Factoring a trinomial whose leading coefficient is one. Includes what factoring a trinomial means, the two conditions the pair of numbers must satisfy, how the signs of b and c determine the signs of the pair, a systematic search through the factor pairs of the constant, checking by multiplying back, and using factoring to solve a border problem.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 10 — Polynomials and Factoring
Factoring x^2 + bx + c
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 10.5 Factoring x^2 + bx + c §10.5, pp. 595-602 — the lesson these objectives are drawn from
Warm-up
Lesson 10.2 expanded products of binomials. This lesson runs that backwards, and the expansion tells you what to look for.
Discussion prompt
Expand the product of x plus p and x plus q, leaving p and q as letters. What do the coefficients of the answer tell you?
Hint: Use FOIL and collect the middle terms.
Answer:
\[ (x + p)(x + q) = x^2 + (p + q)x + pq \]
The middle coefficient is the sum of the two numbers and the constant is their product. So to factor a trinomial you need a pair of numbers doing both jobs at once — adding to b and multiplying to c. That single line is the whole method.
Concept
To factor a trinomial of the form x squared plus b x plus c means to write it as a product of two binomials. The two constants in those binomials must add to b and multiply to c.
factor a trinomial — To write a trinomial as the product of two binomials. For x squared plus b x plus c, this means finding numbers p and q with p plus q equal to b and p times q equal to c.
The signs of b and c narrow the search before it starts.
Figure (svg): The expansion that factoring has to reverse
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 595-595
Section
Section 1
Concept
Expanding the product of x plus p and x plus q gives x squared plus the quantity p plus q times x, plus p q. So factoring means finding a pair whose sum is b and whose product is c.
\[ (x + p)(x + q) = x^2 + (p + q)x + pq \]
Both conditions must hold for the same pair.
Figure (svg): The expansion that factoring has to reverse
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 595-595 — the development of the p and q conditions and Example 1
Picture it
One pair, two jobs.
Figure (svg): The expansion that factoring has to reverse
A pair that satisfies only one condition is no use at all. Two and four multiply to eight and also add to six, and it is the coincidence of both that makes them the right pair.
Worked example
This is Example 1 from the textbook.
\[ \text{Factor } x^2 + 6x + 8. \]
Identify b and c
Why: Read them off the trinomial.
\[ b = 6, \; c = 8 \]
List the factor pairs of c
Why: One and eight, two and four.
\[ 1, 8 \text{ and } 2, 4 \]
Check their sums
Why: Nine, then six.
\[ 9 \text{ and } 6 \]
Take the pair that works
Why: Two and four.
\[ (x + 2) (x + 4) \]
Figure (svg): The systematic search for a pair of numbers
\[ x^2 + 6x + 8 = (x + 2)(x + 4) \]
Verify: multiply the answer back
Why: The four products are x squared, four x, two x and eight, and the middle two combine to six x. Expanding is quick and confirms the factorisation exactly, so no factoring answer needs to be left uncertain.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 595-595
Faded example
Read them off the expansion.
Fill in the blanks
For x squared plus bx plus c, the two numbers must add to b and multiply to c.
Why: The sum matches the middle term and the product matches the constant, which is exactly what expanding the answer produces. Getting the two the wrong way round is a surprisingly durable error.
Worked example
Guided Practice 1 to 4.
\[ \text{Factor } x^2 + 4x + 3, \; x^2 + 5x + 6, \; x^2 + 8x + 7 \text{ and } x^2 + 7x + 6. \]
Take the first
Why: One and three multiply to three.
\[ (x + 1) (x + 3) \]
Take the second
Why: Two and three multiply to six.
\[ (x + 2) (x + 3) \]
Take the third
Why: One and seven multiply to seven.
\[ (x + 1) (x + 7) \]
Take the fourth
Why: One and six also multiply to six.
\[ (x + 1) (x + 6) \]
Figure (svg): One worked trinomial for each of the four sign combinations
\[ (x+1)(x+3), \; (x+2)(x+3), \; (x+1)(x+7), \; (x+1)(x+6) \]
Verify: compare the second and fourth
Why: Both have a constant of six but different middle terms, and they use different factor pairs of six — two and three for a sum of five, one and six for a sum of seven. The constant alone never decides the answer, which is why the sum has to be checked too.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 595-595
Trap
\[ x^2 + 6x + 8 = (x + 1)(x + 8) \]
Pick a pair that multiplies to eight
Why: One times eight is eight, so the constant is right.
Expanding gives x squared plus nine x plus eight, and the middle term should be six x. Both conditions have to hold for the same pair, and this one satisfies only the second.
\[ x^2 + 6x + 8 = (x + 2)(x + 4) \]
Check the sum as well as the product before committing
Why: Two conditions, one pair.
Multiplying back catches this in a single line, which is why the check is worth doing every time.
Matching
Sum and product both.
Match the pairs
Why: The second and third trinomials share a constant but need different pairs, because their middle terms differ. Checking the sum is what separates two candidates that both give the right product.
Elimination
Factoring x squared + 6x + 8.
Eliminate the wrong options
Which product expands to the trinomial?
Survives elimination: A
Why: Only one pair satisfies both conditions at once. Options B and C are instructive because each gets exactly one condition right, which is what makes a partial check misleading.
Socratic
Lesson 10.3's squared binomial doubled its middle term.
Discussion prompt
Explain why the middle coefficient here is p plus q rather than twice something. Then say what happens when p and q happen to be equal.
Hint: Count the cross products and compare them.
Answer:
The two cross products are p x and q x, and those are different quantities unless p and q happen to be equal, so they add to give p plus q rather than doubling either one. The area model shows this directly: the two rectangles have different widths and so different areas.
When p and q are equal the two rectangles become congruent and their sum is twice one of them, which is exactly Lesson 10.3's square of a binomial. So that pattern is not a separate rule but the special case of this one where the two numbers coincide — and the trinomial it produces is x squared plus two p x plus p squared.
Section
Section 2
Concept
When the constant term is positive, the two numbers must have the same sign, since only two positives or two negatives multiply to a positive. The sign of b then says which.
That halves the search before it begins.
Figure (svg): How the signs of b and c decide the signs of the two numbers
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 596-596 — Example 2 and its Study Tip on a positive constant term
Picture it
Two questions in order.
Figure (svg): How the signs of b and c decide the signs of the two numbers
The top two rows are this section and the bottom two are the next. Deciding the signs first turns a vague hunt into a short list of candidates.
Worked example
This is Example 2 from the textbook.
\[ \text{Factor } x^2 - 5x + 6. \]
Read the signs
Why: c positive means the same sign.
List the pairs
Why: Negative one and negative six, negative two and negative three.
\[ -1, -6 \text{ and } -2, -3 \]
Check the sums
Why: Negative seven, then negative five.
\[ -7 \text{ and } -5 \]
Take the pair that works
Why: Negative two and negative three.
\[ (x - 2) (x - 3) \]
Figure (svg): How the signs of b and c decide the signs of the two numbers
\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]
Verify: multiply it back
Why: The cross products are negative three x and negative two x, adding to negative five x, and the constant is negative two times negative three, which is positive six. Both negatives were needed: one of each sign would have given a negative constant.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 596-596
Sorting
Look only at the constant term.
Sort into buckets
Sort each trinomial by the sign pattern of its pair.
The middle term was never consulted for this sort, only the constant. That is the first of the two sign questions, and it is settled before any numbers are chosen.
Worked example
Guided Practice 5 to 8, one of them a perfect square.
\[ \text{Factor } x^2 - 5x + 4, \; x^2 - 4x + 4, \; x^2 - 8x + 7 \text{ and } x^2 - 7x + 12. \]
Take the first
Why: Negative one and negative four.
\[ (x - 1) (x - 4) \]
Take the second
Why: Negative two twice.
\[ (x - 2) (x - 2) \]
Take the third
Why: Negative one and negative seven.
\[ (x - 1) (x - 7) \]
Take the fourth
Why: Negative three and negative four.
\[ (x - 3) (x - 4) \]
Figure (svg): One worked trinomial for each of the four sign combinations
\[ (x-1)(x-4), \; (x-2)^2, \; (x-1)(x-7), \; (x-3)(x-4) \]
Verify: notice the repeated factor in the second
Why: The pair was negative two and negative two, so the factors are identical and the answer is a squared binomial. That is the pattern of Lesson 10.3 arrived at from the other direction, and it means the equation would have one solution rather than two.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 596-596
Trap
\[ x^2 - 5x + 6 = (x + 2)(x - 3) \]
Use two and three with one sign each
Why: They add to negative one and multiply to negative six, near enough.
The constant comes out negative six rather than positive six, so the signs are wrong before the middle term is even considered. A positive constant forbids a mixed pair entirely.
\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]
Decide the sign pattern from c before choosing numbers
Why: Positive c means matching signs.
Checking the constant alone would have caught it without expanding fully.
Faded example
c positive, b negative.
Fill in the blanks
x^2 - 5x + 6: \quad \text-2 -3 \text___ ___ \;\Longrightarrow\; (x - 2)(x - 3)
Why: Two negatives multiply to a positive six and add to a negative five, satisfying both conditions. Writing the numbers with their signs before building the brackets keeps the factors correct.
Prediction
The same constant term of six.
Predict first
How would factoring x squared plus 5x plus 6 differ from x squared minus 5x plus 6?
Correct: The same pair of numbers, but both positive.
\[ x^2 + 5x + 6 = (x+2)(x+3) \qquad x^2 - 5x + 6 = (x-2)(x-3) \]
Why: The constant is positive in both cases, so the pair shares a sign either way, and six is still two times three. Only the shared sign flips, giving two and three rather than negative two and negative three, so the factors become x plus two and x plus three. Changing the sign of b in this situation reflects both solutions across nought without changing their sizes.
Socratic
It seems like a rule to memorise.
Discussion prompt
Explain why the two numbers must share a sign when the constant is positive. Then say what the sign of b adds to that.
Hint: Think about the sign of a product.
Answer:
The constant is the product of the two numbers, and a product is positive only when both factors are positive or both are negative — a mixed pair always gives a negative. So a positive constant rules out mixed pairs entirely, before any particular numbers are considered.
The sign of b then settles which of the two matching cases applies, since the sum of two positives is positive and the sum of two negatives is negative. Two yes-or-no questions, asked in that order, fix both signs completely — and neither of them requires knowing what the numbers actually are.
Section
Section 3
Concept
When the constant term is negative, the two numbers must have different signs. The sign of b tells you which of them is the larger in size.
The sum of a mixed pair is a difference in disguise.
Figure (svg): How the signs of b and c decide the signs of the two numbers
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 596-597 — Examples 3 and 4 and the Study Tip on a negative constant term
Picture it
Opposite signs, and which is bigger.
Figure (svg): How the signs of b and c decide the signs of the two numbers
With a mixed pair the sum is really a subtraction, so the numbers can be far apart in size while their sum is small. That is why the factor pairs must still be checked one at a time.
Worked example
This is Example 3 from the textbook.
\[ \text{Factor } x^2 - 11x - 12. \]
Read the signs
Why: c negative means opposite signs.
List the candidates
Why: One and twelve, with signs to assign.
\[ -1, 12 \text{ or } 1, -12 \]
Check the sums
Why: Eleven, then negative eleven.
\[ 11 \text{ and } -11 \]
Take the pair that works
Why: One and negative twelve.
\[ (x + 1) (x - 12) \]
Figure (svg): How the signs of b and c decide the signs of the two numbers
\[ x^2 - 11x - 12 = (x + 1)(x - 12) \]
Verify: check that the larger number carries b's sign
Why: The middle coefficient is negative, and twelve — the larger of the two — is the one carrying the minus. That correspondence holds for every mixed pair, and it decides which of the two sign assignments to try first.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 596-596
Faded example
The larger number follows b.
Fill in the blanks
x^2 - 11x - 12: \quad \text-12 1 \text- ___, \text___ (x + 1)(x ___ 12)
Why: The larger number takes the sign of the middle coefficient, which puts the minus on the twelve. Swapping the two signs would give a middle term of positive eleven instead.
Worked example
This is Example 4 from the textbook.
\[ \text{Factor } x^2 + 17x - 18. \]
Read the signs
Why: Opposite signs, larger one positive.
\[ b \text{ is positive} \]
Try the outermost pair
Why: One and eighteen.
\[ -1, 18 \]
Check the sum
Why: Seventeen.
\[ 17 \;\checkmark \]
Write the factors
Why: The signs go with the numbers.
\[ (x - 1) (x + 18) \]
Figure (svg): One worked trinomial for each of the four sign combinations
\[ x^2 + 17x - 18 = (x - 1)(x + 18) \]
Verify: notice why the outermost pair was tried first
Why: A large middle coefficient needs two numbers far apart in size, since their sum is really a difference. One and eighteen are the furthest apart of the factor pairs of eighteen, so they were the natural first guess and they worked immediately.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 597-597
Error analysis
The student factored a trinomial with a negative constant term.
Annotate
On: \( \begin{aligned} x^2 - 11x - 12 &= (x - 1)(x + 12) \\ \text{check: } (x-1)(x+12) &= x^2 + 11x - 12 \end{aligned} \)
What makes this worth studying is that the student did check — and then did not act on the result. A check that disagrees is information, and the fix here was a single sign swap rather than a fresh search.
Translation
All four sign cases.
Match the pairs
Why: One example of each sign pattern, and the pattern is visible in the brackets: matching signs when the constant is positive and mixed signs when it is negative. Working one of each is worth more than working four of the same kind.
Hypothesis
For x squared + 17x - 18, the pair is -1 and 18.
Predict first
Why does a large middle coefficient suggest trying widely separated factors first?
Correct: Because a mixed pair's sum is really a difference, so a big sum needs a big gap.
\[ 18 - 1 = 17 \qquad 9 - 2 = 7 \qquad 6 - 3 = 3 \]
Why: Adding a positive and a negative amounts to subtracting their sizes, so the result is small when they are close together and large when they are far apart. Since seventeen is nearly as large as eighteen itself, the pair must be about as far apart as the factor pairs of eighteen allow — which points straight at one and eighteen. That reasoning turns the search into an educated first guess rather than a list to work through.
Socratic
The two numbers are added, not subtracted.
Discussion prompt
Explain why adding two numbers of opposite signs behaves like subtracting their sizes. Then say what that means for how many pairs you need to try.
Hint: Write out one example.
Answer:
Adding a negative is the same as subtracting a positive, so one and negative twelve add to one minus twelve, which is negative eleven — the sizes have been subtracted and the sign taken from the larger. Nothing new is happening; it is Chapter 2's rule for adding signed numbers.
It means the achievable sums are the differences of the factor pairs, and those differences shrink as the pair moves closer together. So a large target sum points to the widest pair and a small one to the closest pair, and you can usually start from the right end of the list rather than the top of it.
Section
Section 4
Concept
The candidates are the factor pairs of the constant term, and there are never many. Once a pair satisfies both conditions there is nothing to gain from listing the rest.
Multiplying back confirms the answer in one line.
Figure (svg): A factorisation checked by expanding it again
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 597-597 — the Study Tip on stopping once the correct pair is found, and Example 5
Picture it
Opposite directions, one easy.
Figure (svg): A factorisation checked by expanding it again
Because the reverse operation is straightforward, a guess costs almost nothing to test. That makes it worth trying the most likely pair immediately rather than working through the list in order.
Worked example
This is Example 5 from the textbook.
\[ \text{Factor } x^2 - 2x - 8 \text{ and check the result.} \]
Read the signs
Why: c negative means opposite signs.
List the factor pairs of eight
Why: One and eight, two and four.
\[ 1, 8 \text{ and } 2, 4 \]
Find the difference of two
Why: Four minus two is two.
\[ 2 \text{ and } -4 \]
Write and check
Why: Expand the answer.
\[ (x + 2) (x - 4) \]
Figure (svg): A factorisation checked by expanding it again
\[ x^2 - 2x - 8 = (x + 2)(x - 4) \]
Verify: expand to confirm
Why: The four products are x squared, negative four x, two x and negative eight, and the middle terms combine to negative two x. Graphing both expressions would show a single curve, which is the calculator check the textbook uses.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 597-597
Faded example
The candidates are finite.
Fill in the blanks
\text2 c = 8: \quad \text4 1, 8 \text___ ___, ___
Why: Eight has just two factor pairs, so at most two sums need checking for each sign assignment. Knowing the list is short is what makes the method feel like a search rather than a gamble.
Worked example
The Study Tip's point, made explicit.
\[ \text{How many pairs must be tested to factor } x^2 + 17x - 18? \]
List all factor pairs of eighteen
Why: Three of them.
\[ 1, 18; \; 2, 9; \; 3, 6 \]
Note the sign options
Why: Each pair has two sign assignments.
Use the size argument
Why: Seventeen is large, so try the widest pair.
\[ 1 \text{ and } 18 \]
Test it
Why: It works immediately.
Figure (svg): The systematic search for a pair of numbers
\[ (x - 1)(x + 18) \]
Verify: say what would have happened working top-down
Why: Even the exhaustive route needs at most six checks, each of which is a single addition. The search is always short; thinking about which end to start from just makes a short search shorter.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 597-597
Trap
Try random pairs of numbers until something looks right.
Guess and see
Why: The answer is a pair of numbers, so guessing seems reasonable.
Without listing the factor pairs there is no way to know when to stop or whether a trinomial simply does not factor. A guess that fails tells you nothing about what to try next.
List the factor pairs of c, assign signs from the rules, then check sums.
Make the search finite and ordered
Why: There are only a handful of candidates.
An exhausted list also proves a trinomial does not factor, which guessing never can.
Sorting
Look for a pair with the right sum and product.
Sort into buckets
Sort each trinomial by whether it factors over the integers.
The three that fail have discriminants that are not perfect squares, which is Lesson 9.7's test applied to the same question. Exhausting the short list is what proves the failure rather than merely suggesting it.
Elimination
You have tried several pairs without success.
Eliminate the wrong options
What actually establishes that no integer factorisation exists?
Survives elimination: A
Why: The list of candidates is finite and short, so exhausting it is a genuine proof rather than a strong suspicion. Computing the discriminant is a faster route to the same conclusion, and it agrees with the exhaustive search every time.
Socratic
It would not be in most of algebra.
Discussion prompt
Explain why trying a likely pair is a sound strategy for factoring. Then say what makes it different from guessing at an answer in general.
Hint: How expensive is a wrong guess?
Answer:
A candidate can be tested in one line by expanding, so a wrong guess costs almost nothing and a right one finishes the problem. The reverse operation being easy is what makes the forward search worth attempting boldly rather than cautiously.
It differs from ordinary guessing because the candidate list is finite, generated systematically, and verifiable. Guessing at an answer with no way to check and no way to enumerate possibilities leaves you no better off when it fails; here a failed test eliminates a candidate and the list is short enough to exhaust.
Section
Section 5
Concept
Factoring turns a quadratic equation into the factored form of Lesson 10.4. Once the equation is written as a product equal to nought, each factor gives a solution.
This is an exact alternative to the quadratic formula.
Figure (svg): A garden with a border of uniform width around it
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 595-602 — the lesson opener on the width of a garden border
Picture it
Each dimension gains twice x.
Figure (svg): A garden with a border of uniform width around it
The border runs along both sides of each dimension, so the total is the garden plus two widths. Adding only one is the commonest setup error in border problems.
Worked example
Combining this lesson with Lesson 10.4.
\[ \text{Solve } x^2 - 2x - 8 = 0. \]
Confirm the zero
Why: Already in standard form.
\[ x ^{2} - 2 x - 8 = 0 \]
Factor
Why: Two and negative four.
\[ (x + 2) (x - 4) = 0 \]
Set each factor to nought
Why: Two equations.
\[ x + 2 = 0 \text{ or } x - 4 = 0 \]
Solve
Why: Flip the signs.
\[ x = -2, \; 4 \]
Figure (svg): A factorisation checked by expanding it again
\[ x = -2 \text{ and } x = 4 \]
Verify: compare with the quadratic formula
Why: The discriminant is four plus thirty-two, which is thirty-six, so the formula gives two plus or minus six, over two — that is four and negative two. The two methods agree, and factoring was quicker because the discriminant happened to be a perfect square.
Faded example
Twice the width, each way.
Fill in the blanks
\text10 x: \quad (6 + 2x)(___ + 2x)
Why: Both dimensions gain two x because the border appears on both sides. The coefficient of two is what distinguishes a border problem from a simple extension on one edge.
Worked example
The lesson opener's situation, with dimensions stated.
\[ \text{A } 10 \text{ by } 6 \text{ garden gets a uniform border. If the total area is } 96, \text{ how wide is it?} \]
Write the dimensions
Why: Each gains twice the width.
\[ (10 + 2 x) (6 + 2 x) \]
Set up the equation
Why: The total area is ninety-six.
\[ 4 x ^{2} + 32 x + 60 = 96 \]
Simplify to standard form
Why: Divide by four after subtracting.
\[ x ^{2} + 8 x - 9 = 0 \]
Factor and solve
Why: Nine and negative one.
\[ x = 1 \text{ or } x = -9 \]
Figure (svg): A garden with a border of uniform width around it
\[ x = 1 \]
Verify: check the answer in the picture
Why: A border one unit wide makes the whole region twelve by eight, an area of ninety-six — exactly as required. The solution of negative nine is discarded because a border cannot have a negative width, which is the same judgement made about times in Chapter 9.
Trap
\[ (10 + x)(6 + x) = 96 \]
Add the border width to each dimension
Why: The border adds x, so x was added.
The border runs along both ends of each dimension, so each gains two widths rather than one. This version gives x squared plus sixteen x minus thirty-six equals nought, whose positive solution is about two — and a border of that width would make the region fourteen by ten, an area of a hundred and forty.
\[ (10 + 2x)(6 + 2x) = 96 \]
Add twice the width to each dimension
Why: One width on each side.
Sketching the picture and marking both border strips makes the doubling impossible to forget.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Factoring | The quadratic formula | |
|---|---|---|
| Works when | the discriminant is a perfect square | always |
| Speed | faster when it works | steady but longer |
| Tells you if there is no solution | only by exhausting the list | immediately, from the discriminant |
Factoring is the faster method on exactly the problems textbooks tend to set, and the formula is the one that never fails. Trying to factor briefly and falling back to the formula is the usual sensible order.
Elimination
A border problem gives x equal to 1 or -9.
Eliminate the wrong options
Which should be reported?
Survives elimination: A
Why: The variable stands for a width, which must be positive, so the negative solution is outside the situation's domain. Saying why it was rejected is part of the answer rather than an optional remark.
Socratic
The formula always works.
Discussion prompt
Say when you would factor rather than use the quadratic formula, and when you would not bother trying. Then say how you would decide quickly.
Hint: Think about the discriminant.
Answer:
Factor when the coefficients are small integers and the constant has few factor pairs, since the search is then over in seconds and gives exact answers with no arithmetic. Do not bother when the numbers are large or awkward, or when a brief look at the factor pairs finds nothing promising.
The quickest decision is to compute the discriminant first: if it is a perfect square the trinomial factors over the integers, and if it is not, no amount of searching will help. That takes one subtraction and turns the choice between methods into a fact rather than a hunch — which is one more use for Lesson 9.7.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Signs of b and c | The two numbers | Example |
|---|---|---|
| b positive, c positive | both positive | x^2 + 6x + 8 = (x+2)(x+4) |
| b negative, c positive | both negative | x^2 - 5x + 6 = (x-2)(x-3) |
| c negative | opposite signs | the larger takes the sign of b |
The constant decides same or different, and the middle coefficient decides which sign dominates. Two questions in that order settle every case in the table.
Pattern
To factor any trinomial whose leading coefficient is one, these five moves cover it.
Steps two and three cost nothing and eliminate half the candidates, which is what turns a hunt into a short, ordered search.
OpenStax Elementary Algebra 2e, §7.2 Factor Trinomials of the Form x2+bx+c §7.2
Check
Both conditions.
Check your understanding
Factor x squared + 6x + 8.
Answer: A
Why: Two and four multiply to eight and add to six, satisfying both conditions at once.
Check
Read the sign of c first.
Check your understanding
Factor x squared - 5x + 6.
Answer: A
Why: A positive constant needs matching signs, and a negative middle term makes both negative. Negative two and negative three add to negative five and multiply to six.
Check
Mixed signs, larger follows b.
Check your understanding
Factor x squared - 11x - 12.
Answer: A
Why: A negative constant needs opposite signs, and the larger number takes the sign of b, which is negative. One and negative twelve add to negative eleven.
Real world
This is the garden question from the lesson opener. A rectangular garden is to be surrounded by a border of uniform width, and the total area is fixed by how much paving is available.
Discussion prompt
A garden measures 10 by 6 and the finished region including the border must have area 96. Write an equation for the border's width, solve it by factoring, and say which solution you kept.
Hint: Each dimension gains twice the width.
Answer:
\[ (10 + 2x)(6 + 2x) = 96 \;\to\; 4x^2 + 32x + 60 = 96 \;\to\; x^2 + 8x - 9 = 0 \]
Dividing through by four before factoring keeps the numbers small, and nine and negative one give a sum of eight and a product of negative nine, so the factors are x plus nine and x minus one.
The solutions are one and negative nine, and only the first is a possible width. Checking it against the picture confirms the setup: a border of width one makes the region twelve by eight, an area of exactly ninety-six. Rejecting the negative solution is a modelling judgement, and dividing by four first was the step that made the factoring easy.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Factor x squared + 17x - 18.
Correct: (x - 1)(x + 18).
\[ (x - 1)(x + 18) = x^2 + 18x - x - 18 = x^2 + 17x - 18 \]
Why: The constant is negative, so the two numbers have opposite signs, and the middle coefficient is positive, so the larger of the two carries the plus. Negative one and eighteen add to seventeen and multiply to negative eighteen, which is exactly what is needed. The first option has the signs the right way round in size but swapped in assignment, giving a middle term of negative seventeen x — a single sign swap away and completely wrong. The last two options use factor pairs that are too close together: a mixed pair's sum is really a difference, so a target of seventeen demands the widest available pair rather than a middling one, which is why one and eighteen was the sensible first guess.
Explain it
They keep trying random pairs of numbers and losing track of what they have tried.
Discussion prompt
In no more than four sentences, give them a systematic method. Then tell them the two questions that halve the work before any numbers are chosen.
Hint: The candidates are the factor pairs of the constant.
Answer:
A usable answer: list the factor pairs of the constant term, since those are the only candidates, then check which pair adds to the middle coefficient. There are rarely more than three or four pairs, so the list is short and you know when you have finished.
Before any of that, ask two questions. Is the constant positive or negative — that decides whether the two numbers share a sign or differ — and is the middle coefficient positive or negative, which decides which sign dominates.
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 asking about c first and b second, in that order, every time. A large constant is fixed by listing its factor pairs and using the size of b to decide which end of the list to start from. Checking both conditions is fixed by multiplying the answer back, which takes one line. Border problems are fixed by sketching the picture and adding twice the width to each dimension. 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 x plus p times x plus q in full, and circle the two places where p and q appear in the answer, writing beside each what condition it imposes. Underneath, build the four-row sign table from the signs of b and c, and write one fully worked trinomial beside each row so that all four cases appear. In the middle, factor a trinomial by writing out the factor pairs of its constant in a column with their sums beside them, marking the pair that works and striking through the ones you did not need to test. Beneath that, draw the area model for one of your factorisations, label all four regions, and mark which two rectangles produce the middle term. In the lower half, take a border problem: sketch a rectangle inside a rectangle, mark the border width on both sides of each dimension, write the area equation, simplify it by dividing out any common factor, factor it, and say which solution the situation allows. Finally, in the margin, write the one-line test for whether a trinomial factors over the integers at all.
Every factorisation on your page should be checked by multiplying back. If any check disagrees, the fix is usually a swapped pair of signs rather than a wrong pair of numbers, so try that before starting the search again.
Recap
Five things, and the second and third are the same question asked about different signs.
| If the question says | Your first move is |
|---|---|
| Factor a trinomial | Read b and c, then decide the signs |
| The constant is positive | Both numbers share a sign |
| The constant is negative | Opposite signs; the larger follows b |
| Solve by factoring | Get a zero on one side first |
| Check your factorisation | Multiply the two binomials back |
Lesson 10.6 removes the restriction that the leading coefficient is one. With a coefficient in front of the squared term the search gets longer, because the first terms of the binomials have to be chosen as well as the last ones.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.5 Factoring x^2 + bx + c §10.5, pp. 595-602 — everything on these slides traces back here
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