Get Ready: Module 3 — Analyzing Structure
Everyone starts somewhere. Before we analyze the structure of polynomials, let's build the floor we'll stand on — one skill at a time.
If the words factor, zero, or polynomial feel slippery right now, that is completely okay. This page rebuilds the handful of skills you need before Module 3 will make sense — explained plainly, worked out fully, with practice you can check yourself.
Skills to build first
Six small skills. Read the “why,” follow the worked example, then try one or two yourself. Answers are hidden until you're ready to peek.
The Distributive Property
What it is: when a number sits outside a set of parentheses, it multiplies everything inside: \(a(b+c)=ab+ac\). Why you need it: every time you multiply factors like \((x-3)(x+2)\), you are distributing. It is the engine behind expanding polynomials in this module.
- Multiply the outside number by the first term: \(3 \cdot 2x = 6x\).
- Multiply the outside number by the second term: \(3 \cdot 5 = 15\).
- Add the results: \(3(2x+5) = 6x + 15\).
a) Expand \(4(x - 7)\).
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- \(4 \cdot x = 4x\)
- \(4 \cdot (-7) = -28\)
- So \(4(x-7) = 4x - 28\).
b) Expand \(-2(3x + 4)\).
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Exponent Rules
What it is: shortcuts for multiplying powers. The big one: when you multiply same-base powers, you add the exponents — \(x^m \cdot x^n = x^{m+n}\). Why you need it: multiplying polynomials means multiplying things like \(x \cdot x = x^2\). Knowing the rules keeps your powers correct.
- Multiply the numbers (coefficients): \(2 \cdot 4 = 8\).
- Multiply the variables by adding exponents: \(x^3 \cdot x^2 = x^{3+2} = x^5\).
- Put it together: \(2x^3 \cdot 4x^2 = 8x^5\).
a) Simplify \(x^4 \cdot x^5\).
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b) Simplify \(3x^2 \cdot 5x\). (Remember: a plain \(x\) means \(x^1\).)
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Adding, Subtracting & Multiplying Polynomials
What it is: a polynomial is just a sum of terms like \(3x^2\), \(-5x\), and \(4\). You combine like terms (same variable, same power) to add or subtract, and you distribute to multiply. Why you need it: Module 3 constantly switches between the factored and expanded forms of a polynomial — and expanding is multiplying.
- First terms: \(x \cdot x = x^2\).
- Outer terms: \(x \cdot 2 = 2x\).
- Inner terms: \(3 \cdot x = 3x\).
- Last terms: \(3 \cdot 2 = 6\).
- Combine the like terms \(2x + 3x = 5x\): \((x+3)(x+2) = x^2 + 5x + 6\).
a) Add \((2x^2 + 3x - 1) + (x^2 - 5x + 4)\).
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b) Multiply \((x - 4)(x + 1)\).
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Factoring
What it is: factoring is multiplying in reverse — turning \(x^2 + 5x + 6\) back into \((x+3)(x+2)\). For \(x^2 + bx + c\), you look for two numbers that multiply to \(c\) and add to \(b\). Why you need it: the factors of a polynomial reveal its zeros — the heart of “analyzing structure.”
- Find two numbers that multiply to \(12\) and add to \(7\).
- Test pairs that multiply to 12: \(1\&12\), \(2\&6\), \(3\&4\). Which adds to 7? \(3 + 4 = 7\). ✓
- Write the factors using those numbers: \(x^2 + 7x + 12 = (x + 3)(x + 4)\).
- Check by expanding: \((x+3)(x+4) = x^2 + 7x + 12\). ✓
a) Factor \(x^2 + 6x + 8\).
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b) Factor \(x^2 - 5x + 6\). (Both numbers will be negative — why?)
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Evaluating Functions
What it is: “evaluate \(f(x)\) at \(x = 2\)” just means substitute \(2\) everywhere you see \(x\), then simplify. \(f(2)\) is read “f of 2.” Why you need it: to check whether a number is a zero of a polynomial, you plug it in and see if you get \(0\).
- Replace every \(x\) with \(2\): \(f(2) = (2)^2 - 3(2) + 1\).
- Do the powers and products: \((2)^2 = 4\) and \(3(2) = 6\).
- Simplify left to right: \(4 - 6 + 1 = -1\). So \(f(2) = -1\).
a) For \(f(x) = 2x + 5\), find \(f(3)\).
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b) For \(g(x) = x^2 + x - 6\), find \(g(-3)\). (Watch the negatives!)
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Finding Zeros / \(x\)-Intercepts
What it is: a zero is an \(x\)-value that makes the function equal \(0\) — and that's exactly where the graph crosses the \(x\)-axis (an \(x\)-intercept). To find them, set the factored form equal to zero and use the Zero Product Property: if \(A \cdot B = 0\), then \(A = 0\) or \(B = 0\). Why you need it: this connects factors → zeros → graph, which is the whole story of Module 3.
- Set the function equal to zero: \((x - 3)(x + 5) = 0\).
- By the Zero Product Property, set each factor to zero: \(x - 3 = 0\) or \(x + 5 = 0\).
- Solve each: \(x = 3\) or \(x = -5\).
- The zeros (and \(x\)-intercepts) are \(x = 3\) and \(x = -5\).
a) Find the zeros of \(f(x) = (x + 1)(x - 4)\).
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b) Find the zeros of \(f(x) = x^2 - 7x + 10\). (Hint: factor first.)
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- Factor: two numbers multiplying to \(10\), adding to \(-7\) are \(-2\) and \(-5\), so \(f(x) = (x-2)(x-5)\).
- Set each factor to zero: \(x = 2\) or \(x = 5\).
- Zeros: \(x = 2\) and \(x = 5\).
Quick Readiness Check
Six short questions across all six skills. Try each, then reveal the answer.
1. Expand \(5(x - 2)\).
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2. Simplify \(x^3 \cdot x^4\).
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3. Multiply \((x + 2)(x + 5)\).
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4. Factor \(x^2 + 8x + 15\).
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5. For \(f(x) = x^2 - 4\), find \(f(2)\).
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6. Find the zeros of \((x - 6)(x + 2)\).
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If these feel comfortable, you're ready for the module. If a couple tripped you up, scroll back to that skill card — no rush, and no shame. Every architect drafts a few rough lines first.
When you're ready →
Take your new floor into the module. You can always come back here.