openfreebooks-bookgenerator/content/math/polynomials-expansion-factorisation/_index.md

+++
title = "Polynomials: expansion & factorisation"
description = "Expand products, collect like terms, and factorise expressions using common factors and identities."
template = "chapter.html"
weight = 1
[extra]
subject = "math"
chapter_id = "polynomials-expansion-factorisation"
strand = "Number & Algebra"
+++

<div class="book-prose">
  <p>
    An <strong>algebraic expression</strong> combines numbers, variables, and operations. A
    <strong>polynomial</strong> in one variable \(x\) is a sum of terms of the form
    \(a x^n\) where \(a\) is a constant and \(n\) is a non-negative integer. Examples:
    \(3x^2 - 5x + 1\), \(x^3 + 2x\), and \(7\).
  </p>

  <h2 id="vocabulary" class="book-prose__heading">Terms, coefficients, and degree</h2>
  <p>
    In \(5x^2 - 3x + 4\), the <strong>terms</strong> are \(5x^2\), \(-3x\), and \(4\). The
    <strong>coefficient</strong> of \(x^2\) is \(5\); the coefficient of \(x\) is \(-3\). The
    <strong>degree</strong> of the polynomial is the highest power of \(x\) — here, \(2\).
    <strong>Like terms</strong> have the same variable part (same power of \(x\)); only their
    coefficients may differ.
  </p>

  <h2 id="adding-subtracting" class="book-prose__heading">Adding and subtracting polynomials</h2>
  <p>
    Line up like terms and add or subtract their coefficients:
    \[ (3x^2 + 2x - 1) + (x^2 - 5x + 4) = 4x^2 - 3x + 3. \]
    When subtracting, distribute the minus sign to every term in the bracket:
    \[ (2x^2 + x) - (x^2 - 3x + 2) = 2x^2 + x - x^2 + 3x - 2 = x^2 + 4x - 2. \]
  </p>

  <h2 id="expanding" class="book-prose__heading">Expanding products</h2>
  <p>
    Use the <strong>distributive law</strong>: each term in the first bracket multiplies each term in
    the second. For two binomials,
    \[ (a+b)(c+d) = ac + ad + bc + bd. \]
    Example:
    \[ (x+3)(x-2) = x^2 - 2x + 3x - 6 = x^2 + x - 6. \]
  </p>
  <p>
    A binomial times a trinomial works the same way — multiply each of the two terms by each of the
    three:
    \[ (x+1)(x^2 - x + 2) = x^3 - x^2 + 2x + x^2 - x + 2 = x^3 + x + 2. \]
  </p>

  <h2 id="collecting" class="book-prose__heading">Collecting like terms</h2>
  <p>
    After expanding, combine like terms so the answer is simplified. From above,
    \(x^2 - 2x + 3x - 6\) becomes \(x^2 + x - 6\). Constants, \(x\)-terms, and \(x^2\)-terms are
    collected separately.
  </p>

  <h2 id="factorising-overview" class="book-prose__heading">Factorising</h2>
  <p>
    <strong>Factorisation</strong> is the reverse of expansion: write an expression as a product of
    simpler factors. Choose a method by inspecting the shape of the expression:
  </p>
  <div class="book-table-wrap">
    <table class="book-table">
      <thead>
        <tr>
          <th scope="col">Situation</th>
          <th scope="col">Method</th>
        </tr>
      </thead>
      <tbody>
        <tr>
          <td>Every term shares a common factor</td>
          <td>Highest common factor (HCF)</td>
        </tr>
        <tr>
          <td>Four terms, pairs share factors</td>
          <td>Grouping</td>
        </tr>
        <tr>
          <td>Matches \(a^2-b^2\) or \((a\pm b)^2\)</td>
          <td>Standard identities</td>
        </tr>
        <tr>
          <td>Quadratic in \(x\): \(ax^2+bx+c\)</td>
          <td>Two-number search or ac-method</td>
        </tr>
      </tbody>
    </table>
  </div>

  <h2 id="hcf" class="book-prose__heading">Highest common factor</h2>
  <p>
    Factor out the largest factor common to every term:
    \[ 6x^2 + 9x = 3x(2x + 3), \qquad 4x^3 - 8x^2 = 4x^2(x - 2). \]
    Always check whether the bracket can be factorised further.
  </p>

  <h2 id="grouping" class="book-prose__heading">Factorising by grouping</h2>
  <p>
    With four terms, split into two pairs and factor each pair:
    \[ ax + ay + bx + by = a(x+y) + b(x+y) = (a+b)(x+y). \]
    Example:
    \[ 2x^2 + 6x - x - 3 = 2x(x+3) - 1(x+3) = (2x-1)(x+3). \]
  </p>

  <h2 id="identities" class="book-prose__heading">Standard identities</h2>
  <div class="book-callout book-callout--formula" role="note">
    <p class="book-callout__label">Identities</p>
    <p class="book-formula">
      \[ a^2 - b^2 = (a+b)(a-b) \]
      \[ (a+b)^2 = a^2 + 2ab + b^2 \]
      \[ (a-b)^2 = a^2 - 2ab + b^2 \]
    </p>
  </div>
  <p>
    Recognising these forms speeds up factorisation. For example,
    \(x^2 - 9 = x^2 - 3^2 = (x+3)(x-3)\) and
    \(x^2 + 6x + 9 = (x+3)^2\).
  </p>

  <h2 id="quadratic-trinomials" class="book-prose__heading">Quadratic trinomials</h2>
  <p>
    To factorise a <strong>monic</strong> quadratic \(x^2 + bx + c\), find two numbers that
    <em>multiply</em> to \(c\) and <em>add</em> to \(b\):
    \[ x^2 + 5x + 6 = (x+2)(x+3) \quad\text{since } 2\times 3 = 6,\; 2+3 = 5. \]
  </p>
  <p>
    For a <strong>non-monic</strong> quadratic \(ax^2 + bx + c\) with \(a \neq 1\), use the
    <strong>ac-method</strong>: find two numbers that multiply to \(ac\) and add to \(b\), split the
    middle term, then factor by grouping:
    \[ 2x^2 + 7x + 3 = 2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3). \]
  </p>
  <p>
    This chapter stops at factorising expressions. Solving equations of the form \(ax^2+bx+c=0\) is
    covered in the chapter on quadratic equations.
  </p>

  <h2 id="try-it-yourself" class="book-prose__heading">Try it yourself</h2>
  <p>
    Use the area model to see how \((a+b)(c+d)\) splits into four partial products. Then practise
    choosing a factorisation method and checking your factors in the explorer below.
  </p>
</div>

<div class="math-widget-mount" data-widget="area-model" data-pagefind-ignore></div>

<div class="math-widget-mount" data-widget="factorisation-checker" data-pagefind-ignore></div>

<div class="book-prose">
  <h2 id="history" class="book-prose__heading">History</h2>
  <p>
    Long before modern symbols, people solved problems that required multiplying lengths and
    collecting areas — the same ideas behind expanding and factorising. Babylonian scribes (around
    1800&nbsp;BCE) recorded rectangle problems on clay tablets: given area and side relationships,
    find unknown lengths. Their recipes are early cousins of “expand, then simplify.”
  </p>

  <figure class="book-figure">
    <img
      class="book-figure__img"
      src="/chapters/math/polynomials-expansion-factorisation/polynomials-history.png"
      alt="Illustrated timeline: Babylonian area on clay tablets, al-Khwarizmi's algebra, Viète's difference of squares, Enlightenment science, and modern expand–factorise steps"
      width="960"
      height="640"
      loading="lazy"
      decoding="async"
    />
    <figcaption class="book-figure__caption">
      From geometric area puzzles to symbolic algebra: how expansion and factorisation became
      standard tools in school mathematics.
    </figcaption>
  </figure>

  <p>
    In the 9th century, <strong>al-Khwarizmi</strong> organised methods for equations and products
    in Arabic texts that later spread to Europe. Renaissance mathematicians such as
    <strong>Viète</strong> introduced letters for unknowns and constants, making it easier to see
    patterns like \(a^2 - b^2\). By the 17th–18th centuries, polynomial algebra was central to
    calculus and science; factorisation became a routine step before solving equations or
    simplifying formulas.
  </p>

  <h2 id="derivation" class="book-prose__heading">Derivation</h2>

  <h3 id="area-distributive" class="book-prose__heading">Area proof of \((a+b)(c+d)\)</h3>
  <p>
    Draw a rectangle with width \(a+b\) and height \(c+d\). The total area is \((a+b)(c+d)\). Split
    the rectangle into four smaller rectangles with areas \(ac\), \(ad\), \(bc\), and \(bd\). The
    sum of partial areas equals the whole:
    \[ (a+b)(c+d) = ac + ad + bc + bd. \]
    This is the distributive law in two dimensions — the same picture as in the area-model widget
    above.
  </p>

  <h3 id="derive-identities" class="book-prose__heading">Deriving standard identities</h3>
  <p>
    Set \(c = a\) and \(d = b\) in the area formula for a square of side \(a+b\):
    \[ (a+b)^2 = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2. \]
    For a square of side \(a-b\) (with \(a &gt; b\)), remove a strip of width \(b\) from a square of
    side \(a\); the remaining L-shape rearranges to show
    \[ (a-b)^2 = a^2 - 2ab + b^2. \]
    Difference of squares: a large square of side \(a\) minus a small square of side \(b\) leaves two
    congruent rectangles, which rearrange to a rectangle of dimensions \((a+b)\) by \((a-b)\):
    \[ a^2 - b^2 = (a+b)(a-b). \]
  </p>

  <h3 id="grouping-walkthrough" class="book-prose__heading">Grouping walkthrough</h3>
  <p>
    Factorise \(3x^2 + 12x - 2x - 8\).
  </p>
  <ol>
    <li>Group: \((3x^2 + 12x) + (-2x - 8)\).</li>
    <li>Factor each pair: \(3x(x+4) - 2(x+4)\).</li>
    <li>Common bracket \((x+4)\): \((3x-2)(x+4)\).</li>
  </ol>

  <h2 id="checkpoints" class="book-prose__heading">Checkpoints</h2>
  <p>
    Pause and reason before continuing. Discuss with a partner or write your work in a notebook.
  </p>

  <div class="book-callout" role="note">
    <p class="book-callout__label">Checkpoint</p>
    <p>
      In \(4x^3 - 2x^2 + 5\), what is the degree? Which terms are like terms with \(4x^3\)?
    </p>
  </div>

  <div class="book-callout" role="note">
    <p class="book-callout__label">Checkpoint</p>
    <p>
      Expand \((2x-1)(x+4)\) without skipping steps. How many terms do you get before collecting?
    </p>
  </div>

  <div class="book-callout" role="note">
    <p class="book-callout__label">Checkpoint</p>
    <p>
      How would you factorise \(x^2 - 16\)? Which identity applies?
    </p>
  </div>

  <div class="book-callout" role="note">
    <p class="book-callout__label">Checkpoint</p>
    <p>
      Expand \(x(x+1) - (x+1)\). What common factor appears? What is the factorised form?
    </p>
  </div>

  <h2 id="applications" class="book-prose__heading">Applications</h2>

  <h3 id="geometry-area" class="book-prose__heading">Geometry: area and perimeter</h3>
  <p>
    A rectangle has sides \((x+3)\) cm and \((2x-1)\) cm. Its area is
    \[ (x+3)(2x-1) = 2x^2 + 5x - 3 \quad\text{cm}^2. \]
    The perimeter is \(2\bigl((x+3)+(2x-1)\bigr) = 2(3x+2) = 6x+4\) cm. Expansion links algebraic
    descriptions of sides to measurable quantities.
  </p>

  <h3 id="revenue-profit" class="book-prose__heading">Revenue and pricing</h3>
  <p>
    If a shop sells \(n\) items at price \((50 - 0.5n)\) dollars each, revenue is
    \(n(50 - 0.5n) = 50n - 0.5n^2\). Expanding a product of two linear factors often appears before
    finding maximum revenue (a quadratic in \(n\)).
  </p>

  <h3 id="number-pattern" class="book-prose__heading">Number patterns</h3>
  <p>
    Notice that \(11^2 = 121\), \(111^2 = 12321\). In general, expanding \((10^k + 10^{k-1} +
    \cdots + 1)^2\) explains palindromic squares — a playful link between identities and arithmetic
    patterns.
  </p>

  <h2 id="question-bank" class="book-prose__heading">Question bank</h2>
  <p>
    Work each question, then open the solution to check your reasoning. Difficulty increases
    through the sets.
  </p>

  <h3 id="questions-easy" class="book-prose__heading">Easy</h3>

  <details class="book-question">
    <summary class="book-question__prompt">Expand \(3(x+2)\).</summary>
    <div class="book-question__solution">
      <p><strong>Answer:</strong> \(3x + 6\).</p>
      <p><strong>Steps:</strong> \(3 \cdot x + 3 \cdot 2 = 3x + 6\).</p>
    </div>
  </details>

  <details class="book-question">
    <summary class="book-question__prompt">Simplify \(2x + 5x - 1\).</summary>
    <div class="book-question__solution">
      <p><strong>Answer:</strong> \(7x - 1\).</p>
      <p><strong>Explanation:</strong> Combine like terms \(2x + 5x = 7x\).</p>
    </div>
  </details>

  <details class="book-question">
    <summary class="book-question__prompt">Factorise \(6x + 9\).</summary>
    <div class="book-question__solution">
      <p><strong>Answer:</strong> \(3(2x + 3)\).</p>
      <p><strong>Steps:</strong> HCF is \(3\); \(6x \div 3 = 2x\), \(9 \div 3 = 3\).</p>
    </div>
  </details>

  <h3 id="questions-intermediate" class="book-prose__heading">Intermediate</h3>

  <details class="book-question">
    <summary class="book-question__prompt">Expand \((x+4)(x-2)\).</summary>
    <div class="book-question__solution">
      <p><strong>Answer:</strong> \(x^2 + 2x - 8\).</p>
      <p><strong>Steps:</strong> \(x^2 - 2x + 4x - 8 = x^2 + 2x - 8\).</p>
    </div>
  </details>

  <details class="book-question">
    <summary class="book-question__prompt">Factorise \(x^2 - 9\).</summary>
    <div class="book-question__solution">
      <p><strong>Answer:</strong> \((x+3)(x-3)\).</p>
      <p><strong>Explanation:</strong> Difference of squares: \(x^2 - 3^2\).</p>
    </div>
  </details>

  <details class="book-question">
    <summary class="book-question__prompt">Factorise \(x^2 + 6x + 9\).</summary>
    <div class="book-question__solution">
      <p><strong>Answer:</strong> \((x+3)^2\).</p>
      <p><strong>Explanation:</strong> Perfect square: \(3 \times 3 = 9\), \(3 + 3 = 6\).</p>
    </div>
  </details>

  <h3 id="questions-difficult" class="book-prose__heading">Difficult</h3>

  <details class="book-question">
    <summary class="book-question__prompt">Factorise \(2x^2 + 6x - x - 3\) by grouping.</summary>
    <div class="book-question__solution">
      <p><strong>Answer:</strong> \((2x-1)(x+3)\).</p>
      <p><strong>Steps:</strong> \(2x(x+3) - 1(x+3) = (2x-1)(x+3)\).</p>
    </div>
  </details>

  <details class="book-question">
    <summary class="book-question__prompt">Factorise \(2x^2 + 7x + 3\).</summary>
    <div class="book-question__solution">
      <p><strong>Answer:</strong> \((2x+1)(x+3)\).</p>
      <p><strong>Steps:</strong> \(ac = 6\); split \(7x = 6x + x\); \(2x(x+3) + 1(x+3) = (2x+1)(x+3)\).</p>
    </div>
  </details>

  <h3 id="questions-hardcore" class="book-prose__heading">Hardcore</h3>

  <details class="book-question">
    <summary class="book-question__prompt">Expand and then factorise \(x(x+2) - (x+2)\).</summary>
    <div class="book-question__solution">
      <p><strong>Answer:</strong> \((x+2)(x-1)\) after factorising (expansion gives \(x^2 + x - 2\)).</p>
      <p><strong>Steps:</strong> Factor common \((x+2)\): \((x+2)(x-1)\). Check: \((x+2)(x-1) = x^2 + x - 2\).</p>
    </div>
  </details>

  <details class="book-question">
    <summary class="book-question__prompt">A rectangle has sides \((x+3)\) cm and \((2x-1)\) cm. Find its area in expanded form.</summary>
    <div class="book-question__solution">
      <p><strong>Answer:</strong> \((2x^2 + 5x - 3)\) cm².</p>
      <p><strong>Calculation:</strong> \((x+3)(2x-1) = 2x^2 - x + 6x - 3 = 2x^2 + 5x - 3\).</p>
    </div>
  </details>

  <h2 id="references" class="book-prose__heading">References</h2>
  <ul>
    <li>al-Khwarizmi, <em>Al-Kitāb al-mukhtaṣar fī ḥisāb al-jabr wal-muqābala</em> (c. 820 CE) — systematic treatment of algebra.</li>
    <li>HKDSE Compulsory Part curriculum — Number and Algebra strand (polynomials, expansion, factorisation).</li>
  </ul>
</div>