If a few are still fuzzy, that's totally fine — drill them below, then come back.

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Practice More (Free)

Want more reps? These free sites drill each skill. Khan Academy has video lessons; IXL gives endless practice questions.

Fraction operations & simplifying

Khan Academy IXL

Factoring

Khan Academy IXL

Square roots & radicals

Khan Academy IXL

Exponent rules

Khan Academy IXL

Domain (allowed \(x\)-values)

Khan Academy IXL

Solving equations

Khan Academy IXL
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When You're Ready →

No rush. When the skills above feel solid, step up into the module.

Module 4 Overview & Visual Lab

Head to the module page, where the interactive Visual Lab lets you drag sliders and watch asymptotes, holes, domain, and range appear in real time.

Module 4 Overview Open the Visual Lab

Course Syllabus

See how Module 4 fits into the full Algebra II journey, plus the policies and pacing behind the course.

Syllabus

Foundations review built in-house for Dr. Ijezie's STEM Studio to support TEA Bluebonnet Learning — Secondary Mathematics (Edition 1, adapted from Carnegie Learning), licensed CC BY-NC 4.0. Non-commercial classroom use.