# Do not expand it **Teaches:** `factored-form-is-readable`, `maximize-the-square-instead` · **Assumes:** `species-of-the-unknown`, `gross-vs-exact`, `square-is-never-negative` · **~30 min** --- ## What you are holding Brahmagupta's rule handed you the squared area of the field family, in this form: $$(\text{area})^2 \;=\; x^2(500-x)^2, \qquad 0 \le x \le 500$$ Your instinct — trained by every worksheet you have ever done — is to multiply it out. Do it once, so that we can talk about it: $$x^2(500-x)^2 \;=\; x^2\big(250000 - 1000x + x^2\big) \;=\; x^4 - 1000x^3 + 250000x^2$$ Now get out the card from `quad-01-species`, and let us settle which of these two expressions is the useful one. Not which is *simpler* — they have the same number of symbols, roughly. Which one **answers questions.** ## The comparison Take each form and write down every fact you can read off it with no further work. ### $x^2(500-x)^2$ - **It is a square.** It is $\big(x(500-x)\big)^2$ — a thing multiplied by itself. Therefore it is never negative, anywhere, and no sign analysis is required to know that. - **Its square root is $x(500-x)$**, on this domain, because both factors are non-negative there. So the area itself is available immediately, and it is a quantity you already understand completely from `quad-02-parts-of-500`. - **It is zero exactly at $x = 0$ and $x = 500$**, because a product is zero exactly when a factor is. Those are the two ways the field can collapse: the parcels flatten into a line and there is no land. - **Each zero is doubled.** $x$ appears twice and $(500-x)$ appears twice. Near $x=0$ the area is about $500x$, so the *squared* area is about $250000x^2$ — it dies away like a square, not like a line. The field does not merely shrink to nothing at the ends; it shrinks to nothing unusually fast. - **Its largest value is $62500^2$, at $x=250$** — by the argument in the next section. ### $x^4 - 1000x^3 + 250000x^2$ - It is a fourth-degree polynomial. That is the honest length of the second list. Everything on the first list is *still true* of the second expression — they are the same quantity — but none of it is **legible** there. You would have to factor it back to find out, which means the expansion cost you work and returned nothing. > **Do this.** Factor $x^4 - 1000x^3 + 250000x^2$ back to $x^2(500-x)^2$, by hand. Take > $x^2$ out first and look hard at what is left. Then say what you had to recognize in > order to finish. Answer: you had to complete a square. The road back is longer than the > road out. ## Why the old grammar makes this sharper In `quad-01-species` you saw that before the 17th century each power of the unknown was a different kind of thing, and that quantities of different kinds could not be combined. Read the two forms in that grammar. The factored form is *a square, times a square*. One species. It is the kind of thing you get by multiplying an area by an area, and everything about it is describable in one breath. The expanded form is a **square of squares**, less a thousand **cubes**, plus two hundred fifty thousand **squares**. Three species, and the first of them has no natural name at all — the sixteenth-century word for it, *zenzizenzic*, is a construction meaning "the square of squares," invented because there was nothing to point at. So a mathematician of that period did not decline to expand out of laziness. Expanding took a well-formed statement about areas and turned it into a sentence that the language could not license, in exchange for nothing. **The pedagogical rule and the historical constraint coincide**, and that is why the rule is worth trusting: *do not expand a product until you know what the expansion will tell you.* ## The move that makes the fourth degree collapse You want the largest field. That means maximizing the area. But Brahmagupta gave you the *square* of the area, and you would rather not take a square root of a general expression. Here is the step, and it is one line: > On non-negative quantities, squaring preserves order. If $0 \le u \le v$ then > $u^2 \le v^2$, and conversely. So the $x$ that makes $(\text{area})^2$ largest is > **exactly** the $x$ that makes the area largest. Therefore you never need to take the square root at all. Maximize $\big(x(500-x)\big)^2$ by maximizing $x(500-x)$ — a second-degree problem you solved two lessons ago, whose answer is $x = 250$ — and you are done. The fourth-degree question has been answered without ever confronting a fourth-degree expression. Students skip this step in silence, and it is worth stopping on, because the conclusion is only true because of the non-negativity. Squaring does *not* preserve order in general: $-5 < 2$ but $25 > 4$. The reason the argument works here is that an area cannot be negative, and we checked that. > **Do this.** Someone maximizes $\big(x(500-x)\big)^2$ over the whole number line instead > of $[0,500]$, and reports that there is no maximum. Are they right? Now have them > maximize $x(500-x)$ over the whole number line. Are the two answers consistent? What > does this show about the "maximize the square instead" step? ## A rule you can carry The habit generalizes past this problem, and it is one of the few genuinely transferable things in elementary algebra. **A product tells you its zeros and its sign. A sum tells you neither.** So when a quantity arrives factored, the questions "where is it zero," "where is it positive," and "which factor is doing the work" are already answered, and expanding erases all three answers. Conversely, a sum tells you the value at $x=0$ and the behavior for enormous $x$, which a product hides. Neither form is better. **The form is a choice, and the choice is made by the question.** The failure mode this is aimed at is real and common: a student who expands automatically, because expanding feels like progress, and who has therefore destroyed the structure before reading it. Multiplying out is not simplification. It is a translation into a language optimized for different questions. --- ## Exit 1. Without expanding: where is $x^2(500-x)^2$ zero, where is it positive, and where is it negative? Justify each in one sentence. 2. $(\text{area})^2 = 1{,}600{,}000{,}000$. Find $x$. Do not expand anything. *(There is more than one answer.)* 3. Explain why $x^2(500-x)^2$ has a maximum but $x^4 - 1000x^3 + 250000x^2$, considered over all real $x$, does not. Same expression — resolve the contradiction. 4. Read $(x-3)(x+5)(x-3)$ and say, without expanding, all of: its zeros, which zero is doubled, its sign for very large $x$, and its sign just to the left and just to the right of $x=3$. Then check the last one by evaluating at $x = 2.9$ and $x = 3.1$. 5. Give an expression where expanding is clearly the *right* move, and say what question you were asking that made it right. Answers: [`ANSWERS.md`](ANSWERS.md). ## Related - Where the expression came from: `quad-03-assessors-rule`. - Why the old grammar refused the expansion: `quad-01-species`. - Seeing the doubled zeros as tangency: `quad-06-the-turn`.