19 Polynomials and factoring

When the curve is the point

You kick a ball straight up. For a moment it keeps going — then it slows, stops, and falls back down. The height isn’t a straight line; it curves up and back. That curve has an equation.

You’re fencing off a rectangular garden and you’ve got 40 metres of fencing. You want the biggest possible area. If you make it long and thin, the area shrinks. If you make it square, the area is as large as it can be. Finding that optimum — that maximum — is a polynomial problem.

Your school sells tickets to a performance. At a lower price more people buy, but revenue per ticket drops. At a higher price fewer people buy. Somewhere in the middle is the price that brings in the most money. That relationship between price and total revenue follows a curve — a polynomial.

Straight lines describe constant rates. Polynomials describe the rest.

19.1 What the notation is saying

Here is a polynomial:

\[ax^2 + bx + c\]

Read it aloud: a times x-squared, plus b times x, plus c.

The letters $a$, $b$, and $c$ are coefficients — the numbers multiplying each power of $x$. They’re just fixed numbers; you’ll swap them for actual values in any particular problem. The letter $x$ is the unknown — the variable you’re either solving for, or letting run freely while you watch what the output does. When you write $y = ax^2 + bx + c$, $y$ is the output — the value the polynomial produces for each value of $x$.

Each piece — $ax^2$, $bx$, $c$ — is called a term. Count the terms and you know you have three. The $c$ term has no $x$ at all; it’s a plain number, a constant.

The degree of a polynomial is the highest power of $x$ present. In $ax^2 + bx + c$ the highest power is 2, so the degree is 2. Degree-2 polynomials are called quadratics. Degree-1 polynomials — like $3x + 7$ — are the linear equations from Vol 2. Degree-3 polynomials are called cubics, and so on.

Why does the $x^2$ term make the graph curve? Because $x^2$ grows faster than $x$. When $x = 2$, $x^2 = 4$. When $x = 10$, $x^2 = 100$. The squared term pulls the output away from the straight line. Take $a = 0$ and you’re back to a line. Leave $a$ nonzero and the whole shape bends.

A specific example: $y = x^2 - 5x + 6$. Here $a = 1$, $b = -5$, $c = 6$. The negative $b$ means the straight-line part is pulling down, and the positive $c$ shifts the whole thing up by 6 units. The coefficient $a = 1$ means the parabola opens upward — a positive $a$ always curves upward, a negative $a$ always curves downward.

19.2 The method

19.2.1 Part a: Expanding

Before you can factor, it helps to know how to expand — to go from the factored form back to the full polynomial. This makes the reverse step (factoring) easier to recognise.

The distributive law says: multiply everything inside the brackets by everything outside.

\[3(x + 4) = 3x + 12\]

When you have two brackets multiplied together, you apply the same idea twice — every term in the first bracket multiplies every term in the second. This is sometimes called FOIL (First, Outer, Inner, Last), though the name matters less than the idea: don’t miss any pair.

\[(x + 3)(x + 2)\]

First terms: $x \cdot x = x^2$
Outer terms: $x \cdot 2 = 2x$
Inner terms: $3 \cdot x = 3x$
Last terms: $3 \cdot 2 = 6$

Add them up: $x^2 + 2x + 3x + 6 = x^2 + 5x + 6$.

The pattern is worth noticing. With $(x + 3)(x + 2)$, the number in the middle ($5$) is the sum of 3 and 2. The constant term ($6$) is their product. That observation is what makes factoring work in reverse.

19.2.2 Part b: Factoring

Factoring means breaking a polynomial back into the brackets that produced it. There are three main methods; apply them in this order.

Method 1: Common factor. Check whether every term shares a factor you can pull out.

\[6x^2 + 9x = 3x(2x + 3)\]

The common factor here is $3x$. Dividing both terms by $3x$ gives the bracket. Always check for a common factor first — it simplifies what comes next.

Method 2: Difference of squares. When you have two perfect squares separated by a minus sign, it factors as:

\[a^2 - b^2 = (a + b)(a - b)\]

For example:

\[x^2 - 25 = (x + 5)(x - 5)\]

Check it by expanding: $(x + 5)(x - 5) = x^2 - 5x + 5x - 25 = x^2 - 25$. Yes.

This works because the middle terms cancel. It doesn’t work if the sign between the squares is positive — $x^2 + 25$ doesn’t factor over the real numbers.

Method 3: Factoring a trinomial $ax^2 + bx + c$. When $a = 1$, look for two numbers that multiply to $c$ and add to $b$.

\[x^2 - 5x + 6\]

You need two numbers that multiply to $6$ and add to $-5$. Try: $-2$ and $-3$. Product: $(-2)(-3) = 6$. Sum: $(-2) + (-3) = -5$. Yes.

\[x^2 - 5x + 6 = (x - 2)(x - 3)\]

When $a \neq 1$, try the grouping method (also called the $ac$ method). Multiply $a$ by $c$, find two numbers with that product and sum $b$, then split the middle term and factor in pairs.

\[2x^2 + 7x + 3\]

Multiply $a \cdot c = 2 \cdot 3 = 6$. Find two numbers that multiply to $6$ and add to $7$: those are $1$ and $6$.

Split the middle term:

\[2x^2 + x + 6x + 3\]

Factor in pairs:

\[x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1)\]

Both terms contain the factor $(2x + 1)$, so it can be pulled out — the same distributive law used in expanding, run backwards.

Check by expanding: $(x + 3)(2x + 1) = 2x^2 + x + 6x + 3 = 2x^2 + 7x + 3$. Yes.

19.2.3 Part c: Solving quadratic equations

A quadratic equation is when you set a quadratic equal to zero:

\[ax^2 + bx + c = 0\]

There are two methods.

Solving by factoring. If you can factor the left side, you can use the fact that if two things multiply to zero, at least one of them must be zero.

\[x^2 - 5x + 6 = 0\]

\[(x - 2)(x - 3) = 0\]

Either $(x - 2) = 0$ or $(x - 3) = 0$. So $x = 2$ or $x = 3$.

Check both: $4 - 10 + 6 = 0$. Yes. $9 - 15 + 6 = 0$. Yes.

Solving with the quadratic formula. Not every quadratic factors neatly. For any quadratic $ax^2 + bx + c = 0$, the solutions are always:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Read it: x equals negative b, plus or minus the square root of b-squared minus four-a-c, all over two-a.

The $\pm$ sign means there are two candidate answers — one where you add the square root term, one where you subtract it (whether there are two, one, or no real answers depends on the discriminant — that’s the next section).

The piece under the square root — $b^2 - 4ac$ — is called the discriminant, often written $\Delta$ (delta). It tells you how many solutions exist before you do any arithmetic:

$\Delta > 0$: two real solutions (the parabola crosses the x-axis twice)
$\Delta = 0$: one solution (the parabola just touches the x-axis at one point)
$\Delta < 0$: no real solutions (the parabola doesn’t reach the x-axis)

19.2.4 Part d: Reading the graph

Every quadratic $y = ax^2 + bx + c$ draws a parabola — a symmetric U-shape (or inverted U if $a < 0$).

The roots are where the parabola crosses the x-axis — the values of $x$ where $y = 0$. These are exactly the solutions to $ax^2 + bx + c = 0$. Factoring gives you the roots. The roots are the x-intercepts of the graph.

The vertex is the turning point — the very top of a downward parabola (the maximum) or the very bottom of an upward one (the minimum). The x-coordinate of the vertex is:

\[x_v = -\frac{b}{2a}\]

Read it: negative b over two a. Substitute this back into the equation to find the y-coordinate — the maximum or minimum value itself.

For $y = x^2 - 5x + 6$: $x_v = -(-5)/(2 \cdot 1) = 5/2 = 2.5$. Then $y = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25$. The vertex is at $(2.5, -0.25)$ — the minimum point, halfway between the two roots at $x = 2$ and $x = 3$.

Code

{
  // Guard: skip if a is too close to zero
  const a = Math.abs(qa) < 0.1 ? 0.5 : qa;
  const b = qb, c = qc;

  const svgW = 500, svgH = 360;
  const margin = { top: 50, right: 30, bottom: 40, left: 50 };
  const W = svgW - margin.left - margin.right;
  const H = svgH - margin.top - margin.bottom;

  const xMin = -6, xMax = 6;
  const xs = d3.range(xMin, xMax + 0.05, 0.05);
  const ys = xs.map(x => a * x * x + b * x + c);
  const yMin = Math.min(...ys, -10), yMax = Math.max(...ys, 10);

  const xScale = d3.scaleLinear([xMin, xMax], [0, W]);
  const yScale = d3.scaleLinear([yMin, yMax], [H, 0]);

  const disc = b * b - 4 * a * c;

  // Roots
  let roots = [];
  if (disc > 0) {
    roots = [
      (-b + Math.sqrt(disc)) / (2 * a),
      (-b - Math.sqrt(disc)) / (2 * a)
    ].filter(r => r >= xMin && r <= xMax);
  } else if (Math.abs(disc) < 1e-9) {
    const r = -b / (2 * a);
    if (r >= xMin && r <= xMax) roots = [r];
  }

  // Vertex
  const vx = -b / (2 * a);
  const vy = a * vx * vx + b * vx + c;

  const svg = d3.create("svg")
    .attr("width", svgW).attr("height", svgH)
    .style("font-family", "inherit");

  const g = svg.append("g")
    .attr("transform", `translate(${margin.left},${margin.top})`);

  // Axes
  if (yScale(0) >= 0 && yScale(0) <= H) {
    g.append("line")
      .attr("x1", 0).attr("y1", yScale(0))
      .attr("x2", W).attr("y2", yScale(0))
      .attr("stroke", "#9ca3af").attr("stroke-width", 1);
  }
  if (xScale(0) >= 0 && xScale(0) <= W) {
    g.append("line")
      .attr("x1", xScale(0)).attr("y1", 0)
      .attr("x2", xScale(0)).attr("y2", H)
      .attr("stroke", "#9ca3af").attr("stroke-width", 1);
  }

  // Grid lines
  for (let xi = xMin; xi <= xMax; xi++) {
    g.append("line")
      .attr("x1", xScale(xi)).attr("y1", 0)
      .attr("x2", xScale(xi)).attr("y2", H)
      .attr("stroke", "#f3f4f6").attr("stroke-width", 1);
    g.append("text")
      .attr("x", xScale(xi)).attr("y", yScale(0) + 14)
      .attr("text-anchor", "middle").attr("font-size", "11px").attr("fill", "#9ca3af")
      .text(xi === 0 ? "" : xi);
  }

  // Curve
  const line = d3.line()
    .x(x => xScale(x))
    .y(x => yScale(a * x * x + b * x + c))
    .curve(d3.curveCatmullRom);

  g.append("path")
    .datum(xs)
    .attr("d", line)
    .attr("fill", "none")
    .attr("stroke", "#2563eb")
    .attr("stroke-width", 2.5);

  // Roots
  roots.forEach(r => {
    g.append("circle")
      .attr("cx", xScale(r)).attr("cy", yScale(0))
      .attr("r", 5)
      .attr("fill", "#dc2626").attr("stroke", "#fff").attr("stroke-width", 1.5);
  });

  // Vertex
  if (vx >= xMin && vx <= xMax) {
    g.append("circle")
      .attr("cx", xScale(vx)).attr("cy", yScale(vy))
      .attr("r", 6)
      .attr("fill", "#f59e0b").attr("stroke", "#fff").attr("stroke-width", 1.5);
  }

  // "No real roots" label
  if (disc < 0) {
    g.append("text")
      .attr("x", W / 2).attr("y", H / 2)
      .attr("text-anchor", "middle").attr("font-size", "13px").attr("fill", "#dc2626")
      .text("no real roots (Δ < 0)");
  }

  // Info panel at top
  const eqStr = `y = ${a}x² ${b >= 0 ? "+" : ""}${b}x ${c >= 0 ? "+" : ""}${c}`;
  svg.append("text")
    .attr("x", margin.left).attr("y", 22)
    .attr("font-size", "13px").attr("font-weight", "600").attr("fill", "#111827")
    .text(eqStr);

  svg.append("text")
    .attr("x", margin.left + 220).attr("y", 22)
    .attr("font-size", "13px").attr("fill", disc >= 0 ? "#059669" : "#dc2626")
    .text(`Δ = ${disc.toFixed(2)}  (${disc > 0 ? "2 roots" : disc === 0 ? "1 root" : "no real roots"})`);

  // Vertex label
  if (vx >= xMin && vx <= xMax) {
    svg.append("text")
      .attr("x", margin.left + xScale(vx) + 8)
      .attr("y", margin.top + yScale(vy) - 4)
      .attr("font-size", "11px").attr("fill", "#d97706")
      .text(`vertex (${vx.toFixed(2)}, ${vy.toFixed(2)})`);
  }

  return svg.node();
}

19.3 Why this works

Where the quadratic formula comes from

The formula isn’t magic — it’s derived by solving $ax^2 + bx + c = 0$ directly, using a technique called completing the square.

The key move: rewrite $x^2 + \frac{b}{a}x$ as a perfect square minus a correction term.

\[x^2 + \frac{b}{a}x = \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\]

Substitute this into $ax^2 + bx + c = 0$, divide through by $a$, then isolate the squared bracket:

\[\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}\]

Take the square root of both sides (giving $\pm$), then subtract $\frac{b}{2a}$:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

That’s it. The formula is just completing the square, done once in general so you don’t have to do it every time.

19.4 Worked examples

Example 1 (sci) — height of a thrown ball

A ball is thrown upward from ground level with an initial speed of 14 m/s. Ignoring air resistance, its height in metres after $t$ seconds is:

\[h(t) = 14t - 4.9t^2\]

When does it land?

Landing means $h = 0$:

\[14t - 4.9t^2 = 0\]

Factor out a common term of $t$:

\[t(14 - 4.9t) = 0\]

Either $t = 0$ (the moment it’s thrown) or $14 - 4.9t = 0$. Solving the second:

\[4.9t = 14 \implies t = \frac{14}{4.9} \approx 2.86 \text{ s}\]

The ball lands after about 2.86 seconds.

The vertex — the highest point — is at $t = 14/(2 \times 4.9) = 14/9.8 \approx 1.43$ s, at height $h(1.43) = 14(1.43) - 4.9(1.43)^2 \approx 20.02 - 10.02 \approx 10$ m.

Code

{
  const h0 = ballH0, v = ballV;
  // h(t) = h0 + v*t - 4.9*t^2
  // Landing: positive root of -4.9t^2 + v*t + h0 = 0
  const disc = v * v + 4 * 4.9 * h0;
  const tLand = (v + Math.sqrt(disc)) / (2 * 4.9);
  const tPeak = v / (2 * 4.9);
  const hPeak = h0 + v * tPeak - 4.9 * tPeak * tPeak;

  const svgW = 500, svgH = 320;
  const margin = { top: 40, right: 30, bottom: 40, left: 55 };
  const W = svgW - margin.left - margin.right;
  const H = svgH - margin.top - margin.bottom;

  const tMax = tLand * 1.05;
  const tArr = d3.range(0, tLand + 0.01, tLand / 200);
  const hArr = tArr.map(t => Math.max(0, h0 + v * t - 4.9 * t * t));

  const xScale = d3.scaleLinear([0, tMax], [0, W]);
  const yMax = hPeak * 1.15 + 1;
  const yScale = d3.scaleLinear([0, yMax], [H, 0]);

  const svg = d3.create("svg")
    .attr("width", svgW).attr("height", svgH)
    .style("font-family", "inherit");

  const g = svg.append("g")
    .attr("transform", `translate(${margin.left},${margin.top})`);

  // Axes
  g.append("line").attr("x1", 0).attr("y1", H).attr("x2", W).attr("y2", H)
    .attr("stroke", "#6b7280").attr("stroke-width", 1.5);
  g.append("line").attr("x1", 0).attr("y1", 0).attr("x2", 0).attr("y2", H)
    .attr("stroke", "#6b7280").attr("stroke-width", 1.5);

  // Axis labels
  g.append("text").attr("x", W / 2).attr("y", H + 32)
    .attr("text-anchor", "middle").attr("font-size", "12px").attr("fill", "#374151")
    .text("time t (seconds)");
  g.append("text")
    .attr("transform", "rotate(-90)")
    .attr("x", -H / 2).attr("y", -40)
    .attr("text-anchor", "middle").attr("font-size", "12px").attr("fill", "#374151")
    .text("height h (metres)");

  // Curve
  const lineGen = d3.line()
    .x((_, i) => xScale(tArr[i]))
    .y(h => yScale(h));

  g.append("path")
    .datum(hArr)
    .attr("d", lineGen)
    .attr("fill", "none")
    .attr("stroke", "#2563eb")
    .attr("stroke-width", 2.5);

  // Peak marker
  g.append("circle")
    .attr("cx", xScale(tPeak)).attr("cy", yScale(hPeak))
    .attr("r", 6).attr("fill", "#f59e0b").attr("stroke", "#fff").attr("stroke-width", 1.5);

  // Landing marker
  g.append("circle")
    .attr("cx", xScale(tLand)).attr("cy", yScale(0))
    .attr("r", 6).attr("fill", "#dc2626").attr("stroke", "#fff").attr("stroke-width", 1.5);

  // Labels
  g.append("text")
    .attr("x", xScale(tPeak) + 8).attr("y", yScale(hPeak) - 8)
    .attr("font-size", "12px").attr("fill", "#d97706")
    .text(`peak: ${hPeak.toFixed(2)} m at t=${tPeak.toFixed(2)} s`);

  g.append("text")
    .attr("x", xScale(tLand) - 8).attr("y", yScale(0) + 18)
    .attr("text-anchor", "end").attr("font-size", "12px").attr("fill", "#dc2626")
    .text(`lands at t=${tLand.toFixed(2)} s`);

  // Title
  svg.append("text")
    .attr("x", margin.left).attr("y", 24)
    .attr("font-size", "13px").attr("font-weight", "600").attr("fill", "#111827")
    .text(`h(t) = ${h0} + ${v}t − 4.9t²`);

  return svg.node();
}

Example 2 (eng) — maximum area from fixed fencing

You have 40 m of fencing and want to enclose a rectangular garden with the largest possible area. What dimensions give the maximum area?

Let the width be $x$ metres. Since the total perimeter is 40 m, the two widths use $2x$ metres, leaving $40 - 2x$ for the two lengths. Each length is $(40 - 2x)/2 = 20 - x$ metres.

Area:

\[A = x(20 - x) = 20x - x^2\]

This is a downward parabola ($a = -1$). Maximum at:

\[x_v = -\frac{20}{2 \times (-1)} = 10 \text{ m}\]

Length $= 20 - 10 = 10$ m. Area $= 10 \times 10 = 100$ m². A square encloses the most area for a given perimeter.

Example 3 (biz) — maximum revenue from a pricing function

A small business sells handmade candles. Market research shows that at a price of $p$ dollars each, the number sold per week is $q = 60 - 2p$. Revenue is:

\[R = p \cdot q = p(60 - 2p) = 60p - 2p^2\]

This is a downward parabola ($a = -2$). Maximum revenue at:

\[p_v = -\frac{60}{2 \times (-2)} = -\frac{60}{-4} = 15\]

Price of $15. Revenue: $R = 15 \times (60 - 30) = 15 \times 30 = 450$ per week.

To confirm it’s a maximum and not a minimum: $a = -2 < 0$, so the parabola opens downward. The vertex is the top. Any price above or below $15 gives less revenue.

Example 4 (comp) — character jump arc

A game character jumps. The height of the character above the ground in metres at time $t$ seconds after the jump is:

\[h(t) = -5t^2 + 10t\]

Find the maximum height and the time the character lands.

The parabola opens downward ($a = -5$), so the vertex is the maximum.

Vertex time: $t_v = -10/(2 \times -5) = 1$ second.

Maximum height: $h(1) = -5(1)^2 + 10(1) = -5 + 10 = 5$ metres.

Landing: set $h = 0$.

\[-5t^2 + 10t = 0\] \[t(-5t + 10) = 0\]

$t = 0$ (take-off) or $t = 2$ seconds (landing).

Evaluating a polynomial at a specific value — like checking the height at $t = 1$ or $t = 2$ — is a calculation every game engine runs many times per second to determine where characters and objects are.

19.5 Where this goes

The natural next step is Vol 3, Chapter 4: Functions and relations. A quadratic is a function — it takes an $x$ value and produces a $y$ value, one output for each input. Once you have the function concept, the parabola becomes one member of a large family: linear functions, quadratic functions, exponential functions, trigonometric functions. The machinery you’ve built here — reading coefficients, finding roots, locating vertices — applies across all of them.

The bigger horizon is calculus. The vertex formula $x_v = -b/(2a)$ gives you the maximum or minimum, but it only works for quadratics. What about a cubic, or a more complicated curve? That’s what differential calculus (Vol 5) answers: a general method for finding the exact peak or trough of any smooth curve. The quadratic vertex is the simplest case of a much more powerful idea. When you reach it, what you learned here will already be familiar.

Where this shows up

A structural engineer calculating deflection in a beam under a distributed load solves a polynomial equation.
A physicist modelling projectile trajectory — range, maximum height, time of flight — works with quadratics throughout.
A pricing analyst finding the revenue-maximising price point solves $dR/dp = 0$, which for a linear demand function produces a quadratic.
A computer graphics programmer evaluating a Bezier spline is computing a polynomial at a parameter value thousands of times per frame.
A data scientist fitting a parabola to experimental data (polynomial regression of degree 2) uses the same $ax^2 + bx + c$ form.

19.6 Exercises

These are puzzles. Each has a clean answer. The interesting part is setting up the equation first, then solving it.

A ball is thrown upward from a cliff 20 m above the ground with an initial upward speed of 10 m/s. Its height above the ground after $t$ seconds is $h(t) = 20 + 10t - 4.9t^2$. When does it hit the ground?

Code

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  container.style.cssText = "border:1px solid #e5e7eb; border-radius:8px; padding:1rem 1.25rem; margin:0.75rem 0; font-family:inherit;";
  const stepsDiv = document.createElement("div");
  stepsDiv.style.marginBottom = "0.75rem";
  function renderSteps() {
    stepsDiv.innerHTML = "";
    for (let i = 0; i < current; i++) {
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      row.style.cssText = "display:grid; grid-template-columns:200px 1fr; gap:0.35rem 1rem; align-items:baseline; padding:0.35rem 0; border-top:1px solid #f3f4f6;";
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      opCell.style.cssText = "font-size:0.85em; color:#6b7280; font-style:italic;";
      opCell.textContent = s.op;
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      eqCell.innerHTML = katex.renderToString(s.eq, { throwOnError: false, displayMode: false });
      row.appendChild(opCell);
      row.appendChild(eqCell);
      if (s.note) {
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  function updateButtons() {
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    counter.textContent = current === 0 ? "Click to reveal steps" : `Step ${current} of ${totalSteps}`;
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  container.appendChild(controls);
  renderSteps();
  updateButtons();
  return container;
}

Code

makeStepperHTML(1, [
  { op: "Set height equal to zero", eq: "20 + 10t - 4.9t^2 = 0", note: "The ball hits the ground when h = 0." },
  { op: "Rearrange into standard form", eq: "-4.9t^2 + 10t + 20 = 0", note: null },
  { op: "Identify coefficients (the quadratic formula handles any sign of a)", eq: "a = -4.9,\\quad b = 10,\\quad c = 20", note: "No need to rearrange further — the formula works with a negative leading coefficient." },
  { op: "Compute the discriminant", eq: "\\Delta = 10^2 - 4(-4.9)(20) = 100 + 392 = 492", note: "Positive discriminant means two real solutions; we take the one giving positive t." },
  { op: "Apply the quadratic formula", eq: "t = \\frac{-10 \\pm \\sqrt{492}}{2 \\times (-4.9)} = \\frac{-10 \\pm 22.18}{-9.8}", note: null },
  { op: "Take the root that gives positive t", eq: "t = \\frac{-10 - 22.18}{-9.8} = \\frac{-32.18}{-9.8} \\approx 3.28 \\text{ s}", note: "The other root gives t ≈ −1.24 s, which is before the ball was thrown." },
  { op: "Check", eq: "h(3.28) = 20 + 10(3.28) - 4.9(3.28)^2 \\approx 20 + 32.8 - 52.7 \\approx 0.1 \\approx 0 \\checkmark", note: null }
])

Factor $x^2 - 7x + 12$ completely, then write down the two roots of $x^2 - 7x + 12 = 0$.

Code

makeStepperHTML(2, [
  { op: "Write the trinomial", eq: "x^2 - 7x + 12", note: null },
  { op: "Find two numbers that multiply to 12 and add to -7", eq: "(-3) \\times (-4) = 12 \\quad\\text{and}\\quad (-3) + (-4) = -7", note: null },
  { op: "Write the factored form", eq: "x^2 - 7x + 12 = (x - 3)(x - 4)", note: null },
  { op: "Set each factor to zero", eq: "x - 3 = 0 \\implies x = 3 \\qquad x - 4 = 0 \\implies x = 4", note: null },
  { op: "Check", eq: "(3)^2 - 7(3) + 12 = 9 - 21 + 12 = 0 \\checkmark \\qquad (4)^2 - 7(4) + 12 = 16 - 28 + 12 = 0 \\checkmark", note: null }
])

A rectangular garden has an area of 48 m². Its length is 2 m more than its width. Find the dimensions.

Code

makeStepperHTML(3, [
  { op: "Let width = x, then length = x + 2", eq: "x(x + 2) = 48", note: null },
  { op: "Expand", eq: "x^2 + 2x = 48", note: null },
  { op: "Rearrange into standard form", eq: "x^2 + 2x - 48 = 0", note: null },
  { op: "Find two numbers that multiply to -48 and add to 2", eq: "8 \\times (-6) = -48 \\quad\\text{and}\\quad 8 + (-6) = 2", note: null },
  { op: "Factor", eq: "(x + 8)(x - 6) = 0", note: null },
  { op: "Solve — take the positive root", eq: "x = 6 \\text{ (width)}, \\quad x + 2 = 8 \\text{ (length)}", note: "x = -8 is rejected since a width cannot be negative." },
  { op: "Check", eq: "6 \\times 8 = 48 \\checkmark", note: null }
])

Use the quadratic formula to solve $2x^2 - 3x - 2 = 0$.

Code

makeStepperHTML(4, [
  { op: "Identify coefficients", eq: "a = 2,\\quad b = -3,\\quad c = -2", note: null },
  { op: "Compute the discriminant", eq: "\\Delta = (-3)^2 - 4(2)(-2) = 9 + 16 = 25", note: null },
  { op: "Apply the quadratic formula", eq: "x = \\frac{3 \\pm \\sqrt{25}}{4} = \\frac{3 \\pm 5}{4}", note: null },
  { op: "Find both solutions", eq: "x = \\frac{3 + 5}{4} = 2 \\qquad\\text{or}\\qquad x = \\frac{3 - 5}{4} = -\\frac{1}{2}", note: null },
  { op: "Check x = 2", eq: "2(4) - 3(2) - 2 = 8 - 6 - 2 = 0 \\checkmark", note: null },
  { op: "Check x = -1/2", eq: "2\\left(\\frac{1}{4}\\right) - 3\\left(-\\frac{1}{2}\\right) - 2 = \\frac{1}{2} + \\frac{3}{2} - 2 = 0 \\checkmark", note: null }
])

A candle company finds its weekly profit in dollars is $P(p) = -3p^2 + 48p - 165$, where $p$ is the price per candle. At what price is profit maximised, and what is that profit?

Code

makeStepperHTML(5, [
  { op: "Identify coefficients", eq: "a = -3,\\quad b = 48,\\quad c = -165", note: "The negative a confirms this is a downward parabola — it has a maximum." },
  { op: "Find the price at the vertex", eq: "p_v = -\\frac{b}{2a} = -\\frac{48}{2(-3)} = -\\frac{48}{-6} = 8", note: null },
  { op: "Substitute to find maximum profit", eq: "P(8) = -3(64) + 48(8) - 165 = -192 + 384 - 165 = 27", note: null },
  { op: "State the answer", eq: "\\text{Maximum profit: } \\$27 \\text{ per week at a price of } \\$8", note: null },
  { op: "Check", eq: "P(7) = -3(49) + 48(7) - 165 = -147 + 336 - 165 = 24 < 27 \\checkmark", note: "Checking a neighbouring value confirms 8 is the peak." }
])

Factor $4x^2 - 25$ completely.

Code

makeStepperHTML(6, [
  { op: "Recognise the structure", eq: "4x^2 - 25 = (2x)^2 - 5^2", note: "Both terms are perfect squares separated by a minus sign — use the difference of squares identity." },
  { op: "Apply a² - b² = (a + b)(a - b)", eq: "4x^2 - 25 = (2x + 5)(2x - 5)", note: null },
  { op: "Check by expanding", eq: "(2x + 5)(2x - 5) = 4x^2 - 10x + 10x - 25 = 4x^2 - 25 \\checkmark", note: null }
])