Recognising and using the special product patterns. Includes the sum and difference pattern, the square of a binomial in both signs, the area model that explains where the middle term comes from, the classic error of squaring term by term, and using the patterns to write the area of a region as a difference.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 10 — Polynomials and Factoring
Special Products of Polynomials
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 581-587 — the lesson these objectives are drawn from
Warm-up
Lesson 10.2 expanded every product the same way. Some products come out unusually tidy, and this lesson is about spotting those in advance.
Discussion prompt
Expand the product of y plus three and y minus three with FOIL. What is unusual about the answer?
Hint: Look at the two middle products.
Answer:
\[ (y + 3)(y - 3) = y^2 - 3y + 3y - 9 = y^2 - 9 \]
The two cross products were negative three y and positive three y, which cancelled exactly. A product of two binomials normally leaves three terms and this one left two, which is worth recognising before doing the work rather than after.
Concept
Certain pairs of binomials have products that follow a fixed pattern. Recognising such a pair makes the multiplication quicker and, later, makes the reverse process of factoring possible.
special product — A product of binomials that follows a fixed pattern, such as the sum and difference pattern or the square of a binomial pattern.
In each pattern, a and b may be numbers, variables or whole expressions.
Figure (svg): The sum and difference pattern, with the middle terms cancelling
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 581-581
Section
Section 1
Concept
The product of the sum and the difference of two terms is the difference of their squares. The two cross products are opposites, so they add to nought.
\[ (a + b)(a - b) = a^2 - b^2 \]
The result has two terms rather than three.
Figure (svg): The sum and difference pattern, with the middle terms cancelling
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 581-581 — the Sum and Difference Pattern and its worked development
Picture it
The cross products cancel.
Figure (svg): The sum and difference pattern, with the middle terms cancelling
Nothing is lost in the cancelling; the two middle products genuinely sum to nought. That is why the shortcut is exact rather than an approximation.
Worked example
This is Example 1 from the textbook.
\[ \text{Find } (5t + 2)(5t - 2). \]
Identify a and b
Why: The two terms of the binomials.
\[ a = 5 t, \; b = 2 \]
Write the pattern
Why: Difference of squares.
\[ a ^{2} - b ^{2} \]
Substitute
Why: Square each part.
\[ (5 t) ^{2} - 2 ^{2} \]
Simplify
Why: Five squared times t squared.
\[ 25 t ^{2} - 4 \]
Figure (svg): A special product checked by expanding it the long way
\[ (5t + 2)(5t - 2) = 25t^2 - 4 \]
Verify: expand it with FOIL
Why: The four products are twenty-five t squared, negative ten t, positive ten t and negative four, and the middle two cancel to leave twenty-five t squared minus four. The pattern and the long route agree, as they must.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 582-582
Matching
Square each term completely.
Match the pairs
Why: In three of these the leading coefficient had to be squared as well as the variable. Only the first has a leading coefficient of one, which is why it is the least useful example to learn the pattern from.
Worked example
Guided Practice 1 to 6.
\[ \text{Find } (x + 2)(x - 2), \; (n - 3)(n + 3), \; (2x + 1)(2x - 1) \text{ and } (3x - 2)(3x + 2). \]
Take the first two
Why: The order of the brackets does not matter.
\[ x ^{2} - 4, \; n ^{2} - 9 \]
Take the third
Why: Two x squared is four x squared.
\[ 4 x ^{2} - 1 \]
Take the fourth
Why: Three x squared is nine x squared.
\[ 9 x ^{2} - 4 \]
Note the pattern
Why: Each answer is a difference of two squares.
Figure (svg): The sum and difference pattern, with the middle terms cancelling
\[ x^2 - 4, \quad n^2 - 9, \quad 4x^2 - 1, \quad 9x^2 - 4 \]
Verify: check that the coefficient is squared too
Why: In the third, a is two x rather than x, so a squared is four x squared and not two x squared. Squaring the whole term including its coefficient is the step most often done only halfway.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 582-582
Trap
\[ (2x + 1)(2x - 1) = 2x^2 - 1 \]
Square the x and carry the two along
Why: The coefficient was treated as a spectator.
The whole term two x is being squared, so the two is squared as well: two x squared is four x squared. Substituting x equal to one shows it — the original is three times one, which is three, and four minus one is three while two minus one is one.
\[ (2x + 1)(2x - 1) = (2x)^2 - 1^2 = 4x^2 - 1 \]
Write the brackets around the whole term before squaring
Why: a is the entire term, coefficient included.
The Study Tip makes this explicit: a and b may be numbers, variables or whole expressions.
Faded example
Coefficient included.
Fill in the blanks
(5t + 2)(5t - 2) = (5t)^2 - 2^2 = 25t^2 - 4
Why: Squaring five t gives twenty-five t squared, because both factors of the term are squared. Writing the brackets first is what makes that automatic rather than something to remember.
Elimination
The pattern needs a sum and a difference of the same two terms.
Eliminate the wrong options
Which of these is a sum and difference product?
Survives elimination: A
Why: The pattern requires the same two terms in both brackets with one sign flipped. Options B and D belong to the next section's pattern, and option C fits neither and needs ordinary FOIL.
Socratic
It happens every time.
Discussion prompt
Explain why the Outer and Inner products always cancel for a sum and difference. Then say what the result tells you about the graph of such a product.
Hint: Write both cross products out.
Answer:
The Outer product is a times negative b and the Inner is b times a. Those are the same magnitude with opposite signs, so their sum is nought regardless of what a and b are — which is why the cancellation is guaranteed rather than a coincidence of particular numbers.
The result a squared minus b squared has no middle term, so as a quadratic it has b equal to nought and its axis of symmetry is the y-axis. That fits: the product is nought exactly when a equals b or a equals negative b, and those two roots are symmetric about nought, which is precisely the case Lesson 9.2 solved by taking square roots.
Section
Section 2
Concept
The square of a binomial has three terms: the square of the first, twice the product of the two, and the square of the second. Only the middle sign depends on the binomial's sign.
\[ (a + b)^2 = a^2 + 2ab + b^2, \quad (a - b)^2 = a^2 - 2ab + b^2 \]
The last term is a square, so it is never negative.
Figure (svg): The square of a binomial pattern in both signs
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 581-582 — the Square of a Binomial Pattern and Example 2
Picture it
Only the middle term changes.
Figure (svg): The square of a binomial pattern in both signs
The last term being a square is a useful check: an answer ending in a negative constant cannot have come from squaring a binomial.
Worked example
This is Example 2, part a, from the textbook.
\[ \text{Find } (3n + 4)^2. \]
Identify a and b
Why: The two terms.
\[ a = 3 n, \; b = 4 \]
Square the first
Why: Three squared times n squared.
\[ 9 n ^{2} \]
Double the product
Why: Twice three n times four.
\[ 24 n \]
Square the second
Why: Four squared.
\[ 16 \]
Figure (svg): The square of a binomial pattern in both signs
\[ (3n + 4)^2 = 9n^2 + 24n + 16 \]
Verify: substitute one for n
Why: The original is seven squared, which is forty-nine, and the answer gives nine plus twenty-four plus sixteen, also forty-nine. Squaring a number and adding up the coefficients are two very different calculations that must agree.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 582-582
Faded example
The middle term has a factor of two.
Fill in the blanks
(3n + 4)^2 = 9n^2 + 2(3n)(4) + 16 = 9n^2 + 24n + 16
Why: The middle term is twice the product of the two terms, not just their product. Forgetting the two gives twelve n and an answer that fails a substitution check immediately.
Worked example
This is Example 2, part b.
\[ \text{Find } (2x - 7y)^2. \]
Identify a and b
Why: Both terms carry variables.
\[ a = 2 x, \; b = 7 y \]
Square the first
Why: Two squared times x squared.
\[ 4 x ^{2} \]
Double the product, with a minus
Why: Twice two x times seven y.
\[ -28 x y \]
Square the second
Why: Seven squared times y squared.
\[ 49 y ^{2} \]
Figure (svg): The square of a binomial pattern in both signs
\[ (2x - 7y)^2 = 4x^2 - 28xy + 49y^2 \]
Verify: check the last term's sign
Why: Negative seven y squared is positive forty-nine y squared, because squaring removes the sign. Only the middle term carries the minus, which is exactly what distinguishes the two versions of the pattern.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 582-582
Error analysis
The student squared a binomial containing a subtraction.
Annotate
On: \( \begin{aligned} (4x - 1)^2 &= (4x)^2 - 2(4x)(1) - 1^2 \\ &= 16x^2 - 8x - 1 \end{aligned} \)
The reliable way to avoid this is to notice that the last term of any squared binomial is a perfect square and therefore never negative. An answer ending in a negative constant has either used the wrong pattern or lost a sign, and that single observation catches the error without redoing the work.
Sorting
Sum and difference, or square of a binomial.
Sort into buckets
Sort each product by which pattern fits it.
The two families differ only in whether the second bracket's sign is flipped, and that one difference decides whether the middle term doubles or vanishes. Everything else about the two patterns is identical.
Translation
Square, double, square.
Match the pairs
Why: Every one of these ends in a positive constant, including the two that came from differences. Two of them have a negative middle term, which is the only place the original sign shows up.
Socratic
It would be easy to expect just ab.
Discussion prompt
Explain where the factor of two in the middle term comes from. Then say why the sum and difference pattern has no middle term at all.
Hint: Count the cross products in each case.
Answer:
Expanding the square by FOIL gives two cross products, the Outer and the Inner, and for a square they are both a times b. Two copies of the same quantity add to twice it, which is exactly the two a b in the pattern — the factor of two counts the cross products rather than appearing from nowhere.
For a sum and difference the same two cross products appear, but one carries a minus sign, so instead of doubling they cancel. Both patterns come from the same four products, and the sign in the second bracket decides whether the middle contributions reinforce each other or destroy each other.
Section
Section 3
Concept
A square of side a plus b can be divided into a square of area a squared, a square of area b squared, and two rectangles each of area a b. Those four pieces are the four terms of the pattern.
The two rectangles are what produce the factor of two.
Figure (svg): A square of side a plus b divided into four regions
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 583-583 — the area model for the square of a binomial pattern
Picture it
Two of them are identical.
Figure (svg): A square of side a plus b divided into four regions
Because the two rectangles are congruent, their combined area is two a b rather than two different quantities. The picture makes the doubling obvious in a way the algebra does not.
Worked example
Rebuilding the rule rather than recalling it.
\[ \text{Use the area model to expand } (x + 5)^2. \]
Draw the large square
Why: Side x plus five.
\[ \text{area } (x + 5)^2 \]
Name the corner squares
Why: One of side x, one of side five.
\[ x^2 \text{ and } 25 \]
Name the two rectangles
Why: Each five by x.
\[ 5x \text{ each} \]
Add the four pieces
Why: Two rectangles are identical.
\[ x ^{2} + 10 x + 25 \]
Figure (svg): A square of side a plus b divided into four regions
\[ (x + 5)^2 = x^2 + 10x + 25 \]
Verify: check with a number
Why: Take x equal to two: the square has side seven and area forty-nine, and the four pieces are four, ten, ten and twenty-five, which total forty-nine. The picture and the algebra agree piece by piece.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 583-583
Faded example
Two squares and two rectangles.
Fill in the blanks
(a + b)^2 = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2
Why: Both rectangles have the same area, which is why they combine into two a b rather than staying separate. Writing them out before combining is what makes the factor of two visible.
Worked example
Comparing the correct area with the missing pieces.
\[ \text{How much area does } x^2 + 25 \text{ leave out of } (x + 5)^2? \]
Write the correct area
Why: All four pieces.
\[ x ^{2} + 10 x + 25 \]
Write the naive answer
Why: The two corner squares only.
\[ x ^{2} + 25 \]
Subtract
Why: The difference is the rectangles.
\[ 10 x \]
Interpret
Why: Two rectangles of area five x.
Figure (svg): Two columns contrasting the correct square of a binomial with the common error
\[ (x^2 + 10x + 25) - (x^2 + 25) = 10x \]
Verify: test how large the omission is
Why: At x equal to ten the correct area is two hundred and twenty-five and the naive answer is a hundred and twenty-five, so a hundred square units are missing — nearly half the square. The error is not a small one, which is worth knowing since it looks so harmless on paper.
Trap
The picture is just an illustration, so the pattern still has to be memorised.
Learn the three terms by rote and skip the diagram
Why: The formula is short enough to remember.
Rote memory fails under pressure, and the commonest failure is dropping the middle term — which is precisely the part the picture makes impossible to forget.
Sketch the square, name the four pieces, and read the pattern off.
Rebuild the pattern from the picture whenever you are unsure
Why: It takes about ten seconds.
A rule you can reconstruct is worth more than one you can only recall.
Hypothesis
Writing the square as a squared plus b squared.
Predict first
In terms of the picture, what does that answer leave out?
Correct: The two rectangles, whose total area is 2ab.
\[ (a + b)^2 - (a^2 + b^2) = 2ab \]
Why: Adding only the two corner squares accounts for the top-left and bottom-right pieces and ignores the other two entirely. Those two rectangles are not small: for a square of side x plus five with x equal to ten, they account for a hundred of the two hundred and twenty-five square units. Seeing the omission as a visible chunk of the picture is what makes the error hard to repeat.
Elimination
In the area model for a squared binomial.
Eliminate the wrong options
Which pieces combine into the middle term?
Survives elimination: A
Why: Each rectangle has area a b and there are two of them, so together they contribute two a b. That is the only place in the picture where a factor of two can come from.
Socratic
Areas cannot be negative.
Discussion prompt
Say why the area model is drawn for a sum rather than a difference. Then say how the difference pattern can still be got from it.
Hint: What would a negative side length mean?
Answer:
A picture needs positive lengths, so a minus b cannot be drawn as a side directly. The model is therefore stated for a sum, where every piece is a genuine area and every term of the pattern is positive.
The difference pattern follows by replacing b with negative b in the sum pattern: the squared terms are unaffected because squaring removes the sign, and the middle term picks up the minus. So one picture supports both rules, and the sign change is handled algebraically rather than geometrically — which is the usual division of labour between a diagram and a formula.
Section
Section 4
Concept
Squaring a sum is not the same as squaring each term. The square of a binomial always has three terms, and dropping the middle one changes the value.
\[ (a + b)^2 \ne a^2 + b^2 \]
One substitution separates the two versions.
Figure (svg): Two columns contrasting the correct square of a binomial with the common error
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 581-582 — the Square of a Binomial Pattern, which has three terms
Picture it
The check settles it.
Figure (svg): Two columns contrasting the correct square of a binomial with the common error
The two versions differ by two a b, which is usually the largest part of the answer for small values. There is no range of numbers where the shortcut happens to be safe.
Worked example
One number is enough to disprove it.
\[ \text{Is } (x + 5)^2 \text{ equal to } x^2 + 25? \text{ Test } x = 1. \]
Evaluate the left side
Why: One plus five, squared.
\[ 36 \]
Evaluate the right side
Why: One plus twenty-five.
\[ 26 \]
Compare
Why: They differ by ten.
Identify the gap
Why: Twice one times five.
\[ 2 a b = 10 \]
Figure (svg): Two columns contrasting the correct square of a binomial with the common error
\[ 36 \ne 26, \quad \text{difference } 10 = 2(1)(5) \]
Verify: try a second value
Why: At x equal to three the left side is sixty-four and the right is thirty-four, a gap of thirty — which is twice three times five. The gap is always two a b, so it grows with x and never closes.
Sorting
Products yes, sums no.
Sort into buckets
Sort each statement by whether it is true for all values.
Every true statement involves a product and every false one involves a sum or a difference. That is the whole distinction, and it is worth stating in those terms rather than as a list of cases.
Worked example
Looking for exceptions.
\[ \text{For which values is } (x + 5)^2 = x^2 + 25? \]
Set the two equal
Why: Subtract the common terms.
\[ 10 x = 0 \]
Solve
Why: Divide by ten.
\[ x = 0 \]
Interpret
Why: Only when one term is nought.
Conclude
Why: Never in general.
Figure (svg): Two columns contrasting the correct square of a binomial with the common error
\[ 10x = 0 \;\Longrightarrow\; x = 0 \]
Verify: say why that single case is not a defence
Why: Two expressions that agree at one point are not the same expression, in the same way that two lines crossing at a point are not the same line. Agreement everywhere is what equality of expressions means, and here it fails for every value but one.
Trap
\[ (x + 5)^2 = x^2 + 5^2 = x^2 + 25 \]
Apply the exponent to each term, as with a product
Why: Lesson 8.1 let an exponent reach both factors of a product.
It did, and a sum is not a product. The power of a product rule applies to multiplication only, and the middle term two a b is exactly what a sum adds that a product does not.
\[ (x + 5)^2 = x^2 + 10x + 25 \]
Expand the square as a product of two identical binomials
Why: Then all four products appear.
The same distinction blocked splitting a radical over a sum in Lesson 9.3, and for the same reason.
Faded example
How far wrong the shortcut is.
Fill in the blanks
(x + 5)^2 - (x^2 + 25) = 10x, \quad \text30 x = 3 \text___ ___
Why: The gap is the whole middle term, so it grows without bound as x does. There is no size of x for which the shortcut is even approximately safe.
Prediction
You suspect a squared binomial was expanded wrongly.
Predict first
What single check exposes it soonest?
Correct: Substitute x equal to one into both forms.
\[ (1 + 5)^2 = 36 \quad \text{against} \quad 1 + 25 = 26 \]
Why: At x equal to one the left side is a small square and the right side is the sum of the coefficients, so both are computed in a couple of seconds and any missing term shows up as a mismatch. Re-reading tends to reproduce the same misreading, graphing is far slower, and the last term's sign catches a different error — a negative constant — but not a missing middle term, which leaves the signs perfectly plausible.
Socratic
It fails for sums but works for products.
Discussion prompt
Explain why squaring a product does distribute over its factors. Then say what a sum lacks that makes the same move fail.
Hint: Write the square out as a repeated multiplication.
Answer:
Squaring x y means x y times x y, and multiplication can be rearranged freely, so the factors regroup as x times x times y times y — which is x squared times y squared. Nothing is created or lost, because every factor simply appears twice.
Squaring x plus y means x plus y times x plus y, and expanding that needs the distributive property, which produces four products rather than two. The two cross products have nowhere to go under the naive rule, and they are exactly the middle term. A sum has no rearrangement that turns a repeated addition into a product of squares, which is why the move works in one case and not the other.
Section
Section 5
Concept
When a region is described as one shape with another removed, both areas can often be written as special products. Expanding each with a pattern and subtracting is faster than expanding by FOIL.
Squared terms often cancel in the subtraction.
Figure (svg): A square with a smaller rectangle removed from it
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 583-583 — Example 3, Find the Area of a Figure, and its verbal model
Picture it
Both pieces are special products.
Figure (svg): A square with a smaller rectangle removed from it
The large square is a squared binomial and the removed rectangle is a sum and difference, so both expansions are one line each. The subtraction then collapses the quadratic terms.
Worked example
This is Example 3 from the textbook.
\[ \text{Find the area of the region left when } (x + 1)(x - 1) \text{ is removed from } (x + 3)^2. \]
Write the verbal model
Why: Blue equals whole minus removed.
\[ A = (x + 3) ^{2} - (x + 1) (x - 1) \]
Expand the square
Why: Square of a binomial.
\[ x ^{2} + 6 x + 9 \]
Expand the rectangle
Why: Sum and difference.
\[ x ^{2} - 1 \]
Subtract
Why: Flip both signs.
\[ 6 x + 10 \]
Figure (svg): A square with a smaller rectangle removed from it
\[ (x^2 + 6x + 9) - (x^2 - 1) = 6x + 10 \]
Verify: test with a number
Why: Take x equal to four: the square is seven by seven, an area of forty-nine, and the removed rectangle is five by three, an area of fifteen, leaving thirty-four. The answer gives twenty-four plus ten, which is also thirty-four.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 583-583
Faded example
Two patterns in one problem.
Fill in the blanks
(x + 3)^2 - (x + 1)(x - 1) = (x^2 + 6x + 9) - (x^2 - 1) = 6x + 10
Why: The minus in front of the bracket turns the negative one into a positive one, which is why the constant grows rather than shrinks. That sign flip is the step the trap above skips.
Worked example
A quadratic minus a quadratic.
\[ \text{Why has the difference of two quadratic areas no } x^2 \text{ term?} \]
Note both leading terms
Why: Both are x squared.
\[ x^2 \text{ and } x^2 \]
Subtract them
Why: They are identical.
\[ 0 \]
Look at what survives
Why: The linear and constant parts.
\[ 6 x + 10 \]
Interpret
Why: The region is a frame of fixed thickness.
Figure (svg): A square with a smaller rectangle removed from it
\[ x^2 - x^2 = 0 \]
Verify: check that a linear answer is sensible
Why: The removed rectangle grows almost as fast as the square does, so what is left between them widens only slowly. A frame of roughly constant thickness has an area proportional to its perimeter, which grows linearly rather than quadratically — so a linear answer is exactly what the picture predicts.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 583-583
Trap
\[ (x^2 + 6x + 9) - (x^2 - 1) = x^2 + 6x + 9 - x^2 - 1 = 6x + 8 \]
Change the sign of the x squared term and copy the rest
Why: Only the first term looked affected.
The negative one also flips, becoming positive one, so the constant is nine plus one rather than nine minus one. Testing with x equal to four gives thirty-four for the correct answer and thirty-two for this one.
\[ = x^2 + 6x + 9 - x^2 + 1 = 6x + 10 \]
Flip every sign inside the subtracted bracket
Why: Lesson 10.1's rule, unchanged.
Checking the constant term alone would have caught it, since the constants are the easiest part to recompute.
Elimination
A square with a rectangle removed.
Eliminate the wrong options
Which two patterns expand the two areas?
Survives elimination: A
Why: Recognising which pattern each piece fits turns two four-product expansions into two one-line expansions. FOIL would give the same answers, which is worth knowing, but the patterns are why this problem is short.
Hypothesis
Each parent passes on one of two forms of a gene.
Predict first
Why are the mixed pairings twice as likely as either pure one?
Correct: Because a mixed pairing can arise in two ways, one from each parent order.
\[ (a + b)^2 = a^2 + 2ab + b^2 \]
Why: The four cells of the inheritance square are equally likely, but two of them describe the same mixed pairing reached by different routes — the first form from one parent or from the other. Counting both gives twice the chance, which is exactly the two in the middle term of the square of a binomial pattern. The algebra and the biology are counting the same four outcomes.
Socratic
FOIL would give the same answers.
Discussion prompt
Give a reason for learning these patterns other than saving time. Then say where in the next few lessons they become essential.
Hint: Think about running the process backwards.
Answer:
Recognising a pattern lets you run it backwards. Seeing x squared minus nine and knowing it must have come from x plus three times x minus three is factoring, which cannot be done by expanding — expansion only goes one way, and a memorised pattern goes both.
Lesson 10.7 is entirely about that reverse direction, factoring the special products, and Lesson 12.5 uses the square of a binomial pattern backwards to complete the square and derive the quadratic formula. Both depend on recognising a trinomial as a squared binomial on sight, which is a skill that expanding alone never builds.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Sum and difference | Square of a binomial | |
|---|---|---|
| Form | (a + b)(a - b) | (a + b)^2 or (a - b)^2 |
| The cross products | cancel | add, giving 2ab |
| Number of terms | two | three |
One flipped sign in the second bracket is the entire difference between the two columns. Everything in the bottom two rows follows from it.
Pattern
To expand any product of two binomials efficiently, these five moves cover it.
Step one takes a second and decides between a one-line answer and a four-product expansion, which is worth far more than the second it costs.
Check
Square the whole term.
Check your understanding
What is (3x + 4)(3x - 4)?
Answer: A
Why: The pattern gives the square of three x minus the square of four, and three x squared is nine x squared.
Check
The middle term is doubled.
Check your understanding
What is (x + 5) squared?
Answer: A
Why: The middle term is twice the product of x and five, which is ten x, and the last term is five squared.
Check
Squaring removes a sign.
Check your understanding
What is (4x - 1) squared?
Answer: A
Why: Negative one squared is positive one, so only the middle term carries the minus sign.
Real world
This is the tiger question from the lesson opener. Each parent carries two forms of a coat-colour gene and passes one of them on at random, so the offspring's pair can be worked out with a two-by-two square.
Discussion prompt
If both parents carry one of each form, write the four equally likely outcomes and say how the square of a binomial pattern describes them. Then say why the mixed outcome is twice as likely as either pure one.
Hint: Four cells, and two of them match.
Answer:
\[ (a + b)^2 = a^2 + 2ab + b^2 \]
The four cells are the pure first form, two mixed pairings, and the pure second form — exactly the four terms of the pattern, with the coefficients giving the counts.
The mixed outcome is twice as likely because it can happen two ways: the first form from one parent or from the other, and those are genuinely different events with the same result. That is the same two that the area model produced from two congruent rectangles, and it is why a pattern from algebra predicts a result in genetics without either subject knowing about the other.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
What is (2x - 7y) squared?
Correct: 4x squared - 28xy + 49y squared.
\[ (2x - 7y)^2 = (2x)^2 - 2(2x)(7y) + (7y)^2 \]
Why: Squaring a binomial gives three terms: the square of two x, twice the product of the two terms with the binomial's sign, and the square of seven y. The last term is a square and so is positive even though the binomial was a difference, which rules out the first and fourth options immediately. The middle term is twice two x times seven y, which is twenty-eight x y and not fourteen, so the doubling matters. The first option is the sum and difference pattern, which applies when the two brackets have opposite signs — here both brackets are identical, so the cross products add rather than cancel.
Explain it
They wrote that x plus five, squared, is x squared plus twenty-five.
Discussion prompt
In no more than four sentences, explain what is missing and why. Then give them a check they can run in five seconds.
Hint: Draw the square.
Answer:
A usable answer: squaring x plus five means multiplying x plus five by itself, which gives four products rather than two. The two cross products are each five x, so the answer is x squared plus ten x plus twenty-five — and the ten x is what got left out.
The check is to put x equal to one: the left side is six squared, which is thirty-six, and their answer gives twenty-six. Any missing term shows up as a mismatch, and it takes about five seconds.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Choosing the pattern is fixed by asking whether the second bracket's sign is flipped. Squaring the whole term is fixed by writing brackets round it first. The doubling is fixed by rebuilding the area model, where two congruent rectangles make it unavoidable. The last term's sign is fixed by remembering it is a square and therefore never negative. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write both special product patterns, and beside each write one worked example in which the leading coefficient is greater than one so that the squaring of the coefficient is visible. Underneath, expand one sum and difference product by FOIL in full, striking through the two cross products where they cancel, so the shortcut is derived rather than asserted. In the middle, draw the area model for a squared binomial with all four regions labelled, shade the two congruent rectangles in one colour, and write beside them that they are the source of the factor of two. Beneath that, write the false rule that squares a sum term by term, work out both sides at three values of the variable, and write the gap each time to show that it is always twice the product. In the lower half, draw a square with a smaller rectangle removed, label both sets of dimensions, write the verbal model in words, and find the remaining area using one pattern for each piece, marking every sign that flips in the subtraction. Finally, in the margin, write the one-second test for deciding which pattern a product fits.
Your three gap calculations should all come out as twice the product of the two terms. If one of them does not, recheck the arithmetic rather than concluding the shortcut sometimes works.
Recap
Five things, and the fourth is the error the other four exist to prevent.
| If the question says | Your first move is |
|---|---|
| The brackets differ only in a sign | Use a squared minus b squared |
| A binomial is squared | Square, double, square |
| The answer ends in a negative constant | Suspect a lost sign; squares are positive |
| A coefficient sits in front | Square it along with the variable |
| A region has a piece removed | Expand each with a pattern, then subtract |
Lesson 10.4 begins the reverse journey. Instead of multiplying binomials together, it starts from a product that already equals nought and asks what that forces the factors to be — which is the idea behind every factoring method in the rest of the chapter.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.3 Special Products of Polynomials §10.3, pp. 581-587 — everything on these slides traces back here
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