Multiplying polynomials. Includes using the distributive property twice on a product of binomials, the FOIL pattern and what its four letters name, multiplying longer polynomials in a vertical format, distributing horizontally so that every term meets every term, and writing an area as a product of two binomials.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 10 — Polynomials and Factoring
Multiplying 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 10.2 Multiplying Polynomials §10.2, pp. 575-580 — the lesson these objectives are drawn from
Warm-up
Lesson 2.6 multiplied a polynomial by a monomial. This lesson multiplies two polynomials, using the same property twice.
Discussion prompt
Expand three times the quantity two x plus three. Then say what would change if the three in front were replaced by x plus one.
Hint: Distribute to each term.
Answer:
\[ 3(2x + 3) = 6x + 9 \]
Each term inside the brackets was multiplied by the three. If the three became x plus one, then each term inside would have to be multiplied by the whole of x plus one — which means distributing a second time. That is the entire content of this lesson.
Concept
To multiply two polynomials, apply the distributive property so that every term of one is multiplied by every term of the other, then combine like terms.
FOIL pattern — A way of remembering the four products in a product of two binomials: the First terms, the Outer terms, the Inner terms and the Last terms.
For longer polynomials, count the products rather than using FOIL.
Figure (svg): The distributive property applied twice to a product of binomials
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 575-576
Section
Section 1
Concept
First distribute the whole second binomial to each term of the first. Then distribute each of those terms across the second binomial. Finally combine like terms.
\[ (x + 4)(x + 5) = x(x + 5) + 4(x + 5) \]
Two applications of one property from Lesson 2.6.
Figure (svg): The distributive property applied twice to a product of binomials
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 575-575 — the development of the double distribution and Example 1
Picture it
Whole bracket first, then term by term.
Figure (svg): The distributive property applied twice to a product of binomials
The first line does the work and the rest is arithmetic. Writing that line out is what keeps every pairing accounted for before any multiplying starts.
Worked example
This is Example 1 from the textbook.
\[ \text{Find } (x + 2)(x - 3). \]
Distribute the whole bracket
Why: To each term of the first.
\[ x(x - 3) + 2(x - 3) \]
Distribute again
Why: Four products in all.
\[ x(x) + x(-3) + 2(x) + 2(-3) \]
Multiply
Why: Watch the two negatives.
\[ x ^{2} - 3 x + 2 x - 6 \]
Combine like terms
Why: The two x terms.
\[ x ^{2} - x - 6 \]
Figure (svg): The distributive property applied twice to a product of binomials
\[ (x + 2)(x - 3) = x^2 - x - 6 \]
Verify: substitute a value into both forms
Why: At x equal to four the original is six times one, which is six, and the answer is sixteen minus four minus six, also six. One substitution checks every one of the four products at once.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 575-575
Faded example
First round, before any multiplying.
Fill in the blanks
(x + 4)(x + 5) = x(x + 5) + 4(x + 5)
Why: Each term of the first bracket receives the whole of the second bracket, which stays intact at this stage. Splitting it too early is what leads to missed products.
Worked example
Guided Practice 1 to 3, including one with a coefficient.
\[ \text{Find } (x + 1)(x + 2), \; (x - 2)(x + 4) \text{ and } (2x + 1)(x - 2). \]
Take the first
Why: Both signs positive.
\[ x ^{2} + 3 x + 2 \]
Take the second
Why: The middle terms are four x and negative two x.
\[ x ^{2} + 2 x - 8 \]
Take the third
Why: The leading coefficient is two.
\[ 2 x ^{2} - 4 x + x - 2 \]
Combine the third
Why: Negative four x plus x.
\[ 2 x ^{2} - 3 x - 2 \]
Figure (svg): A rectangle split into four parts to show a product of binomials
\[ x^2 + 3x + 2, \quad x^2 + 2x - 8, \quad 2x^2 - 3x - 2 \]
Verify: check the constant terms
Why: The constant of each product is the product of the two constants: one times two is two, negative two times four is negative eight, and one times negative two is negative two. That is a fast partial check, and it catches sign errors in the last product immediately.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 575-575
Trap
\[ (x + 2)(x - 3) = x^2 - 6 \]
Multiply the first terms together and the last terms together
Why: Two brackets, two pairs, so two products.
That leaves out the two cross products entirely. Substituting x equal to four gives six for the original and ten for this answer, so the two are not the same expression at all.
\[ (x + 2)(x - 3) = x^2 - 3x + 2x - 6 = x^2 - x - 6 \]
Multiply every term of one bracket by every term of the other
Why: Two terms times two terms gives four products.
The area model makes the omission visible: two of the four rectangles were being ignored.
Elimination
Expanding a product of two binomials.
Eliminate the wrong options
What is (x + 2)(x - 3)?
Survives elimination: A
Why: The four products are x squared, negative three x, two x and negative six, and the two middle ones combine to negative x. Substituting x equal to four gives six, which only the correct answer reproduces.
Matching
Four products each time.
Match the pairs
Why: The constant term of each answer is the product of the two constants, and the leading coefficient is the product of the two leading coefficients. Those two ends can be checked without doing the middle at all.
Socratic
Once was enough in Lesson 2.6.
Discussion prompt
Explain why multiplying two binomials needs the distributive property twice while multiplying by a monomial needs it once. Then say how many times it would be needed for two trinomials.
Hint: Count the terms doing the distributing.
Answer:
A monomial is a single term, so distributing it across the other bracket finishes the job in one pass. A binomial has two terms, and each of them must reach every term of the other bracket, so the first pass splits the product into two smaller products and a second pass expands each of those.
For two trinomials the same logic gives three smaller products after the first pass, each needing its own expansion — so three second-round distributions and nine products in total. The pattern is that the number of products is the number of terms in one factor times the number in the other, which is the counting check used later in this lesson.
Section
Section 2
Concept
In a product of two binomials the four products are those of the First terms, the Outer terms, the Inner terms and the Last terms. Multiply those four, then combine like terms.
It is a memory aid for two binomials only.
Figure (svg): The FOIL pattern with its four products labelled
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 576-576 — the FOIL pattern and Example 2
Picture it
Then two of them combine.
Figure (svg): The FOIL pattern with its four products labelled
The Outer and Inner products are the only ones that are ever like terms, which is why a product of two binomials so reliably comes out as a trinomial.
Worked example
The textbook's illustration of the pattern.
\[ \text{Find } (3x + 4)(x + 5) \text{ using FOIL.} \]
First
Why: Three x times x.
\[ 3 x ^{2} \]
Outer
Why: Three x times five.
\[ 15 x \]
Inner
Why: Four times x.
\[ 4 x \]
Last, then combine
Why: Four times five, and the middle terms.
\[ 3 x ^{2} + 19 x + 20 \]
Figure (svg): The FOIL pattern with its four products labelled
\[ (3x + 4)(x + 5) = 3x^2 + 19x + 20 \]
Verify: check the ends
Why: The leading coefficient is three times one, which is three, and the constant is four times five, which is twenty. Both ends match without doing the middle, and the middle can then be checked with a substitution of x equal to one: seven times six is forty-two, and three plus nineteen plus twenty is also forty-two.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 576-576
Faded example
First, Outer, Inner, Last.
Fill in the blanks
(3x + 4)(x + 5): \quad F = 3x^2, \; O = 15x, \; I = 4x, \; L = 20
Why: The Outer product pairs the first term of the first bracket with the last term of the second. The Inner pairs the other two, and together they give the middle term of the answer.
Worked example
This is Example 2 from the textbook.
\[ \text{Find } (2x + 3)(2x + 1) \text{ using FOIL.} \]
First
Why: Two x times two x.
\[ 4 x ^{2} \]
Outer
Why: Two x times one.
\[ 2 x \]
Inner
Why: Three times two x.
\[ 6 x \]
Last, then combine
Why: Three times one, and the middle terms.
\[ 4 x ^{2} + 8 x + 3 \]
Figure (svg): The FOIL pattern with its four products labelled
\[ (2x + 3)(2x + 1) = 4x^2 + 8x + 3 \]
Verify: substitute one for x
Why: The original gives five times three, which is fifteen, and the answer gives four plus eight plus three, also fifteen. Substituting one is the quickest possible check, since it simply adds up all the coefficients on each side.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 576-576
Error analysis
The student used FOIL on a product with a negative term.
Annotate
On: \( \begin{aligned} (2x - 3)(x + 4) &= 2x^2 + 8x + 3x + 12 \\ &= 2x^2 + 11x + 12 \end{aligned} \)
The reliable fix is to read each binomial as a sum with signed terms — two x plus negative three — so that the sign travels with the number into every product it appears in. Treating the minus as a separate operation between terms is what makes it easy to leave behind.
Sorting
In a product of two binomials.
Sort into buckets
Sort the four FOIL products by whether they are like terms with each other.
That is why a product of two binomials normally has three terms rather than four. When the Outer and Inner products cancel instead of adding, only two terms survive — which is the special case Lesson 10.3 is about.
Prediction
The Outer and Inner products can cancel.
Predict first
What has to be true for a product of two binomials to have no middle term?
Correct: The Outer and Inner products must be opposites.
\[ (x + 3)(x - 3) = x^2 - 3x + 3x - 9 = x^2 - 9 \]
Why: The middle term is the sum of the Outer and Inner products, so it disappears exactly when those two cancel. That happens with a pair like x plus three and x minus three, whose cross products are negative three x and three x. The result is a binomial rather than a trinomial, and Lesson 10.3 gives that pattern a name and uses it heavily.
Socratic
It is not in the list of properties.
Discussion prompt
Say whether FOIL is a mathematical rule in its own right. Then say what someone who only knows FOIL will get wrong.
Hint: Where do the four products come from?
Answer:
It is a reminder rather than a rule. The mathematics is the distributive property applied twice, and FOIL just names the four products that procedure happens to produce when both factors have exactly two terms. Nothing is being claimed beyond what distributing already gives.
Someone who only knows FOIL will try to apply it to a binomial times a trinomial and produce four products where six are needed, quietly losing two terms. That is why the next sections drop the mnemonic and count products instead, and why the counting check — terms times terms — is worth more than the letters.
Section
Section 3
Concept
For polynomials with three or more terms, write each in standard form and multiply in a vertical layout. Each term of the lower polynomial multiplies the whole upper polynomial, and like terms are lined up in columns.
It is the layout used for multi-digit numbers, with exponents in place of place values.
Figure (svg): A trinomial multiplied by a binomial in vertical format
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 576-576 — Example 3, Multiply Polynomials Vertically
Picture it
One row per lower term.
Figure (svg): A trinomial multiplied by a binomial in vertical format
The indentation is doing the same job as in numerical long multiplication: it keeps terms of equal degree in the same column so that the final addition combines the right things.
Worked example
This is Example 3 from the textbook.
\[ \text{Find } (x + 2)(5 + 3x + x^2). \]
Write both in standard form
Why: The trinomial needs reordering.
\[ x^2 + 3x + 5 \text{ and } x + 2 \]
Multiply by the two
Why: Each term of the trinomial.
\[ 2 x ^{2} + 6 x + 10 \]
Multiply by the x
Why: Shifted one column left.
\[ x ^{3} + 3 x ^{2} + 5 x \]
Add the columns
Why: Combine like terms.
\[ x ^{3} + 5 x ^{2} + 11 x + 10 \]
Figure (svg): A trinomial multiplied by a binomial in vertical format
\[ x^3 + 5x^2 + 11x + 10 \]
Verify: substitute one for x
Why: The original gives three times nine, which is twenty-seven, and the answer gives one plus five plus eleven plus ten, also twenty-seven. Six products were formed and all six are accounted for in that total.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 576-576
Faded example
Two rows for a binomial multiplier.
Fill in the blanks
(x + 2)(x^2 + 3x + 5): \quad 2 \cdot \text10 = 2x^2 + 6x + 5x, \quad x \cdot \text___ = x^3 + 3x^2 + ___
Why: Each row is the whole upper polynomial multiplied by one term of the lower one. The second row is shifted because every one of its terms has one more factor of x.
Worked example
Guided Practice 7, with the counting check applied.
\[ \text{Find } (x + 1)(x^2 + 3x + 2) \text{ and check the product count.} \]
Count the expected products
Why: Two terms times three terms.
\[ 6 \]
Multiply by the one
Why: The whole trinomial.
\[ x ^{2} + 3 x + 2 \]
Multiply by the x
Why: Shifted one column.
\[ x ^{3} + 3 x ^{2} + 2 x \]
Add the columns
Why: Combine like terms.
\[ x ^{3} + 4 x ^{2} + 5 x + 2 \]
Figure (svg): Every term of one polynomial multiplied by every term of the other
\[ x^3 + 4x^2 + 5x + 2 \]
Verify: substitute one for x
Why: The original gives two times six, which is twelve, and the answer gives one plus four plus five plus two, also twelve. The six products were formed as three in each row, and the two rows are what make the count come out right.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 576-576
Trap
\[ (x + 2)(5 + 3x + x^2) \text{ multiplied with the trinomial left unsorted} \]
Set up the vertical layout with the terms in the order given
Why: The terms are all there, so the order looked unimportant.
The columns then mix a constant with a squared term, so the final addition combines things that are not like terms. The answer comes out with terms in the wrong places even though every individual product was correct.
\[ x^2 + 3x + 5 \quad \text{on top} \]
Write both polynomials in standard form before drawing the layout
Why: Sorting is what makes the columns meaningful.
This is the same reason standard form mattered for adding in Lesson 10.1.
Elimination
Multiplying a binomial by a trinomial.
Eliminate the wrong options
Before any combining, how many products are formed?
Survives elimination: A
Why: Two terms times three terms gives six products, and counting them is a genuine check on the working. Getting five in your expansion means one pairing was skipped, which the count catches before any combining hides it.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Vertical | Horizontal | |
|---|---|---|
| Needs standard form? | yes, for the columns | not strictly, since terms are grouped by hand |
| Like terms are found by | reading down a column | grouping in brackets |
| Best for | longer polynomials | short products such as two binomials |
Both produce the same products in a different arrangement. The vertical layout does the sorting for you and the horizontal one asks you to do it, which is why the first scales better.
Socratic
The layouts look identical.
Discussion prompt
Explain the correspondence between multiplying polynomials vertically and multiplying multi-digit numbers. Then say what plays the part of carrying.
Hint: What do the columns represent in each case?
Answer:
In a number, the columns are powers of ten and each digit is a coefficient; in a polynomial, the columns are powers of x and each coefficient is written explicitly. Multiplying by each digit and shifting one place is exactly multiplying by each term and shifting one degree, and the final addition combines the columns in both cases.
Carrying is the one thing polynomials do not need, because a coefficient can be any size while a digit cannot exceed nine. Evaluating a polynomial at x equal to ten turns it into a number and the carrying reappears — which is a neat way of seeing that place value is a polynomial in disguise.
Section
Section 4
Concept
Multiplying horizontally means handing the whole second polynomial to each term of the first, expanding each result, and then grouping like terms. Nothing may be left unpaired.
The product of the term counts is how many products to expect.
Figure (svg): Every term of one polynomial multiplied by every term of the other
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 577-577 — Example 4, Multiply Polynomials Horizontally
Picture it
Three terms meeting two.
Figure (svg): Every term of one polynomial multiplied by every term of the other
Drawing the pairings once, for a product you find hard, is a good way to be sure none was skipped. After that the count alone is usually enough.
Worked example
This is Example 4 from the textbook.
\[ \text{Find } (4x^2 + 3x - 1)(2x + 5). \]
Distribute to each term
Why: Three smaller products.
\[ 4 x ^{2}(2 x + 5) + 3 x(2 x + 5) - 1(2 x + 5) \]
Expand each
Why: Six products in all.
\[ 8 x ^{3} + 20 x ^{2} + 6 x ^{2} + 15 x - 2 x - 5 \]
Group like terms
Why: The squared terms and the x terms.
\[ (20 x ^{2} + 6 x ^{2}), \; (15 x - 2 x) \]
Combine
Why: Twenty-six and thirteen.
\[ 8 x ^{3} + 26 x ^{2} + 13 x - 5 \]
Figure (svg): Every term of one polynomial multiplied by every term of the other
\[ 8x^3 + 26x^2 + 13x - 5 \]
Verify: substitute one for x
Why: The original gives six times seven, which is forty-two, and the answer gives eight plus twenty-six plus thirteen minus five, also forty-two. The negative one in the first factor produced both of the negative products, which is where the signs needed care.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 577-577
Faded example
Two groups to combine.
Fill in the blanks
8x^3 + 20x^2 + 6x^2 + 15x - 2x - 5 = 8x^3 + 26x^2 + 13x - 5
Why: The cubic and constant terms had no partners and come through untouched. Only the two middle degrees had pairs, which is typical when a trinomial meets a binomial.
Worked example
Two cheap checks on the same product.
\[ \text{Check } (4x^2 + 3x - 1)(2x + 5) = 8x^3 + 26x^2 + 13x - 5 \text{ without redoing it.} \]
Count the products
Why: Three terms times two.
\[ 6 \text{ expected} \]
Check the leading term
Why: Four x squared times two x.
\[ 8x^3 \;\checkmark \]
Check the constant
Why: Negative one times five.
\[ -5 \;\checkmark \]
Substitute one for x
Why: Both sides.
\[ 42 = 42 \;\checkmark \]
Figure (svg): Every term of one polynomial multiplied by every term of the other
\[ 6 \text{ products}; \; 8x^3; \; -5; \; 42 = 42 \]
Verify: say what each check would have caught
Why: The leading term catches an error in the highest-degree product, the constant catches a sign slip on the last one, and the substitution catches anything in the middle. Together they cost about twenty seconds and cover the whole expansion.
Trap
\[ (4x^2 + 3x - 1)(2x + 5) \;\Longrightarrow\; \text{four products by FOIL} \]
Apply First, Outer, Inner, Last to the two brackets
Why: There are two brackets, so FOIL seemed to apply.
FOIL names the four products of two binomials. Here one factor has three terms, so six products are needed and two of them have nowhere to go in the pattern — the expansion silently loses terms.
\[ 4x^2(2x+5) + 3x(2x+5) - 1(2x+5) \]
Distribute term by term and count the products
Why: Three times two is six.
Counting works for any two polynomials, which is why it outlasts the mnemonic.
Translation
Multiply the term counts.
Match the pairs
Why: The count is always the product of the two term counts, before any combining. Knowing the expected number turns a long expansion into something you can audit rather than merely hope about.
Sorting
FOIL is only for two binomials.
Sort into buckets
Sort each product by whether FOIL applies to it directly.
Half of these fail the FOIL condition, and each of those needs six products rather than four. The count is what tells you which situation you are in.
Socratic
It says nothing about the values.
Discussion prompt
Explain what a product count can and cannot detect. Then name a second cheap check that covers what it misses.
Hint: Think about what kind of error changes the count.
Answer:
The count detects omissions and duplications — a missed pairing or a term multiplied twice — because either changes how many products appear before combining. It cannot detect a wrong sign or a wrong coefficient, since those leave the count intact.
Substituting a small value such as x equal to one covers exactly that gap: it adds up all the coefficients on both sides, so any sign error or arithmetic slip changes the total. Used together, the count catches structural mistakes and the substitution catches numerical ones, and neither takes long enough to skip.
Section
Section 5
Concept
When a length and a width are each given by a binomial, the area is their product. Expanding it gives a polynomial that can be evaluated for any value of the variable.
First degree polynomials often represent lengths and widths.
Figure (svg): A window whose glass is framed on all four sides
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 577-577 — Example 5, Multiply Binomials to Find an Area
Picture it
Two binomial dimensions.
Figure (svg): A window whose glass is framed on all four sides
The ratio fixes the glass at three x by two x, and the frame's fixed additions turn each dimension into a binomial. Both features are needed for the product to be interesting.
Worked example
This is Example 5 from the textbook.
\[ \text{Glass is } 3x \text{ by } 2x; \text{ the frame adds } 10 \text{ to the height and } 6 \text{ to the width. Find the total area.} \]
Write the dimensions
Why: Add the frame to each.
\[ 3x + 10 \text{ and } 2x + 6 \]
Write the area model
Why: Height times width.
\[ A = (3 x + 10) (2 x + 6) \]
Use FOIL
Why: Four products.
\[ 6 x ^{2} + 18 x + 20 x + 60 \]
Combine
Why: The two middle terms.
\[ 6 x ^{2} + 38 x + 60 \]
Figure (svg): A window whose glass is framed on all four sides
\[ A = 6x^2 + 38x + 60 \]
Verify: test with a number
Why: Take x equal to five: the window is twenty-five by sixteen, an area of four hundred. The polynomial gives six times twenty-five plus thirty-eight times five plus sixty, which is a hundred and fifty plus a hundred and ninety plus sixty — also four hundred.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 577-577
Faded example
Glass plus frame, each way.
Fill in the blanks
\text10 = 3x + 6, \quad \text___ = 2x + ___, \quad A = 6x^2 + 38x + 60
Why: The two additions are different because the frame's width is fixed while the glass is taller than it is wide. Each dimension gets its own constant, and the product of those constants is the sixty.
Worked example
Each term of the answer means something.
\[ \text{In } A = 6x^2 + 38x + 60, \text{ what does each term represent?} \]
Take the squared term
Why: Three x times two x.
Take the constant
Why: Ten times six.
Take the middle term
Why: The two cross products.
Check the total
Why: All four pieces of the area model.
\[ 6 x ^{2} + 38 x + 60 \]
Figure (svg): A rectangle split into four parts to show a product of binomials
\[ 6x^2 \text{ glass}, \; 38x \text{ sides}, \; 60 \text{ corners} \]
Verify: check the constant against the diagram
Why: The frame adds ten to the height and six to the width, and ten times six is sixty — the four corner pieces of the frame, which together form a rectangle of that area. Every term of an expanded area model corresponds to a piece of the picture.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 577-577
Trap
\[ A = (3x + 10)(2x) \]
Add the frame to the height only
Why: The ten was mentioned first, so it was the one applied.
The frame surrounds the glass, so it adds to both dimensions. This expression gives six x squared plus twenty x, which at x equal to five is two hundred and fifty rather than the correct four hundred.
\[ A = (3x + 10)(2x + 6) \]
Add the stated amount to each dimension separately
Why: Height gets ten, width gets six.
Testing with a number would have exposed the omission immediately.
Hypothesis
In an expanded area model.
Predict first
In A equal to 6x squared plus 38x plus 60, what does the 60 represent?
Correct: The area of the frame's four corner pieces together.
\[ 10 \times 6 = 60 \]
Why: The constant is the product of the two frame additions, ten and six, which is the area contributed by the parts of the frame that lie beyond the glass in both directions at once. The squared term is the glass and the middle term is the frame's sides. Every term of an expanded area model has a piece of the picture attached to it, which is what makes the area model figure worth drawing.
Elimination
Glass 3x by 2x, frame adding 10 and 6.
Eliminate the wrong options
Which expression is the total area?
Survives elimination: A
Why: Each dimension is the glass plus that dimension's share of the frame. Option B is the most tempting, because adding areas feels natural — but the frame's area grows with the glass, so it cannot be a constant.
Socratic
Glass area plus frame area sounds right.
Discussion prompt
Explain why the total area cannot be found by adding a fixed frame area to the glass area. Then say how the correct answer does account for the frame.
Hint: Does a bigger window need a bigger frame?
Answer:
The frame runs along the sides of the glass, so a taller window needs a longer frame down each side and a wider one needs a longer frame along the top and bottom. Its area therefore depends on the glass's dimensions and cannot be a constant added on.
In the expanded product the frame appears as thirty-eight x plus sixty: the linear part is the sides, which grow with the glass, and the constant is the corners, which do not. Splitting the answer that way shows exactly which part of the frame scales and which part is fixed, and that is information the sum-of-two-areas approach would have thrown away.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Method | Works for | How like terms are found |
|---|---|---|
| FOIL | two binomials only | the Outer and Inner products |
| Vertical | any two polynomials | by reading down a column |
| Horizontal | any two polynomials | by grouping them in brackets |
All three produce the same products, and only the first has a restriction. That restriction is the reason to learn the other two rather than treating FOIL as the method.
Pattern
To multiply any two polynomials, these five moves cover it.
Step two takes two seconds and turns the expansion into something you can audit, which matters most on exactly the long products where a term is easiest to lose.
OpenStax Elementary Algebra 2e, §6.3 Multiply Polynomials §6.3
Check
Four products, then combine.
Check your understanding
What is (x + 2)(x - 3)?
Answer: A
Why: The four products are x squared, negative three x, two x and negative six, and the two middle terms combine to negative x.
Check
Count before you expand.
Check your understanding
How many products are formed when a trinomial is multiplied by a binomial?
Answer: A
Why: Three terms times two terms gives six products before any like terms are combined.
Check
Both dimensions grow.
Check your understanding
Glass is 3x by 2x and a frame adds 10 to the height and 6 to the width. What is the total area?
Answer: A
Why: The dimensions are three x plus ten and two x plus six, and the four products are six x squared, eighteen x, twenty x and sixty.
Real world
This is the window question from the lesson opener. The glass has a height-to-width ratio of three to two, and the frame adds 6 inches to the width and 10 inches to the height.
Discussion prompt
Write a polynomial for the total area of the window including its frame, test it with a value of x, and say what each of the three terms represents in the picture.
Hint: The ratio makes the glass three x by two x.
Answer:
\[ A = (3x + 10)(2x + 6) = 6x^2 + 18x + 20x + 60 = 6x^2 + 38x + 60 \]
Testing with x equal to five gives a window twenty-five inches by sixteen, an area of four hundred square inches, and the polynomial gives a hundred and fifty plus a hundred and ninety plus sixty — also four hundred.
The squared term is the glass, six x squared; the linear term is the frame running along the four sides, whose area grows as the glass does; and the constant sixty is the four corner pieces, ten by six in total, which stay the same size however large the window is. Reading a polynomial back into a picture like that is what makes an area model worth setting up rather than just multiplying.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
What is (2x - 3)(x + 4)?
Correct: 2x squared + 5x - 12.
\[ (2x - 3)(x + 4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12 \]
Why: The minus sign belongs to the three, so both products involving it are negative: the four products are two x squared, eight x, negative three x and negative twelve. The two middle terms combine to positive five x, since eight is larger than three. The first option keeps the three positive throughout, which is the commonest slip and gives a constant of positive twelve — a sign that is easy to check on its own. Substituting x equal to one settles the whole question in one line: the original is negative one times five, which is negative five, and only one of the four options gives that.
Explain it
They used FOIL on a binomial times a trinomial and got four terms.
Discussion prompt
In no more than four sentences, explain why FOIL did not fit and what to do instead. Then give them the check that would have shown something was missing.
Hint: How many products should there have been?
Answer:
A usable answer: FOIL names the four products you get when both factors have two terms. With a trinomial there are three terms in one factor, so six products are needed and the four letters cannot reach them all.
Instead, hand the second polynomial to each term of the first and expand each piece. The check is to count: multiply the number of terms in one factor by the number in the other, and make sure that many products appear before you combine anything.
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: Signs are fixed by reading each binomial as a sum of signed terms, so the minus travels into every product it belongs to. Knowing when FOIL applies is fixed by counting terms: two and two, or it does not. Vertical setup is fixed by writing both polynomials in standard form before drawing anything. Area models are fixed by adding to each dimension separately and testing with a number. 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 expand one product of binomials the long way, writing the first distribution as its own line before splitting it further, so all four products appear separately. Beside it draw the same product as a rectangle cut into four pieces, label each piece with its area, and connect each piece to one of the four products with a line. Underneath, write the FOIL pattern with its four arcs over a product that has a negative term, and mark which two products carry that minus sign. In the middle, multiply a binomial by a trinomial twice, once vertically with the columns ruled in and once horizontally with the like terms bracketed, writing the expected product count above each. Beneath that, draw a framed window, label the glass and both total dimensions, and expand the area, then write beside each term of the answer which part of the picture it is. Finally, in the margin, write the two cheap checks: the product count and the substitution of x equal to one.
Your two methods for the trinomial product should agree term for term. If they do not, compare how many products each one actually formed before combining, since a disagreement almost always means one method lost a pairing.
Recap
Five things, and the second is a nickname for the first.
| If the question says | Your first move is |
|---|---|
| Multiply two binomials | Four products, then combine the middle two |
| Multiply by a trinomial | Count six products; do not use FOIL |
| Use a vertical format | Write both in standard form first |
| Find the total area | Add to each dimension, then multiply |
| Check your expansion | Count products, check both ends, substitute one |
Lesson 10.3 looks at three products that come up so often they are worth recognising on sight. Two of them are squares of binomials and the third is the case where the middle term disappears entirely.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.2 Multiplying Polynomials §10.2, pp. 575-580 — everything on these slides traces back here
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