Numbers & Functions · 4 Polynomials, Factoring, and the Quadratic Formula

4.10 The shape of a polynomial

A quick word on what these machines look like. The degree controls the basic shape: degree 1 is a straight line (one crossing); degree 2 a parabola (up to two); degree 3 an S-like curve (up to three).

x degree 1 line 1 crossing x degree 2 parabola up to 2 crossings x degree 3 cubic up to 3 crossings Degree n → at most n roots → at most n crossings Figure 4.5. Polynomial shapes by degree. A degree-n polynomial can cross the axis at most n times — the highest power is a hard ceiling on the number of roots.

A polynomial of degree n can have at most n roots.

This is your first glimpse of the Fundamental Theorem of Algebra — which, once we have complex numbers, will sharpen “at most n” into “exactly n, if you count them right.” Another reason Chapter 8 matters.

degree 1 degree 2 degree 3 degree 4 degree 5+ NO formula possible (Abel Ruffini) The pattern holds then shatters at 5 Figure 4.6. The solvability ladder. Degrees 1 through 4 each have a formula built from roots; at degree 5 the pattern hits a wall and never recovers (Abel–Ruffini).

4.11 Where we are now

Polynomials are the friendly family: built from only adding and multiplying, accepting every real number, never jamming. We learned to read their degree and leading term, to hunt for their roots, to factor them when the numbers cooperate, and to fall back on the quadratic formula when they don’t. And right at the heart of that formula we found the discriminant, whose negative branch marches us straight back to the oldest pattern in the book: an operation has jammed, and a new kind of number is waiting to be invented. Next we meet the machines that grow.

Things worth thinking about 1. Plain 7 is a polynomial of degree 0 — a machine that ignores its input. What does its graph look like? How many roots does it have? (Careful — when is 7 ever equal to zero?)

2. We said a degree-n polynomial has at most n roots. Imagine a parabola with exactly two roots, then one with exactly one, then one with zero. What’s physically different about where the three sit relative to the axis? 3. The discriminant tells you how many real roots a quadratic has before you finish solving it. When would you want to know “how many roots” without needing the roots themselves?

4. (Looking ahead.) When the discriminant is negative, the parabola “never Polynomials, Factoring, and the Quadratic Formula touches the axis.” Yet we claim it secretly has two roots. How can a curve have roots it never visibly touches? Let the strangeness sit — that discomfort is a new kind of number about to be born.

5. (Deep water.) We can solve degrees 1–4 with formulas, but degree 5 has none — and that was proven. What’s the difference between “nobody has found it yet” and “it’s impossible and here’s why”? Can you think of anything else proven impossible, rather than merely never done?

Problems to work Problem 1 — Factor and read the roots.

Each factors into nice whole-number pieces. Find two numbers that multiply to the constant and add to the middle coefficient, write the factored form, and read off the roots.

(a) x2 −7x + 12 (b) x2 + x −6 (c) x2 −9 (Sneaky (c): what two numbers multiply to −9 and add to 0?)

Problem 2 — Complete the square yourself.

Solve x2 + 8x + 7 = 0 using the four steps from the chapter — don’t use the finished formula. Get the loose number across, find the piece that completes the square, add it to both sides, fold into (something)2, square-root with a ±, and solve. Then check by factoring.

Problem 3 — When factoring fails, turn the crank.

The quadratic x2 −4x + 1 = 0 does not factor nicely — try for a minute and you’ll see. Now use the quadratic formula. Identify a, b, c, plug in, and find both roots. (Your answers will involve a square root that doesn’t simplify — that’s expected, and exactly why factoring failed.)

Problem 4 — Read the fork.

For each quadratic, compute only the discriminant b2 −4ac and use its sign to say — without finding the roots — whether it has two real roots, one repeated, or none.

(a) x2 −6x + 9 (b) x2 + 2x + 5 (c) 2x2 + 3x −1 When you finish (b), pause: you’ve found a quadratic with no real roots. Keep it in your pocket — Chapter 8 will give it the two roots it’s hiding. ⋆Teach it — explain it to someone Problem 5 — Teach it. Explain to someone why (x−2)(x−3) = 0 forces x to be either 2 or 3 — and why the same trick would not let you solve (x −2)(x −3) = 12 by setting each piece equal to 12. (This is a mistake almost everyone makes once. Getting someone else to see why it’s wrong will lock the zero-product idea into your own head for good.)

Problem 6 — Verify a toolkit identity.

Multiply out the right side of (a −b)(a2 + ab + b2) term by term, and watch the middle terms cancel until only a3 −b3 survives. (Hint: you’ll get six terms before any cancelling — write them all down first, then hunt for the pairs that kill each other.)

Problem 7 — Use the toolkit.

Factor each on sight, using a tool from the toolkit. Name which tool you used. (a) x2 −49; (b) x2 + 10x + 25; (c) x3 −8 (hint: 8 = 23, a difference of cubes). Problem 8 — Extend the triangle.

Write the next two rows of Yang Hui’s triangle below 1, 4, 6, 4, 1, using only the build rule. Then use your new row to write down (a + b)5 without multiplying anything.

Going further ⋆Teach it — explain it to someone Problem 9 — Teach it (frozen thought). Explain to a parent or sibling why the quadratic formula isn’t something to be afraid of — it’s just completing-the-square “done once, with letters, so nobody has to do it again.” Walk them through the easy numeric example x2 + 6x + 5 = 0 first, then point at the general version and show it’s the same four moves. The idea to land: a formula is frozen reasoning, and if you ever forget it you can thaw it back out.

Problem 10 — [challenge] A disguised difference of squares.

Factor x4 −16 completely. (Hint: it’s a difference of squares — x4 = (x2)2 and 16 = 42 — but one of the pieces you get is itself a difference of squares. Keep going until nothing factors further over the reals. One factor will stubbornly refuse: x2 + 4. Remember it for Chapter 8.)

Polynomials, Factoring, and the Quadratic Formula Problem 11 — [challenge] Build a quadratic to order.

Find a quadratic x2 + bx + c whose two roots are exactly 5 and −2. (Hint: work backwards — if the roots are 5 and −2, what must the factored form be? Multiply it out.) Then do it again for roots 3 and 3 (a repeated root), and check that your second quadratic has discriminant zero.

Problem 12 — [challenge] Sum and product of roots.

For x2 + bx + c = (x −r)(x −s), multiply the right side out and match coefficients to show that r + s = −b and r s = c. (So the two roots always add to −b and multiply to c — which is exactly why the “find two numbers that add and multiply” factoring game works!) Use this to find, in your head, two numbers that add to 7 and multiply to 10 — and hence factor x2 −7x + 10. Problem 13 — [challenge] The geometric-sum payoff.

Use the identity 1+x+x2+· · ·+xn−1 = xn−1 x−1 to add up 1+2+4+8+· · ·+1024 without adding the eleven numbers one by one. (Hint: this is the doubling chain, so x = 2, and the last term 1024 = 210 tells you n.) Then marvel: a chessboard’s 64 doubling squares total 264 −1 grains — more rice than humanity has ever grown.

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