Algebra 1
Multiplying polynomials
This is the forward direction of the course's main thread: you take (x + 2)(x + 3) and multiply it out to x² + 5x + 6. The next unit hands you x² + 5x + 6 and asks you to find the pieces it came from, so multiplying out now is the very thing you'll learn to undo.
There's one spot where almost everyone loses a piece, the middle term, and the area box in this lesson is built to make that piece impossible to miss.
Start with the simpler case: a single term times a polynomial. The factor outside the parentheses hands a flyer to everyone inside, skipping no one. So 3x(x + 4) gives 3x · x and 3x · 4: $$3x(x+4)=3x\cdot x+3x\cdot 4=3x^2+12x$$ The exponents follow last lesson's product rule: x · x is x², and the coefficient 3 rides along.
Now two binomials, where the middle term lives. Think of (x + 2)(x + 3) as the area of a rectangle whose width is x + 2 and whose height is x + 3. Slice the width into its two parts and the height into its two parts, and the rectangle breaks into four smaller cells. Each cell's area is one little product, and the whole area is all four added up: $$\begin{array}{c|c|c} & x & 3 \\ \hline x & x^2 & 3x \\ \hline 2 & 2x & 6 \end{array}\qquad\Rightarrow\qquad x^2 + \underbrace{3x+2x}_{=\,5x} + 6 = x^2+5x+6$$ There they are: four cells, x², 3x, 2x, and 6. The two in the middle, 3x and 2x, are the ones people drop, and they add up to the 5x that makes this a trinomial instead of just x² + 6.
These same four products, taken in a fixed order, are sometimes called FOIL. The box is worth preferring, because it shows you why a middle term exists and keeps working when the pieces get bigger, where that shortcut runs out.
If the grid isn't your thing, there's a second way to see the same thing: distribute twice. First the x greets both terms in the second parentheses, then the 2 greets both: $$x(x+3)+2(x+3)=x^2+3x+2x+6=x^2+5x+6$$ Same four products, same middle term, no grid. Pick whichever picture you find easier to keep straight. They give the same answer because they are the same work.
New words
none new. This is the distributive property (Unit 2) applied once (monomial × polynomial) or twice (binomial × binomial).
Worked example
Monomial × binomial, distribute: 10.4.w1 $$3x(x+4)=3x^2+12x$$
Binomial × binomial, the area box: 10.4.w2 $$(x+2)(x+3)=x^2+5x+6 \quad(\text{middle term }3x+2x=5x)$$
A negative in one binomial (watch signs): 10.4.w3 $$(x-3)(x-4)=x^2-4x-3x+12=x^2-7x+12$$
Difference of squares, the middle term vanishes: 10.4.w4 $$(x-4)(x+4)=x^2+4x-4x-16=x^2-16$$ The +4x and -4x cancel, and that's why (a-b)(a+b)=a²-b².
Squaring a binomial, the middle term is doubled: 10.4.w5 $$(x+3)^2=(x+3)(x+3)=x^2+3x+3x+9=x^2+6x+9$$ Note that (x+3)²≠x²+9. The middle 6x is the whole point.
Two of those deserve a second look. In the difference-of-squares example, the two middle cells, +4x and −4x, go to zero together, which is the whole reason (x − 4)(x + 4) collapses to just x² − 16.
And in the last one, squaring a binomial, the tempting move is to square each piece and write x² + 9. But (x + 3)² means (x + 3)(x + 3), and running the box gives a middle term of 3x + 3x = 6x. So (x + 3)² is x² + 6x + 9; the 6x is exactly the part that gets dropped, and catching it is what the example is for.
Across all of these, the error to watch for is the missing middle term: writing (x + 2)(x + 3) as x² + 6, or (x + 3)² as x² + 9. It happens when only the first-and-last products get written down.
The box is the cure, because it makes the question concrete. Every part of the width meets every part of the height, so there are four cells, not two; if you only have two terms, ask where the middle cells went. When a sign is in play, fill each cell with its signed product (x · −4 is −4x) and then add, and a substitution check at x = 1 will flag any sign that went astray.
Run the box once on a friendly case to feel it click. Expand (x + 5)(x + 1). The four cells are x², x, 5x, and 5; the middle two, x and 5x, add to 6x; so it's x² + 6x + 5. Four cells, then collect the middle pair, and that's the whole method.
Check yourself
- 10.4.c1 Expand (x + 2)(x + 5) with the area box, and point to where the middle term 7x comes from. (Four cells: x², 5x, 2x, 10; the middle pair 5x + 2x is the 7x, giving x² + 7x + 10.)
- 10.4.c2 Why does (x − 5)(x + 5) have no middle term, but (x − 5)(x − 5) does? (In the first, the middle cells are −5x and +5x, which go to zero, leaving x² − 25; in the second they're −5x and −5x, which add to −10x, giving x² − 10x + 25.)
- 10.4.c3 Suppose someone writes (x + 4)² = x² + 16. Using the box, show them exactly what they dropped. (Running (x + 4)(x + 4) gives middle cells 4x and 4x; they add to 8x, so the answer is x² + 8x + 16, and the dropped piece is the 8x.)
You can now multiply a single term across a polynomial, multiply two binomials with the area box, and reliably produce the middle term, including the cases where it doubles or vanishes. Hold onto each (x + 2)(x + 3) → x² + 5x + 6 you do: the next unit starts from the right-hand side and hunts for the left, and this same box is the tool you'll fill in reverse.
The set below mixes monomial products with binomial products instead of grouping them, harder in the moment but better for remembering. The answers wait at the end of the lesson, and each problem matches one of the worked examples above, so reread the right one whenever you get stuck. The last three are the special patterns: watch for a vanishing middle term in one and a doubled one in the others.
Monomial × polynomial (distribute):
Reveal answerHide to problem 1
3x²+12xReveal answerHide to problem 2
6x²-10xReveal answerHide to problem 3
-4x³+8x²-4xBinomial × binomial (use the area box):
Reveal answerHide to problem 4
x²+5x+6Reveal answerHide to problem 5
x²+6x+5Reveal answerHide to problem 6
x²+5x-14Reveal answerHide to problem 7
x²-7x+12With a leading coefficient:
Reveal answerHide to problem 8
2x²+7x+3Reveal answerHide to problem 9
3x²+13x-10Special patterns (preview of Unit 11):
Reveal answerHide to problem 10
x²-16Reveal answerHide to problem 11
x²+6x+9Reveal answerHide to problem 12
x²-10x+25Substitution spot-checks at x=1: #6: (-1)(8)=-8, and 1+5-14=-8. #9: (1)(6)=6, and 3+13-10=6. #11: (4)²=16, and 1+6+9=16.
You've reached the end of building expressions up. You can simplify with the exponent rules and rebuild the zero and negative cases when memory slips; move a number between standard and scientific form and compute in scientific form; name and add and subtract polynomials with the subtraction signs handled; and multiply binomials with the area box, middle term and all. The next unit takes that last box and runs it backward: you'll be handed the finished area and asked to find the edges. Every (x + 2)(x + 3) → x² + 5x + 6 you practiced is one you're about to learn to read from right to left.