Factoring the special products of Lesson 10.3 in reverse. Includes the difference of two squares pattern, the two perfect square trinomial patterns, the test for recognising when a pattern applies, expressions that fit no pattern, removing a constant factor first, and solving an equation whose factored form has a repeated factor.
Subject: Algebra 1 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 1 · Chapter 10 — Polynomials and Factoring
Factoring Special Products
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 10.7 Factoring Special Products §10.7, pp. 609-615 — the lesson these objectives are drawn from
Warm-up
Lesson 10.3 expanded the special products. This lesson recognises the results and runs them backwards.
Discussion prompt
Factor x squared minus nine using the search from Lesson 10.5. What is unusual about the target sum?
Hint: There is no middle term, so what must b be?
Answer:
\[ x^2 - 9 = (x + 3)(x - 3) \]
With no middle term the coefficient b is nought, so the two numbers must add to nought — which forces them to be a number and its opposite. Their product is then the negative of a square, which is exactly what a difference of two squares looks like.
Concept
The special products of Lesson 10.3 can be read backwards as factoring patterns. A difference of two squares and a perfect square trinomial can each be factored on sight.
perfect square trinomial — A trinomial of the form a squared plus or minus two a b plus b squared, which factors as the square of a binomial.
Both patterns need the relevant terms to be squares of integers.
Figure (svg): The difference of two squares pattern
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 609-609
Section
Section 1
Concept
An expression of the form a squared minus b squared factors as the sum of a and b times their difference. Both terms must be squares for the pattern to apply.
\[ a^2 - b^2 = (a + b)(a - b) \]
A sum of two squares does not factor this way.
Figure (svg): The difference of two squares pattern
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 609-610 — the Difference of Two Squares Pattern and Example 1
Picture it
Take the roots, then add and subtract.
Figure (svg): The difference of two squares pattern
The whole procedure is to identify a and b as square roots of the two terms. Everything else is copying them into the pattern.
Worked example
This is Example 1, parts a to c, from the textbook.
\[ \text{Factor } m^2 - 4, \; 4p^2 - 25 \text{ and } 9q^2 - 64. \]
Take the first
Why: Four is two squared.
\[ (m + 2) (m - 2) \]
Take the second
Why: Four p squared is two p, squared.
\[ (2 p + 5) (2 p - 5) \]
Take the third
Why: Nine q squared is three q, squared.
\[ (3 q + 8) (3 q - 8) \]
Note what a is each time
Why: The whole term, coefficient included.
\[ m, \; 2 p, \; 3 q \]
Figure (svg): The difference of two squares pattern
\[ (m+2)(m-2), \; (2p+5)(2p-5), \; (3q+8)(3q-8) \]
Verify: expand one of them
Why: Two p plus five times two p minus five gives four p squared minus ten p plus ten p minus twenty-five, and the middle terms cancel to leave four p squared minus twenty-five. The cancellation is what makes the answer a binomial.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 610-610
Sorting
Both terms must be squares, and it must be a difference.
Sort into buckets
Sort each expression by whether it factors as a difference of two squares.
Two of the failures have a non-square constant and one is a sum. Those are the only two ways this pattern can fail, and both can be checked at a glance.
Worked example
Guided Practice 1 to 8, sorted by whether the pattern applies.
\[ \text{Factor } x^2 - 16, \; r^2 - 20, \; 4y^2 - 49 \text{ and } 16q^2 - 45. \]
Take the first
Why: Sixteen is four squared.
\[ (x + 4) (x - 4) \]
Take the second
Why: Twenty is not a square.
Take the third
Why: Both terms are squares.
\[ (2 y + 7) (2 y - 7) \]
Take the fourth
Why: Forty-five is not a square.
Figure (svg): Two columns separating expressions that fit the pattern from those that do not
\[ (x+4)(x-4); \; \text{no}; \; (2y+7)(2y-7); \; \text{no} \]
Verify: check why the failures fail
Why: Twenty and forty-five are not squares of integers, so no integer b exists with b squared equal to them. The first terms were fine in both cases, which shows that both terms have to pass the test rather than just one.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 610-610
Trap
\[ x^2 + 9 = (x + 3)(x + 3) \]
Apply the pattern with both signs positive
Why: The pattern uses a plus and a minus, so two pluses seemed reasonable.
Expanding gives x squared plus six x plus nine, which has an unwanted middle term. A sum of two squares does not factor over the integers at all, and the pattern is stated for a difference only.
x squared plus nine cannot be factored using integers.
Check that the expression is a difference before using the pattern
Why: The minus sign is essential.
Its discriminant is negative thirty-six, so Lesson 9.7 predicts no real roots and therefore no real linear factors.
Faded example
Take the square root of each term.
Fill in the blanks
9q^2 - 64 = (3q)^2 - 8^2 = (3q + 8)(3q - 8)
Why: The coefficient is part of the term being squared, so a is three q rather than q. Taking the root of only the variable would give the wrong factors.
Elimination
Both terms must be perfect squares.
Eliminate the wrong options
Which of these cannot be factored over the integers?
Survives elimination: A
Why: Eight is not the square of an integer, so there is no integer b with b squared equal to eight. Saying an expression does not factor is a complete answer rather than a failure to find one.
Socratic
The difference does.
Discussion prompt
Explain why a squared plus b squared has no factorisation over the integers. Then say how the discriminant confirms it.
Hint: Ask when the expression could be nought.
Answer:
If it factored into two linear factors, then setting it equal to nought would give real solutions — but a sum of two squares is never nought unless both terms are, since squares cannot be negative and so cannot cancel each other. With no real roots there can be no real linear factors.
For x squared plus nine the discriminant is nought minus thirty-six, which is negative thirty-six, and Lesson 9.7 says a negative discriminant means no real solutions. That is exactly the same conclusion by a different route, and it explains why the pattern is stated only for a difference.
Section
Section 2
Concept
A trinomial factors as the square of a binomial when its first and last terms are squares and its middle term is twice the product of their roots. The middle sign becomes the sign in the binomial.
\[ a^2 \pm 2ab + b^2 = (a \pm b)^2 \]
The last term is positive in both versions.
Figure (svg): The two perfect square trinomial patterns
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 609-610 — the Perfect Square Trinomial Pattern and Example 2
Picture it
Only the middle sign.
Figure (svg): The two perfect square trinomial patterns
The final term is a square in both, so it is always positive. A trinomial ending in a negative constant is never a perfect square.
Worked example
This is Example 2 from the textbook.
\[ \text{Factor } x^2 + 4x + 4, \; a^2 + 18a + 81 \text{ and } 16y^2 - 24y + 9. \]
Take the first
Why: Roots x and two; twice their product is four x.
\[ (x + 2) ^{2} \]
Take the second
Why: Roots a and nine; twice their product is eighteen a.
\[ (a + 9) ^{2} \]
Take the third
Why: Roots four y and three.
\[ 2(4 y) (3) = 24 y \]
Use the negative version
Why: The middle term is subtracted.
\[ (4 y - 3) ^{2} \]
Figure (svg): The two-part test for a perfect square trinomial
\[ (x+2)^2, \; (a+9)^2, \; (4y-3)^2 \]
Verify: expand the third
Why: Four y minus three, squared, gives sixteen y squared minus twenty-four y plus nine. The nine came out positive even though the binomial had a minus, which is the check that the last term is always a square.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 610-610
Faded example
Twice the product of the roots.
Fill in the blanks
16y^2 - 24y + 9: \quad \text3 4y \text24y ___, \; 2(4y)(___) = ___
Why: The computed value matches the middle term, so the pattern applies. Had it come out as something else, the trinomial would not be a perfect square whatever its ends looked like.
Worked example
Guided Practice 12 to 14.
\[ \text{Factor } 4b^2 + 4b + 1, \; 25m^2 - 10m + 1 \text{ and } 9a^2 + 30a + 25. \]
Take the first
Why: Roots two b and one.
\[ 2(2b)(1) = 4b \;\checkmark \]
Write it
Why: The middle term is positive.
\[ (2 b + 1) ^{2} \]
Take the second
Why: Roots five m and one.
\[ (5 m - 1) ^{2} \]
Take the third
Why: Roots three a and five.
\[ (3 a + 5) ^{2} \]
Figure (svg): The two-part test for a perfect square trinomial
\[ (2b+1)^2, \; (5m-1)^2, \; (3a+5)^2 \]
Verify: check the third's middle term
Why: Twice three a times five is thirty a, which matches. That check is the whole difference between a perfect square trinomial and one that merely has square ends, so it is worth doing rather than assuming.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 610-610
Error analysis
The student factored a trinomial whose ends were both squares.
Annotate
On: \( \begin{aligned} 9a^2 + 20a + 25 &= (3a + 5)^2 \\ \text{check: } (3a + 5)^2 &= 9a^2 + 30a + 25 \end{aligned} \)
Passing the first half of the test and failing the second is exactly the situation the two-part check exists for. Square ends are suggestive and not sufficient, and the middle term is what actually decides.
Matching
Roots first, then the middle sign.
Match the pairs
Why: Every last term is positive, including in the two that came from differences. The middle sign is the only place the binomial's sign shows up.
Elimination
Both parts of the test must pass.
Eliminate the wrong options
Which trinomial factors as the square of a binomial?
Survives elimination: A
Why: Only one passes both parts of the test. Each wrong option fails in a different way, which is why running both checks rather than one is worth the few seconds.
Socratic
Even when the binomial has a minus.
Discussion prompt
Explain why a perfect square trinomial always ends in a positive constant. Then say what a negative constant tells you immediately.
Hint: What is the sign of b squared?
Answer:
The last term is b squared, and squaring any real number gives something positive or nought, so it can never be negative. That is true whether the binomial was a sum or a difference, since the minus affects only the cross term.
A negative constant therefore rules out the perfect square pattern instantly, before any roots are computed. It also suggests a different structure: a negative constant means the two factors have opposite signs, which is the mixed-sign case of Lesson 10.5 rather than a repeated factor.
Section
Section 3
Concept
A pattern applies only when its conditions are met. For a difference of squares both terms must be squares; for a perfect square trinomial the ends must be squares and the middle must be twice the product of their roots.
Saying an expression does not factor is a valid answer.
Figure (svg): Two columns separating expressions that fit the pattern from those that do not
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 610-611 — Example 1, part d, and Example 3, part c, where no pattern applies
Picture it
One failed condition is enough.
Figure (svg): Two columns separating expressions that fit the pattern from those that do not
The right-hand column is not a list of harder problems; it is a list of expressions with no integer factorisation. Recognising that quickly saves a great deal of fruitless searching.
Worked example
This is Example 1, part d, from the textbook.
\[ \text{Can } a^2 - 8 \text{ be factored over the integers?} \]
Check the form
Why: Two terms, subtracted.
Check the first term
Why: a squared is a square.
\[ \;\checkmark \]
Check the second term
Why: Eight is not a square.
\[ \;\times \]
Conclude
Why: The pattern does not apply.
Figure (svg): Two columns separating expressions that fit the pattern from those that do not
\[ a^2 - 8 \text{ does not factor over the integers} \]
Verify: say what it would factor into otherwise
Why: Over the reals it would be a plus the root of eight times a minus the root of eight, but the root of eight is irrational, so there is no integer factorisation. The question asks about integers, and the answer to that is a clear no.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 610-610
Sorting
Count the terms first.
Sort into buckets
Sort each expression by which pattern to try.
Counting the terms decides which test to run before anything else. Two terms point at one pattern and three at the other, and the tests themselves then confirm or reject it.
Worked example
This is Example 3, part c, from the textbook.
\[ \text{Factor } 4x^2 + 24x + 44 \text{ as far as possible.} \]
Remove the common factor
Why: All three are divisible by four.
\[ 4(x ^{2} + 6 x + 11) \]
Test the trinomial
Why: One is a square; eleven is not.
\[ \;\times \]
Try the search instead
Why: No pair multiplies to eleven and adds to six.
\[ 1 \text{ and } 11 \to 12 \]
Conclude
Why: The common factor is all that comes out.
\[ 4(x ^{2} + 6 x + 11) \]
Figure (svg): Two columns separating expressions that fit the pattern from those that do not
\[ 4(x^2 + 6x + 11) \]
Verify: check with the discriminant
Why: For x squared plus six x plus eleven the discriminant is thirty-six minus forty-four, which is negative eight — negative, so there are no real roots and therefore no real linear factors. The search and the discriminant agree.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 611-611
Trap
\[ a^2 - 8 = (a + 2\sqrt{2})(a - 2\sqrt{2}) \]
Use the pattern with an irrational b
Why: The pattern works for any a and b.
It does mathematically, but the question asks for factoring over the integers and the root of eight is irrational. In this course such an expression is reported as not factorable rather than factored with radicals.
a squared minus eight cannot be factored using integers.
State the conclusion with the reason
Why: Eight is not the square of an integer.
Knowing that the reals would allow it is worth knowing and is not what was asked.
Prediction
For a difference of two terms.
Predict first
If one term is not a perfect square, what follows?
Correct: The difference of squares pattern does not apply.
\[ 50 - 98x^2 = 2(25 - 49x^2) \quad \text{but} \quad a^2 - 8 \text{ has no common factor} \]
Why: The pattern requires both terms to be squares, so failing on either one rules it out immediately. With only two terms and no common factor there is usually nothing else to try, so the expression does not factor over the integers — as with a squared minus eight and r squared minus twenty. Checking a common factor first is worth doing, since that can sometimes turn non-squares into squares.
Hypothesis
Before trying any pattern or search.
Predict first
What single calculation predicts whether a trinomial factors over the integers?
Correct: The discriminant, which must be a perfect square.
\[ x^2 + 6x + 11: \; 36 - 44 = -8 \qquad x^2 + 4x + 4: \; 16 - 16 = 0 \]
Why: A trinomial factors over the integers exactly when b squared minus four a c is a perfect square, since that is when the quadratic formula returns rational values. For x squared plus six x plus eleven it is negative eight, so nothing will work; for a perfect square trinomial it is exactly nought, which is why those have a repeated factor. One subtraction settles what could otherwise be a long fruitless search.
Socratic
The two facts arrive from different directions.
Discussion prompt
Explain why a trinomial with discriminant nought must be a perfect square. Then say what that means about its graph.
Hint: How many roots does it have?
Answer:
A discriminant of nought means the quadratic formula returns one repeated value, so the trinomial has a single root counted twice — and a polynomial with a repeated root factors with that factor repeated. Two copies of the same linear factor is precisely a squared binomial.
Graphically the curve touches the axis at its vertex rather than crossing it, which is Lesson 9.7's middle case and Lesson 10.4's repeated factor. All three descriptions — zero discriminant, repeated factor, touching graph — are the same situation, and being able to move between them is worth more than any one of them alone.
Section
Section 4
Concept
Removing a constant factor often turns coefficients that are not squares into ones that are. Always take out the common factor before testing for a pattern.
Fifty and ninety-eight are not squares; twenty-five and forty-nine are.
Figure (svg): A constant factored out before a pattern is applied
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 611-611 — Example 3, Factor Out a Constant First
Picture it
Take the two out first.
Figure (svg): A constant factored out before a pattern is applied
Without the first step the expression looks like nothing in particular. With it, the difference of squares is unmistakable, which is why the step is not optional.
Worked example
This is Example 3, part a, from the textbook.
\[ \text{Factor } 50 - 98x^2. \]
Remove the common factor
Why: Both terms are even.
\[ 2(25 - 49 x ^{2}) \]
Test the remainder
Why: Twenty-five and forty-nine are squares.
\[ 5 ^{2} - (7 x) ^{2} \]
Apply the pattern
Why: Sum times difference.
\[ (5 + 7 x) (5 - 7 x) \]
Include the constant
Why: Keep the two in front.
\[ 2(5 + 7 x) (5 - 7 x) \]
Figure (svg): A constant factored out before a pattern is applied
\[ 50 - 98x^2 = 2(5 + 7x)(5 - 7x) \]
Verify: expand the whole thing
Why: Five plus seven x times five minus seven x is twenty-five minus forty-nine x squared, and doubling gives fifty minus ninety-eight x squared. Neither fifty nor ninety-eight is a square, so the pattern was genuinely invisible before the common factor came out.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 611-611
Faded example
The squares appear after the factor comes out.
Fill in the blanks
50 - 98x^2 = 2(25 - 49x^2) = 2(5 + 7x)(5 - 7x)
Why: Dividing both terms by two turns fifty into twenty-five and ninety-eight into forty-nine, both perfect squares. The pattern was there all along and the common factor was concealing it.
Worked example
This is Example 3, part b, and Guided Practice 18.
\[ \text{Factor } 3x^2 + 30x + 75 \text{ and } 8n^2 - 24n + 18. \]
Take the first
Why: All three are divisible by three.
\[ 3(x ^{2} + 10 x + 25) \]
Apply the pattern
Why: Roots x and five; twice their product is ten x.
\[ 3(x + 5) ^{2} \]
Take the second
Why: All three are even.
\[ 2(4 n ^{2} - 12 n + 9) \]
Apply the pattern
Why: Roots two n and three.
\[ 2(2 n - 3) ^{2} \]
Figure (svg): A constant factored out before a pattern is applied
\[ 3(x + 5)^2, \qquad 2(2n - 3)^2 \]
Verify: check the second's middle term
Why: Twice two n times three is twelve n, matching the middle term of the bracket. Note that three x squared plus thirty x plus seventy-five is not itself a perfect square trinomial — the three in front prevents it — which is why the removal has to happen first.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 611-611
Trap
\[ 50 - 98x^2 \;\Longrightarrow\; \text{neither term is a square, so it does not factor} \]
Test the pattern on the expression as given
Why: Fifty and ninety-eight are not squares, so the test failed.
The test was run in the wrong order. Removing the common factor of two leaves twenty-five and forty-nine, both of which are squares, and the expression factors perfectly well.
\[ 50 - 98x^2 = 2(25 - 49x^2) = 2(5 + 7x)(5 - 7x) \]
Remove any common factor before testing for a pattern
Why: It can reveal squares that were hidden.
This is the same first step as in Lesson 10.6, and for the same reason.
Translation
Constant first, then a pattern.
Match the pairs
Why: Every one needed the constant removed before its pattern became visible. Two turned out to be differences of squares and two perfect squares, decided by the number of terms.
Elimination
Factoring 3p squared - 36p + 108.
Eliminate the wrong options
What should be done first?
Survives elimination: A
Why: Removing the three leaves p squared minus twelve p plus thirty-six, whose ends are squares and whose middle is twice six p. Option B is the trap: the test is right but run at the wrong moment.
Socratic
The expression has not changed.
Discussion prompt
Explain how a common factor can hide a difference of squares. Then say what kind of common factor is most likely to do this.
Hint: What does multiplying a square by two do?
Answer:
Multiplying a perfect square by a number that is not itself a square produces something that is not a square: twenty-five is a square and fifty is not, forty-nine is a square and ninety-eight is not. The structure survives inside the expression but its outward form no longer passes the test.
Non-square common factors are the ones that do this — twos, threes, fives and so on. A square common factor such as four would leave both terms square and the pattern still visible, though removing it is still worth doing to keep the numbers small and the factorisation complete.
Section
Section 5
Concept
An equation whose factored form is a constant times a squared binomial has a single solution. Setting the repeated factor equal to nought gives it directly.
The constant factor never contributes a solution.
Figure (svg): A perfect square trinomial giving a single solution
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 611-611 — Example 4, Graphical and Analytical Reasoning
Picture it
One solution, at the vertex.
Figure (svg): A perfect square trinomial giving a single solution
The factored form also shows the function is never negative, since a square is never negative and the constant in front is positive. That is more than the standard form tells you at a glance.
Worked example
This is Example 4 from the textbook.
\[ \text{Solve } 2x^2 - 12x + 18 = 0. \]
Remove the common factor
Why: All three are even.
\[ 2(x ^{2} - 6 x + 9) = 0 \]
Recognise the pattern
Why: Roots x and three; twice their product is six x.
\[ 2(x - 3) ^{2} = 0 \]
Set the repeated factor to nought
Why: The two cannot be nought.
\[ x - 3 = 0 \]
Solve
Why: Add three.
\[ x = 3 \]
Figure (svg): A perfect square trinomial giving a single solution
\[ x = 3 \]
Verify: substitute it back
Why: Two times nine is eighteen, minus thirty-six, plus eighteen — that is nought. The graph touches the axis at three and never crosses it, which is what a repeated factor looks like.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 611-611
Faded example
The constant comes out first.
Fill in the blanks
2x^2 - 12x + 18 = 0 \;\to\; 2(x^2 - 6x + 9) = 0 \;\to\; 2(x - 3)^2 = 0 \;\to\; x = 3
Why: Every coefficient is halved when the two comes out, including the constant. The resulting trinomial is then a perfect square, which the original was not.
Worked example
The zero-product property applied carefully.
\[ \text{In } 2(x - 3)^2 = 0, \text{ why does the } 2 \text{ contribute nothing?} \]
List the factors
Why: A constant and a repeated binomial.
\[ 2, \; (x - 3), \; (x - 3) \]
Set each to nought
Why: The property applies to all of them.
Reject the first
Why: Two is never nought.
Keep the second
Why: It gives the only solution.
\[ x = 3 \]
Figure (svg): A perfect square trinomial giving a single solution
\[ 2 \ne 0, \quad \text{so only } x - 3 = 0 \]
Verify: check by dividing the equation by two
Why: Dividing both sides by two leaves x minus three, squared, equal to nought, with the same solution. A non-zero constant factor can always be divided out of an equation without changing its solutions, which is another way of seeing the same thing.
Trap
\[ 2(x - 3)^2 = 0 \;\Longrightarrow\; x = 3 \text{ and } x = 3 \]
List one solution per bracket
Why: The square means two brackets, so two solutions were listed.
Both brackets give the same equation and the same number, so there is one solution, not two. Writing it twice is a repetition rather than a second answer.
\[ 2(x - 3)^2 = 0 \;\Longrightarrow\; x = 3 \]
Count distinct solutions rather than brackets
Why: A repeated factor contributes one.
The graph confirms it: the curve meets the axis at exactly one point.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Feature | From 2(x - 3) squared | From 2x squared - 12x + 18 |
|---|---|---|
| The solution | read off as 3 | found with the formula or by factoring |
| Number of solutions | one, from the repeated factor | one, since the discriminant is zero |
| Sign of the function | never negative | not obvious without work |
The last row is what the factored form adds beyond the solution: a square times a positive constant cannot be negative, so the curve never dips below the axis. Standard form conceals that entirely.
Prediction
The equation negative two times (x minus three) squared equals nought.
Predict first
How would the solution change?
Correct: Not at all; the solution is still 3.
\[ -2(x-3)^2 = 0 \;\Longrightarrow\; (x-3)^2 = 0 \;\Longrightarrow\; x = 3 \]
Why: A non-zero constant factor never contributes a solution, whatever its sign, because it can never be nought. Dividing both sides by negative two leaves the same repeated factor and the same answer. What does change is the graph: a negative constant flips the parabola so it touches the axis from below, but it still touches at exactly the same place.
Socratic
The search would find the same factors.
Discussion prompt
Say what recognising a perfect square trinomial tells you that a successful search does not. Then name a later use of the same recognition.
Hint: Think about roots and about graphs.
Answer:
Recognising the pattern tells you immediately that the equation has one solution rather than two, that its graph touches the axis rather than crossing, and that the expression is never negative. A search that happens to produce two identical factors gives the same factorisation but none of that reading.
The same recognition run backwards is the heart of Lesson 12.5, where a constant is deliberately chosen to make a trinomial a perfect square so that it can be written as a squared binomial and solved by taking roots. Completing the square is exactly this pattern used as a construction rather than as an observation, and it is how the quadratic formula is derived.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Difference of squares | Perfect square trinomial | |
|---|---|---|
| Number of terms | two | three |
| The test | both terms are squares and subtracted | square ends and a doubled middle |
| The factors | a sum times a difference | a repeated factor, so a squared binomial |
Counting the terms decides which test to run, and the test then confirms or rejects the pattern. Both steps are quick and neither can be skipped.
Pattern
To factor any expression using the special patterns, these five moves cover it.
Step one has to come before step three, because a common factor can hide squares that are plainly visible once it is removed.
OpenStax Elementary Algebra 2e, §7.4 Factor Special Products §7.4
Check
Both terms must be squares.
Check your understanding
Factor 9q squared - 64.
Answer: A
Why: Nine q squared is three q squared and sixty-four is eight squared, so the pattern gives the sum times the difference.
Check
Check the middle term.
Check your understanding
Factor 16y squared - 24y + 9.
Answer: A
Why: The roots are four y and three, twice their product is twenty-four y, and the middle term is negative, so the binomial is a difference.
Check
Common factor first.
Check your understanding
Factor 50 - 98x squared completely.
Answer: A
Why: Removing the common factor of two leaves twenty-five minus forty-nine x squared, whose terms are both squares.
Real world
This is the pole-vault question from the lesson opener. A vaulter's height above the ground can be modelled by a quadratic, and the peak of the vault is where its factored form touches the axis.
Discussion prompt
Suppose the height above the peak is modelled by h equal to negative 2t squared plus 12t minus 18. Factor it, find when the height is nought, and say what the factored form tells you about the vault.
Hint: Take out the common factor first.
Answer:
\[ -2t^2 + 12t - 18 = -2(t^2 - 6t + 9) = -2(t - 3)^2 \]
The height is nought when t minus three is nought, so at three seconds — and that is the only time, since the factor is repeated.
The factored form says more than the solution alone: because a square is never negative and the constant in front is negative, the expression is never positive. So this model describes a quantity that reaches nought exactly once and is otherwise below it, which is the signature of a curve touching its peak rather than crossing through. Recognising the perfect square gave all of that at a glance, where the quadratic formula would have given only the single root.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Is 9a squared + 20a + 25 a perfect square trinomial?
Correct: No, because twice 3a times 5 is 30a, not 20a.
\[ 2(3a)(5) = 30a \ne 20a \]
Why: Both ends are perfect squares, with roots three a and five, so the trinomial passes the first half of the test and looks convincing. The second half is what decides: twice the product of the roots is thirty a, and the trinomial offers twenty a, so it is not a perfect square. Expanding the proposed answer confirms it, giving nine a squared plus thirty a plus twenty-five rather than what was asked for. In fact its discriminant is four hundred minus nine hundred, which is negative, so it does not factor over the integers at all. Square ends are suggestive and never sufficient, which is exactly why the test has two parts.
Explain it
They said fifty minus ninety-eight x squared does not factor because neither fifty nor ninety-eight is a square.
Discussion prompt
In no more than four sentences, explain what step they skipped. Then give them the rule about the order of operations here.
Hint: What do both numbers have in common?
Answer:
A usable answer: both fifty and ninety-eight are even, so a factor of two comes out first, leaving twenty-five minus forty-nine x squared. Those two are squares, so the difference of squares pattern applies and the answer is two times five plus seven x times five minus seven x.
The rule is to remove any common factor before testing for a pattern, never after. A non-square factor like two can turn perfectly good squares into numbers that fail the test, so testing first gives the wrong answer.
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: Coefficients are fixed by taking the square root of the whole term rather than just the variable. The perfect square test is fixed by always running both halves, especially the doubled middle term. The constant is fixed by making it the first thing you look for, before any test. Deciding something does not factor is fixed by computing the discriminant, which settles it in one subtraction. 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 as factoring rules, with one worked example beside each in which the leading coefficient is greater than one so the square rooting of a whole term is visible. Underneath, write the two-part test for a perfect square trinomial as two numbered questions, and beside it work an expression that passes the first question and fails the second, showing the arithmetic that rejects it. In the middle, take three expressions that do not factor over the integers, and beside each write the reason and its discriminant, so that the two explanations can be seen to agree. Beneath that, factor two expressions that need a constant removed first, writing the removal as its own line and marking that neither original coefficient was a square while both new ones are. In the lower half, solve an equation whose factored form is a constant times a squared binomial, sketch its graph touching the axis, and write one sentence saying why the function never changes sign. Finally, in the margin, write the order in which the steps must be done and why removing the constant cannot come last.
Every discriminant you compute for a perfect square trinomial should come out as exactly nought. If one does not, the trinomial passed your test but should not have, so recheck the doubled middle term.
Recap
Five things, and the fourth is the step that makes the others possible.
| If the question says | Your first move is |
|---|---|
| Two terms with a minus between | Check whether both are squares |
| Three terms with square ends | Check whether the middle is doubled |
| The coefficients share a factor | Remove it before testing |
| A term is not a perfect square | The pattern does not apply |
| The factored form has a repeated factor | One solution, not two |
Lesson 10.8 handles polynomials of degree three, where the aim is to break a cubic into a monomial factor and a quadratic — and then to apply everything from the last four lessons to what remains.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.7 Factoring Special Products §10.7, pp. 609-615 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 1 — $55/session, free consultation.