University of Free Knowledge
QA 152 · fol. 16

When the Unknown Multiplies Itself

A product equals zero only when a factor does, so factoring x² + bx + c into (x + p)(x + q) — product c, sum b — turns one quadratic equation into two one-step equations. · 12 min

Every equation you have solved so far kept the unknown to the first power: x appeared, perhaps on both sides, but never multiplied by itself. One balancing sequence always cornered it. An equation like x² + 5x + 6 = 0 is different. Subtract, divide as you like — the x² and the x refuse to merge, and balancing stalls. Solving it takes one new principle about the number zero, plus a skill you already own: un-multiplying an expression into factors.

Guess before you learn

Two numbers multiply to give exactly 0. What do you know for certain?

THE DEPTH DIAL — the same idea, younger or deeper
9–12

9–12

To use the property on x² + bx + c = 0, reverse the multiplication. Expanding (x + p)(x + q) gives x² + (p + q)x + pq — the middle coefficient is the sum of p and q, the constant their product. Factoring is the reverse search: find two numbers with product c and sum b.

Geometry confirms the algebra: y = x² + bx + c graphs as a parabola, and the solutions of x² + bx + c = 0 are exactly the x-values where that parabola crosses the x-axis — the inputs where y is 0. Two crossings, one touch, or none: a quadratic may have two, one, or zero real solutions, and factoring finds them whenever the crossings land on rational numbers.

zero-product property

If a product of factors equals 0, at least one factor is 0. The licence to split (x + 2)(x + 3) = 0 into two one-step equations.

-4-2024-505xyy = x² + x − 6x = −3x = 2
PLATE I The parabola y = x² + x − 6 crosses the x-axis at exactly the solutions of x² + x − 6 = 0.

Now the reverse skill. Multiply (x + p)(x + q) and collect terms: x² + (p + q)x + pq. Read that as a recipe in reverse — to factor x² + bx + c, hunt for two numbers whose product is c and whose sum is b. For x² + 5x + 6, the pairs multiplying to 6 are 1 · 6 and 2 · 3; only 2 + 3 makes 5. So x² + 5x + 6 = (x + 2)(x + 3), and an equation nobody could balance becomes two that anyone can.

Solve x² + 5x + 6 = 0 — the steps fade as you master them

1
Find two numbers with product 6 and sum 5
2 and 3
2
Factor the left side
(x + 2)(x + 3) = 0
3
Apply the zero-product property
x + 2 = 0 or x + 3 = 0
4
Solve each one-step equation
x = −2 or x = −3
5
Check x = −2: compute (−2)² + 5(−2) + 6
4 − 10 + 6 = 0, which matches
Retrieval Gate — answer before you continue 0 / 4

1.(x − 4)(x + 1) = 0. The solutions are:

2.(x − 1)(x − 6) = 6. What may you conclude, as written?

3.x² − 5x + 6 = 0 factors as (x − 2)(x − 3) = 0. Give the larger solution.

4.In one sentence: why does the splitting move fail on (x + 1)(x + 2) = 12?

Signs carry information. In x² − 7x + 12, the product must be +12 and the sum −7: both numbers are negative, giving (x − 3)(x − 4). In x² + 2x − 8, the product is −8, so the numbers have opposite signs, and the sum +2 says the positive one is larger: (x + 4)(x − 2). You never memorize the cases. The product decides whether the signs agree; the sum decides which sign leads.

Ink That Thinks — guess first; the answer draws itself.
The parabola y = x² − 2x − 3. In pencil, place its two x-axis crossings and its lowest point.

-2024-4-2024xy
Tap to place each point.
PLATE II Roots in graphite first, then in ink — the factors know where the parabola lands.
EQUATIONPAIR NEEDEDFACTORED FORMSOLUTIONSx² + 7x + 12 = 03 and 4(x + 3)(x + 4) = 0−3 and −4x² − 5x + 6 = 0−2 and −3(x − 2)(x − 3) = 02 and 3x² + 2x − 8 = 04 and −2(x + 4)(x − 2) = 0−4 and 2x² − 9 = 03 and −3(x + 3)(x − 3) = 03 and −3
PLATE III Four factorings: the pair multiplies to c and sums to b; the solutions are the pair with signs flipped.
Why is this true?

Why are the solutions the pair with signs flipped — the pair 2, 3 giving roots −2, −3?

Because the factor x + 2 vanishes when x = −2. The root is whatever makes its factor zero, which is the opposite of the number written inside it.

Retrieval Gate — answer before you continue 0 / 4

1.Solve x² + 2x − 8 = 0. Give the negative solution.

2.Match each quadratic to its factored form — the signs do the sorting.

x² + 7x + 12
x² − 7x + 12
x² + x − 12
x² − x − 12

3.A parabola y = x² + bx + c touches the x-axis at exactly one point. What does that say about x² + bx + c = 0?

4.From memory: state the zero-product property, and why the right side must be 0 before factoring helps.

This folio closes the course, and the arc is worth seeing whole. A letter held a number. Expressions gave the letter grammar; equations made claims about it; balancing recovered it. Lines put pairs of unknowns in a plane, and systems crossed them. And when the unknown finally multiplied itself, zero — of all numbers — handed you the key. Algebra II takes up the quadratics that refuse to factor; geometry takes up the plane. Both will assume everything in these sixteen folios, and you now own all of it.

Note

The Examination Desk — tests, typeset properly — waits at the end of this course. These sixteen folios are its entire syllabus.

Practice — new ink and old, interleaved

1.Check by substitution: evaluate (−3)² + 5(−3) + 6.

2.Match each expression to its simplified form.

3⁴ · 3²
3⁴ / 3²
(3⁴)²
3⁰

3.From memory: how do you factor x² + bx + c, and what must the two numbers satisfy?

4.Which equation is ready for the zero-product property, exactly as written?

5.Match each expression to its simplified form.

4a + 3a
6b − b
9c − 9c
2d + 5

6.Put the moves for solving a quadratic by factoring in working order.

  1. Check a solution in the original equation
  2. Set each factor equal to 0
  3. Factor the left side into (x + p)(x + q)
  4. Arrange the equation as x² + bx + c = 0
  5. Solve the two one-step equations

7.Without looking back: what is a variable, and what does 7m mean?

8.Solve x² − 9x + 20 = 0. Give the smaller solution.

9.You substitute y = x − 2 into 3y + x = 10. Which is correct?

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