Chapter 10 of Algebra 1: Concepts and Skills, built for a visual learner. Polynomials added in power columns, multiplication drawn as area models so no cross-product is lost, the two special products shown as cancelling or doubling middle cells, the zero-product property, factoring as a finite factor-pair search, and factoring completely including grouping.
Subject: Algebra 1 · 60 slides · symbolic lesson
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Title
Algebra 1 · Chapter 10
Multiplying out with area models, and running the whole process backwards to factor
Objectives
This chapter is one idea in two directions: multiplying out and factoring back in. Every picture works both ways.
Figure (svg): An area model of x plus 3 times x plus 2, split into four rectangles labelled with each product
Section
Section 10.1
Concept
A polynomial is a sum of terms, each a number times a power of x. Adding two of them means combining like terms, exactly as in Chapter 2.
Figure (svg): Two polynomials stacked with matching powers lined up in columns, ready to add
degree — The highest power appearing in the polynomial. A degree of 2 makes it quadratic, and it is the degree that decides the shape of the graph.
Worked example
Simplify the expression below.
\[ (5x^2 - 3x + 4) - (2x^2 + x - 7) \]
Distribute the subtraction across every term of the second bracket
Why: The minus sign applies to all three terms inside, not just the first. This is where nearly every error in the section happens.
\[ 5x^2 - 3x + 4 - 2x^2 - x + 7 \]
Combine like terms by power
Why: Three squared terms, three x terms, three constants — each column handled separately.
\[ 3x^2 - 4x + 11 \]
Figure (svg): The subtraction shown as sign flips on each of the three terms of the second polynomial
Verify: test with x equal to 1
Why: The original gives 6 minus negative 4, which is 10. The answer gives 3 minus 4 plus 11, which is also 10.
Error analysis
Find the error in this subtraction.
Annotate
On: \( (4x^2 + x) - (x^2 - 3x) \;\overset{?}{=}\; 4x^2 + x - x^2 - 3x = 3x^2 - 2x \)
Rewrite the subtraction as adding the opposite of every term before combining anything.
Sorting
Sort each pair by whether they can be combined.
Sort into buckets
Which pairs are like terms?
Fill the middle
Fill each blank.
Fill in the blanks
(3x^2 - x + 2) + (x^2 + 4x - 9) = 4x^2 + 3x - 7
Why: Each power is handled in its own column: the squared terms give 4, the x terms give negative 1 plus 4, which is 3, and the constants give 2 minus 9, which is negative 7. Addition needs no sign flips at all, which is what makes it easier than subtraction.
Section
Section 10.2
Concept
Multiplying two brackets means every term in the first multiplies every term in the second. The area model shows exactly why.
Figure (svg): An area model of x plus 3 times x plus 2, split into four rectangles labelled with each product
\[ (x + 3)(x + 2) = x^2 + 5x + 6 \]
Worked example
Expand the product below.
\[ (2x - 3)(x + 4) \]
Draw the four rectangles and fill each one
Why: Two terms by two terms means four products, and drawing the grid guarantees none is missed.
| x | +4 | |
|---|---|---|
| 2x | 2x squared | 8x |
| -3 | -3x | -12 |
Combine the two middle terms
Why: Eight x and negative three x are like terms, so they merge into five x.
\[ 2x^2 + 5x - 12 \]
Figure (svg): An area model of two x minus three times x plus four, with the four products labelled
Verify: test with x equal to 2
Why: The original gives 1 times 6, which is 6. The answer gives 8 plus 10 minus 12, which is also 6.
Prediction
Do not multiply. Just count.
\[ (x + 2)(x^2 + 3x + 1) \]
Predict first
How many individual products will the expansion contain before combining?
Correct: Six — two terms times three terms.
Why: Each of the two terms in the first bracket must multiply each of the three in the second, giving two times three, which is six products. Counting first is a genuinely useful check: if your expansion has fewer than six lines before combining, one has been missed.
Explain it to yourself
Explain the rule rather than quoting it.
Discussion prompt
Why must each term of the first bracket multiply each term of the second?
Hint: What happens if you treat the second bracket as a single object first?
Answer:
Because the distributive property applies twice. The whole second bracket multiplies the first term, then the whole second bracket multiplies the second term.
The area model is that idea drawn: a rectangle whose width is split into two parts and whose height is split into two parts genuinely has four pieces, and the total area is all of them. Nothing is optional.
Error analysis
Find what this expansion left out.
Annotate
On: \( (x + 5)(x + 2) \;\overset{?}{=}\; x^2 + 10 \)
The cross-products are where the middle term comes from. A product of two binomials that has no middle term should always be checked.
Invariant
Watch an expansion happen and name the thing that never changes.
Step through it
One quantity is identical in every frame. Which, and why is that the whole point of a check?
The value is the invariant. That is why substituting one number into both forms catches almost every expansion error in a single line.
Picture it
If the grid feels abstract, here is the same product built from physical pieces.
Figure (svg): Algebra tiles arranged into a rectangle: one large square tile, five strips and six unit squares
Factoring, physically, is asking: can these tiles be rearranged into a rectangle, and what are its sides?
Section
Section 10.3
Concept
Two products come up so often that recognising them saves real time. Both are just area models where something convenient happens.
Figure (svg): An area model showing the difference of squares, with the two middle terms cancelling
\[ (a + b)(a - b) = a^2 - b^2 \qquad (a + b)^2 = a^2 + 2ab + b^2 \]
Worked example
Expand each product, using the pattern rather than the full grid.
\[ (x + 7)(x - 7) \qquad (x + 5)^2 \]
For the first, recognise the difference of squares
Why: The two brackets are identical except for the sign, so the cross-products are exact opposites and cancel.
\[ x^2 - 49 \]
For the second, recognise the perfect square
Why: Squaring a binomial gives the two squares plus twice the cross-product — the two middle terms are identical, so they double rather than cancel.
\[ x^2 + 10x + 25 \]
Figure (svg): An area model of x plus five squared, with the two identical middle rectangles adding to ten x
Verify: test the second with x equal to 1
Why: The original gives 6 squared, which is 36. The answer gives 1 plus 10 plus 25, which is also 36.
Trap
Expand the square below.
\[ (x + 4)^2 \]
Square each term inside the bracket
Why: The exponent sits outside, so it feels like it should apply to each part in turn — the same instinct that works for a product.
\[ (x + 4)^2 \;\to\; x^2 + 16 \]
Test with x equal to 1: the original is 5 squared, which is 25, but this gives 17. The middle term is missing.
Expand it as the product it actually is.
Write it as two brackets and use the area model
Why: Squaring means multiplying by itself, so there are four products, two of which are identical.
\[ (x + 4)(x + 4) = x^2 + 8x + 16 \]
Test with x equal to 1: 1 plus 8 plus 16 is 25, matching. An exponent distributes across multiplication, never across addition.
Discrimination
Sort each product by the pattern it fits. Do not expand.
Sort into buckets
Which special product, if any, does each fit?
Reverse engineer
Special products are most useful in reverse.
Fill in the blanks
x^2 - 81 = (x + 9)(x - 9) \;\text6\; x^2 + 12x + 36 = (x + ___)^2
Why: For the difference of squares, take the square root of each term: 81 has root 9. For the perfect square, the constant's root is 6 and the middle coefficient is twice that, which confirms the pattern fits. If the middle term had not been exactly 12, the expression would not be a perfect square at all.
Section
Section 10.4
Concept
If two things multiply to give zero, at least one of them must be zero. No other number has this property.
Figure (svg): Two factors multiplied to give zero, with arrows showing that at least one of them must itself be zero
\[ (x - 3)(x + 5) = 0 \;\Longrightarrow\; x = 3 \text{ or } x = -5 \]
Worked example
Solve the equation below.
\[ (2x + 1)(x - 4) = 0 \]
Set each factor equal to zero separately
Why: The product is zero, so one of the two factors must be. Each gives its own small equation.
\[ 2x + 1 = 0 \quad \text{or} \quad x - 4 = 0 \]
Solve each one
Why: These are ordinary one-step and two-step equations from Chapter 3.
\[ x = -\tfrac{1}{2} \quad \text{or} \quad x = 4 \]
Figure (svg): A parabola crossing the axis at minus one half and four, matching the two factors
Verify: substitute both roots into the original
Why: At x equal to 4 the second factor is zero, so the product is zero. At x equal to negative one half the first factor is zero, so the product is zero again. Both work.
Counterexample
A classmate says: if a product equals 12, then one factor must be 12 or the other must be.
Discussion prompt
Find a counterexample, and say why zero is genuinely special.
Hint: How many pairs of numbers multiply to give 12?
Answer:
Three times four is 12, and neither factor is 12. Also 2 times 6, and 24 times one half — endlessly many pairs work.
Zero is the only number with the property that a product can only reach it if a factor is already there. Every other target can be hit by countless pairs, which is exactly why solving requires setting the equation to zero first.
\[ (x - 3)(x + 5) = 12 \;\text{ does NOT give } x - 3 = 12 \]
Error analysis
Find the flaw in this reasoning.
Annotate
On: \( (x - 2)(x + 1) = 4 \;\overset{?}{\Longrightarrow}\; x - 2 = 4 \text{ or } x + 1 = 4 \)
Set the equation to zero before factoring. This is the same discipline as the quadratic formula requiring standard form.
Prediction
Read the factors and commit.
\[ (x - 2)^2 = 0 \]
Predict first
How many distinct solutions does this equation have?
Correct: One — the repeated factor gives a single root at x equals 2.
Why: Both factors are identical, so setting each to zero gives the same answer twice. The graph touches the axis at x equal to 2 without crossing, which is exactly the zero-discriminant case from Chapter 9 seen from the factoring side.
Section
Section 10.5
Concept
To factor a quadratic with leading coefficient 1, find two numbers that multiply to c and add to b.
Figure (svg): A search table of factor pairs of twelve with their sums, highlighting the pair that adds to seven
\[ x^2 + 7x + 12 = (x + 3)(x + 4) \]
Worked example
Factor the expression below.
\[ x^2 - 5x + 6 \]
List the factor pairs of the constant
Why: Six is 1 times 6, or 2 times 3. Since the constant is positive and the middle term negative, both numbers must be negative.
Check which pair adds to the middle coefficient
Why: Negative 2 and negative 3 multiply to 6 and add to negative 5, which is what is needed.
\[ (x - 2)(x - 3) \]
Figure (svg): An area model of x minus two times x minus three reassembling into the original quadratic
Verify: expand the answer back out
Why: Multiplying gives x squared minus 3x minus 2x plus 6, which is x squared minus 5x plus 6 — the expression we started with.
Sorting
Before searching, the signs of b and c already tell you a lot. Sort each quadratic.
Sort into buckets
What signs will the two numbers in the factors have?
Faded example
Fill each blank.
Fill in the blanks
x^2 + 2x - 15 = (x + 5)(x - 3)
Why: The constant is negative, so the two numbers have opposite signs, and their product is 15. The pairs are 1 and 15, or 3 and 5. Since they must combine to positive 2, the pair is positive 5 and negative 3. Getting them the wrong way round would give a middle term of negative 2 instead.
Pattern
A short, finite search, always in the same order.
The first line saves the most time and is the one most often skipped.
Socratic
One question, no computation.
\[ (x + p)(x + q) = x^2 + (p + q)x + pq \]
Discussion prompt
Looking at the expanded form above, why must the two numbers multiply to c and add to b?
Hint: Expand the general product and compare it term by term with the target.
Answer:
Expanding the general product shows the constant term is p times q and the middle coefficient is p plus q. So matching against a given quadratic forces exactly those two conditions.
The method is not a trick — it is reading the expansion backwards. That is why it fails when the leading coefficient is not 1: the expansion then has extra factors in both places.
Reverse engineer
Read the pair method in reverse.
Fill in the blanks
(x + 6)(x - 2) = x^2 + 4x - 12
Why: The middle coefficient is the sum of the two numbers, 6 plus negative 2, which is 4. The constant is their product, 6 times negative 2, which is negative 12. Checking a factoring this way takes about five seconds and catches sign errors immediately.
Explain it
A classmate factors by trying every pair, including ones with impossible signs.
Discussion prompt
Explain how the signs of b and c narrow the search before any pairs are tried.
Hint: What must be true of two numbers whose product is positive?
Answer:
If the constant is positive, the two numbers share a sign — and the middle term tells you which one, since two positives add to a positive and two negatives to a negative.
If the constant is negative, the two numbers have opposite signs, and the middle term tells you which of them is larger in size. Either way you have halved or better the number of pairs worth testing before starting.
Section
Section 10.6
Concept
When the squared term has a coefficient, the two numbers must account for it too. The grid method handles this without guesswork.
\[ 2x^2 + 7x + 3 = (2x + 1)(x + 3) \]
Figure (svg): A grid for two x squared plus seven x plus three, with the four cells filled and the factors read off the edges
Worked example
Factor the expression below.
\[ 3x^2 + 10x + 8 \]
Multiply the leading coefficient by the constant
Why: Three times 8 is 24. Now look for two numbers multiplying to 24 and adding to 10.
Find the pair
Why: Four and 6 multiply to 24 and add to 10.
Split the middle term using that pair
Why: Rewrite 10x as 4x plus 6x, which changes nothing but makes grouping possible.
\[ 3x^2 + 4x + 6x + 8 \]
Group in pairs and factor each group
Why: The first pair shares an x and the second shares a 2, and both leave the same bracket behind.
\[ x(3x + 4) + 2(3x + 4) = (3x + 4)(x + 2) \]
Figure (svg): A grid for three x squared plus ten x plus eight, with the split middle term filling the two off-diagonal cells
Verify: expand the answer back out
Why: Multiplying gives 3x squared plus 6x plus 4x plus 8, which is 3x squared plus 10x plus 8 — the original expression.
Ranking
For the quadratic below, rank the moves.
\[ 2x^2 + 11x + 12 \]
Put in order
Why: Multiplying a by c gives the target product of 24, and the pair adding to 11 is 3 and 8. Splitting the middle term into 3x plus 8x lets the four terms group into 2x squared plus 3x and 8x plus 12, which factor to x times 2x plus 3 and 4 times 2x plus 3, giving 2x plus 3 times x plus 4.
Prediction
The first step decides everything. Commit before searching.
\[ 6x^2 - 7x - 5 \]
Predict first
What two numbers are you looking for?
Correct: A pair multiplying to negative 30 and adding to negative 7.
Why: Multiply the leading coefficient by the constant: 6 times negative 5 is negative 30. The pair must still add to the middle coefficient, negative 7. Those numbers are 3 and negative 10. Using just the constant is the habit carried over from the simpler case, and it does not work once a is not 1.
Section
Section 10.7
Concept
Before any other method, check whether every term shares a common factor. Removing it makes everything afterwards smaller.
Figure (svg): A polynomial with a common factor being pulled out to the front of a bracket
\[ 6x^3 + 9x^2 = 3x^2(2x + 3) \]
Worked example
Factor the expression below completely.
\[ 2x^3 - 18x \]
Take out the greatest common factor first
Why: Both terms share a factor of 2x, so pull it out in front.
\[ 2x(x^2 - 9) \]
Recognise what is left as a difference of squares
Why: Nine is a perfect square, so the bracket factors further.
\[ 2x(x + 3)(x - 3) \]
Check that nothing factors further
Why: Each remaining factor is linear or a single term, so the factoring is complete.
Figure (svg): The expression factored in two stages, with the common factor removed first and the difference of squares second
Verify: expand back out
Why: Multiplying x plus 3 by x minus 3 gives x squared minus 9, and 2x times that gives 2x cubed minus 18x — the original expression.
Error analysis
This factoring is correct but incomplete. Find what remains.
Annotate
On: \( 3x^2 - 27 \;\overset{?}{=}\; 3(x^2 - 9) \)
After every factoring step, look at what is left and ask whether it factors again.
Matching
Look for a common factor first, then a pattern.
Match the pairs
Why: The first is a plain difference of squares. The second and fourth are perfect squares differing only in the sign of the middle term, which becomes the sign inside the bracket. The third needs its common factor of 5 removed before the difference of squares becomes visible.
Discrimination
Do not factor. Just name the first move.
Sort into buckets
What is the first thing to do with each expression?
Estimation
Finding the greatest common factor quickly is worth practising on its own.
\[ 12x^4 - 18x^3 + 30x^2 \]
Predict first
What is the greatest common factor of these three terms?
Correct: 6x squared.
\[ 12x^4 - 18x^3 + 30x^2 = 6x^2(2x^2 - 3x + 5) \]
Why: The largest number dividing 12, 18 and 30 is 6, and the lowest power of x present in all three terms is x squared. Taking 12 would fail on the 18, and taking x cubed would fail on the last term — the common factor is limited by the weakest term in each respect.
Section
Section 10.8
Concept
A four-term polynomial often factors by grouping: pair the terms, factor each pair, and hope the same bracket appears twice.
\[ x^3 + 3x^2 + 2x + 6 = x^2(x + 3) + 2(x + 3) = (x + 3)(x^2 + 2) \]
Figure (svg): Four terms grouped into two pairs, each factored to reveal the same bracket
Worked example
Factor the expression below completely.
\[ x^3 - 4x^2 - 9x + 36 \]
Split into two pairs
Why: Take the first two terms together and the last two together, keeping every sign attached.
\[ (x^3 - 4x^2) + (-9x + 36) \]
Factor each pair
Why: The first shares x squared. The second shares negative 9 — taking out the negative is what makes the brackets match.
\[ x^2(x - 4) - 9(x - 4) \]
Take the shared bracket out front
Why: Both terms now contain x minus 4, so it factors out.
\[ (x - 4)(x^2 - 9) \]
Factor what is left
Why: The remaining bracket is a difference of squares.
\[ (x - 4)(x + 3)(x - 3) \]
Figure (svg): The cubic curve crossing the horizontal axis at minus three, three and four, one crossing per factor
Verify: test with x equal to 4
Why: The original gives 64 minus 64 minus 36 plus 36, which is 0. The factored form has a factor of x minus 4, which is also 0 there. They agree.
Fill the middle
Fill each blank.
Fill in the blanks
x^3 + 2x^2 + 5x + 10 = x^2(x + 2) + 5(x + 2) = (x + 2)(x^2 + 5)
Why: The first pair gives x squared times x plus 2, and the second gives 5 times x plus 2. Because both brackets match, x plus 2 comes out front and the leftovers x squared and 5 form the second factor. If the two brackets had not matched, the grouping would have failed and you would try pairing the terms differently.
Edge cases
Push the method until it breaks.
\[ x^3 + 2x^2 + 3x + 8 \]
Discussion prompt
Try grouping this. What goes wrong, and what would you do next?
Hint: Do the two pairs leave the same bracket behind?
Answer:
The first pair gives x squared times x plus 2, and the second gives no useful shared factor beyond 1 — the brackets do not match, so nothing comes out front.
The next move is to try a different pairing, or to reorder the terms. If no pairing works, this particular cubic simply does not factor by grouping, and factoring is not the right tool for it.
That is worth knowing: not every polynomial factors nicely. Recognising when to stop is as useful as knowing how to continue.
Check
Solve it on paper before you click.
Check your understanding
Expand (3x - 2)(x + 5).
Answer: A
Why: The four products are 3x squared, 15x, negative 2x and negative 10. Combining the two middle terms gives 13x, so the answer is 3x squared plus 13x minus 10. Checking with x equal to 1: the original is 1 times 6, which is 6, and the answer gives 3 plus 13 minus 10, which is 6.
Check
Solve it on paper before you click.
Check your understanding
Factor x^2 - 3x - 28 completely.
Answer: A
Why: The constant is negative, so the two numbers have opposite signs, and they must multiply to 28 and combine to negative 3. That pair is negative 7 and positive 4. Expanding gives x squared plus 4x minus 7x minus 28, which is x squared minus 3x minus 28.
Commit first
Answer, then rate your confidence.
\[ 4x^2 - 36 \]
Predict first
What is this factored completely?
Correct: 4(x + 3)(x - 3)
\[ 4x^2 - 36 = 4(x^2 - 9) = 4(x + 3)(x - 3) \]
Why: Taking out the common factor of 4 first leaves x squared minus 9, which is a difference of squares. The second option is technically equal but not fully factored, since each bracket still contains a common factor of 2. The last option stops one step early, leaving a difference of squares unfactored.
Exit ticket
Last commitment of the chapter.
Predict first
Which of these is shakiest right now?
Correct: Whatever you picked is the one to drill first.
Why: The last one costs the most, because skipping it makes every subsequent step harder and often hides a special product entirely. Making it a reflex — look for a common factor before anything else — speeds up the whole chapter.
Connect it up
One page, drawn by you.
Draw it
Draw one area model grid large in the middle. Label the outside edges as the factors and the inside cells as the expanded terms. Then draw arrows in both directions: left to right for multiplying out, right to left for factoring. Around the edges attach: like terms, special products, zero-product property, common factor, grouping.
If the arrows only point one way, redraw them. The whole chapter is one process read in two directions.
Recap
Multiplying out and factoring are the same picture read forwards and backwards, and you can now do both.
| if you remember one thing | it should be |
|---|---|
| about subtraction | the minus reaches every term inside the bracket |
| about multiplying | count the products before you start; two by two means four |
| about the zero-product rule | it only works when the product is exactly zero |
| about factoring | common factor first, and always check whether the leftovers factor again |
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