Factoring a trinomial whose leading coefficient is not one. Includes the four numbers a factorisation must supply, testing trial factors by their Outer and Inner products, using the sign rules to narrow the search, removing a common factor before factoring, solving quadratic equations by factoring, and applying the method to a vertical motion model.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 10 — Polynomials and Factoring
Factoring ax^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 10.6 Factoring ax^2 + bx + c §10.6, pp. 603-608 — the lesson these objectives are drawn from
Warm-up
Lesson 10.5 factored trinomials whose squared term had a coefficient of one. Removing that restriction adds one choice to the search.
Discussion prompt
Expand the product of two x plus one and x plus five. Which parts of the answer did the two leading coefficients affect?
Hint: Look at the first term and the middle term.
Answer:
\[ (2x + 1)(x + 5) = 2x^2 + 10x + x + 5 = 2x^2 + 11x + 5 \]
The two leading coefficients multiplied to give the two in front of x squared, and they also entered the middle term through the Outer product. So changing them changes two of the three coefficients, which is why they now have to be chosen rather than assumed.
Concept
To factor a x squared plus b x plus c, find numbers m and n whose product is a and numbers p and q whose product is c, arranged so that the Outer and Inner products of FOIL add to b.
trial factors — Candidate pairs of binomials formed from factors of a and factors of c. Each is tested by checking whether its Outer and Inner products add to the required middle term.
Lesson 10.5 was the case where m and n are both one.
Figure (svg): The four numbers a factorisation has to supply
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 603-603
Section
Section 1
Concept
The leading coefficients of the two binomials must multiply to a, their constants must multiply to c, and the Outer and Inner products must add to b. All three must hold at once.
\[ ax^2 + bx + c = (mx + p)(nx + q) \]
Only the third condition depends on how the numbers are arranged.
Figure (svg): The four numbers a factorisation has to supply
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 603-603 — the statement of the m, n, p, q conditions and Example 1
Picture it
Two on products, one on a sum.
Figure (svg): The four numbers a factorisation has to supply
The first two conditions choose the numbers and the third chooses their positions. That separation is what makes a table of trials the natural way to organise the work.
Worked example
This is Example 1 from the textbook.
\[ \text{Factor } 2x^2 + 11x + 5. \]
List factors of a
Why: Two is prime.
\[ 1 \text{ and } 2 \]
List factors of c
Why: Five is prime.
\[ 1 \text{ and } 5 \]
Try one arrangement
Why: Outer plus Inner.
\[ (x+1)(2x+5) \to 7x \]
Try the other
Why: Swap the constants.
\[ (2x+1)(x+5) \to 11x \]
Figure (svg): A table of trial factors with their middle terms
\[ 2x^2 + 11x + 5 = (2x + 1)(x + 5) \]
Verify: expand the answer
Why: The four products are two x squared, ten x, x and five, and the middle terms combine to eleven x. Both trials used the same four numbers, so only the arrangement distinguished them.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 603-603
Faded example
Two products and a sum.
Fill in the blanks
The leading coefficients multiply to a, the constants multiply to c, and the Outer and Inner products add to b.
Why: Two conditions choose which numbers to use and the third checks whether they are in the right places. Only the third can fail once the numbers are chosen correctly.
Worked example
Guided Practice 1 to 3.
\[ \text{Factor } 2x^2 + 7x + 3, \; 2x^2 + 5x + 3 \text{ and } 3x^2 + 10x + 3. \]
Take the first
Why: The three pairs with the two x.
\[ (2 x + 1) (x + 3) \]
Take the second
Why: The other arrangement of the same numbers.
\[ (2 x + 3) (x + 1) \]
Take the third
Why: Factors of three on both sides.
\[ (3 x + 1) (x + 3) \]
Compare the first two
Why: Same numbers, different middle terms.
\[ 7x \text{ against } 5x \]
Figure (svg): A table of trial factors with their middle terms
\[ (2x+1)(x+3), \; (2x+3)(x+1), \; (3x+1)(x+3) \]
Verify: check the first two against each other
Why: The first gives six x plus x, which is seven x, and the second gives two x plus three x, which is five x. The four numbers were identical in both, and swapping which constant sits with which leading coefficient changed the middle term — which is the whole reason arrangements must be tested.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 603-603
Trap
\[ 2x^2 + 11x + 5 = (x + 1)(2x + 5) \]
Use factors one and two for a, and one and five for c
Why: All the products come out right.
The leading coefficients multiply to two and the constants to five, so two of the three conditions hold. But the middle term is five x plus two x, which is seven x rather than eleven x, so the arrangement is wrong.
\[ 2x^2 + 11x + 5 = (2x + 1)(x + 5) \]
Test the middle term for each arrangement, not just the products
Why: The third condition is the one that discriminates.
With a leading coefficient of one there was only one arrangement, which is why this issue is new.
Elimination
Factoring 2x squared + 11x + 5.
Eliminate the wrong options
Which product gives a middle term of 11x?
Survives elimination: A
Why: Three of these use the right numbers and only one puts them in the right places. Option D fails an earlier condition, which is why it can be discarded without computing a middle term at all.
Matching
Watch which constant pairs with which coefficient.
Match the pairs
Why: The second and third use identical numbers in swapped positions and produce different trinomials. That pair is worth studying, because it shows exactly what the arrangement contributes.
Socratic
Lesson 10.5 had only one.
Discussion prompt
Explain why a leading coefficient other than one creates several trials from the same set of numbers. Then say how many trials a given pair of pairs produces.
Hint: Ask which constant goes with which leading coefficient.
Answer:
With a leading coefficient of one, both binomials start with x, so the two constants are interchangeable and give the same product either way. Once the leading coefficients differ, each constant can be paired with either of them, and the Outer and Inner products change when they swap.
A pair of factors for a and a pair for c give two distinct arrangements, since swapping both pairs at once just rewrites the same product in the other order. So each combination of factor pairs costs two tests rather than one, which is why the tables in this lesson are longer than any in the last.
Section
Section 2
Concept
The sign rules from Lesson 10.5 still apply and eliminate half the candidates. When the constant is positive and the middle coefficient negative, only negative constants need be tried at all.
There is no need to finish the table once a trial works.
Figure (svg): Sign rules cutting down the number of trials
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 604-604 — Example 2 and its Study Tip on stopping once the correct factors are found
Picture it
Asked before any trials.
Figure (svg): Sign rules cutting down the number of trials
With four numbers to place, the sign rules save more than they did before. Skipping them means testing arrangements that could have been ruled out for free.
Worked example
This is Example 2 from the textbook.
\[ \text{Factor } 6x^2 - 19x + 15. \]
Fix the signs
Why: c positive and b negative.
List the factor pairs
Why: For six and for fifteen.
\[ 1,6 \text{ or } 2,3; \; 1,15 \text{ or } 3,5 \]
Test two arrangements
Why: Both give the wrong middle term.
\[ -21 x, \; - 91 x \]
Test the middle pairs
Why: Two and three with three and five.
\[ (2x-3)(3x-5) \to -19x \]
Figure (svg): Sign rules cutting down the number of trials
\[ 6x^2 - 19x + 15 = (2x - 3)(3x - 5) \]
Verify: expand to confirm
Why: The products are six x squared, negative ten x, negative nine x and fifteen, and the middle terms combine to negative nineteen x. Both constants had to be negative for the fifteen to come out positive.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 604-604
Sorting
Read c, then b.
Sort into buckets
Sort each trinomial by the signs its two constants need.
Every one of these was decided from two signs, with no numbers involved. Doing this first is what stops the trial table from doubling in length.
Worked example
Choosing which trial to attempt first.
\[ \text{Why is } (x - 15)(6x - 1) \text{ a poor first guess for } 6x^2 - 19x + 15? \]
Compute its middle term
Why: Negative x and negative ninety x.
\[ -91 x \]
Compare with the target
Why: Nineteen is much smaller.
Explain the size
Why: Extreme pairings give extreme middle terms.
\[ 1 \text{ with } 15 \]
Choose better
Why: Pair middling factors together.
\[ 2,3 \text{ with } 3,5 \]
Figure (svg): Sign rules cutting down the number of trials
\[ -91x \text{ against a target of } -19x \]
Verify: check the pattern on the other trials
Why: The trial pairing one with one gave negative twenty-one, close to the target, and the extreme pairing gave negative ninety-one. Middle terms grow when a large factor of a meets a large factor of c, so a modest target points at balanced pairings.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 604-604
Error analysis
The student factored a trinomial with a positive constant and a negative middle term.
Annotate
On: \( \begin{aligned} 6x^2 - 19x + 15 &= (2x + 3)(3x - 5) \\ \text{check: } (2x+3)(3x-5) &= 6x^2 - 10x + 9x - 15 \\ &= 6x^2 - x - 15 \end{aligned} \)
The check was done and then ignored, which is the expensive part. A mismatch in the constant term points straight at the signs rather than at the numbers, so the repair here was two sign changes rather than a fresh search.
Faded example
Only two products are needed.
Fill in the blanks
(2x - 3)(3x - 5): \quad \text-10x = -19x, \; \text___ = -9x, \; \text___ = ___
Why: Testing a trial needs only the Outer and Inner products, since the first and last terms are guaranteed by how the numbers were chosen. That makes each test two multiplications and one addition.
Prediction
Pairing a large factor of a with a large factor of c.
Predict first
What happens to the size of the middle term?
Correct: It grows, because one of the two products becomes large.
\[ (x - 15)(6x - 1) \to -91x \qquad (2x - 3)(3x - 5) \to -19x \]
Why: The Outer and Inner products are formed by multiplying across the brackets, so pairing the extremes puts a big factor of a with a big factor of c in one of them. For six x squared minus nineteen x plus fifteen the extreme pairing gave ninety-one while a balanced one gave nineteen. A modest middle coefficient therefore points at balanced pairings, which is a useful way to choose which trial to attempt first.
Socratic
The Study Tip says to.
Discussion prompt
Explain why there is no need to check the remaining trials once one works. Then say what would change if you were asked to show that a trinomial does not factor.
Hint: Can two different factorisations give the same trinomial?
Answer:
A trinomial that factors over the integers does so in essentially one way, so a trial that produces the right middle term is the answer and no other arrangement can also produce it. Continuing would confirm what is already established and cost time for nothing.
Showing that a trinomial does not factor is the opposite situation: every candidate must be tested, because a single untested arrangement could be the one that works. That is why the sign rules matter so much there — they legitimately remove half the table rather than merely postponing it.
Section
Section 3
Concept
If the three coefficients have a common factor, remove it before factoring the trinomial. The remaining trinomial has smaller coefficients and a much shorter list of trials.
Leaving it out at the end is the one risk.
Figure (svg): A common factor removed before the trinomial is factored
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 604-604 — Example 3, Factor with a Common Factor for a, b, and c
Picture it
The common factor comes out first.
Figure (svg): A common factor removed before the trinomial is factored
Six and four have several factor pairs each; three and two have almost none. Removing the two turned a long table into a short one.
Worked example
This is Example 3 from the textbook.
\[ \text{Factor } 6x^2 + 2x - 4. \]
Spot the common factor
Why: All three coefficients are even.
\[ 2 \]
Factor it out
Why: Divide each term by two.
\[ 2(3 x ^{2} + x - 2) \]
Factor the trinomial
Why: c negative, so opposite signs.
\[ (x + 1) (3 x - 2) \]
Include the common factor
Why: Do not lose the two.
\[ 2(x + 1) (3 x - 2) \]
Figure (svg): A common factor removed before the trinomial is factored
\[ 6x^2 + 2x - 4 = 2(x + 1)(3x - 2) \]
Verify: expand the whole answer
Why: The brackets give three x squared plus x minus two, and multiplying by two gives six x squared plus two x minus four. The common factor has to be included in the expansion for the check to work, which is a reminder to include it in the answer.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 604-604
Faded example
All three coefficients share it.
Fill in the blanks
6x^2 + 2x - 4 = 2(3x^2 + x - 2)
Why: Every coefficient is divided by the common factor, including the constant. Dividing only some of them would change the polynomial rather than rewrite it.
Worked example
Guided Practice 8 and 9.
\[ \text{Factor } 6x^2 + 14x + 4 \text{ and } 20x^2 + 5x - 15. \]
Take the first
Why: All three are even.
\[ 2(3 x ^{2} + 7 x + 2) \]
Factor what remains
Why: One and two with one and three.
\[ 2(3 x + 1) (x + 2) \]
Take the second
Why: All three are divisible by five.
\[ 5(4 x ^{2} + x - 3) \]
Factor what remains
Why: Opposite signs, since c is negative.
\[ 5(4 x - 3) (x + 1) \]
Figure (svg): A common factor removed before the trinomial is factored
\[ 2(3x + 1)(x + 2), \qquad 5(4x - 3)(x + 1) \]
Verify: check the second by expanding
Why: Four x minus three times x plus one is four x squared plus four x minus three x minus three, which is four x squared plus x minus three, and five times that is twenty x squared plus five x minus fifteen. Without removing the five first, twenty and fifteen would have given a considerably longer table.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 604-604
Trap
\[ 6x^2 + 2x - 4 = (x + 1)(3x - 2) \]
Factor the trinomial inside the brackets and report that
Why: The hard part was the trinomial, so that felt like the answer.
Expanding this gives three x squared plus x minus two, which is half the original. The two was divided out and never restored, so the answer is a different polynomial.
\[ 6x^2 + 2x - 4 = 2(x + 1)(3x - 2) \]
Write the common factor down immediately and keep it in front
Why: It is part of the factorisation.
Expanding the complete answer is the check that catches this, which is one more reason to do it.
Sorting
Check all three coefficients.
Sort into buckets
Sort each trinomial by whether its coefficients share a factor greater than one.
In the last one the four and eight share a factor but the three does not, which is why nothing can come out. All three coefficients must share the factor, not merely two of them.
Elimination
Factoring 6x squared + 2x - 4.
Eliminate the wrong options
Which is the full factorisation?
Survives elimination: A
Why: The common factor belongs in front and each bracket should have nothing left to remove. Option D is worth noticing: it is a correct expansion but not a complete factorisation, which is a distinction that matters when the answer is used further.
Socratic
It could be done at the end instead.
Discussion prompt
Give two reasons for taking out a common factor before starting the trials. Then say what happens to the number of trials.
Hint: Compare the factor pairs of the old and new coefficients.
Answer:
First, the coefficients shrink, so each has fewer factor pairs and the trial table gets much shorter. Second, the arithmetic in each trial is easier, since the products being compared are smaller numbers.
For six x squared plus two x minus four, the coefficients six and four have several pairs between them; after removing the two, three and two are both prime and have one pair each. The table drops from several rows to two, which is a large saving for a step that costs one line.
Section
Section 4
Concept
To solve a quadratic equation by factoring, move everything to one side, factor the result, and set each factor equal to nought. A leading coefficient other than one usually gives fractional solutions.
The last step is Lesson 10.4, unchanged.
Figure (svg): An equation rearranged, factored and solved
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 605-605 — Example 4, Solve a Quadratic Equation
Picture it
Only the middle line is new.
Figure (svg): An equation rearranged, factored and solved
Each factor with a coefficient produces a fraction, which is normal rather than a sign of error. Whole-number solutions are the exception once the leading coefficient is not one.
Worked example
This is Example 4 from the textbook.
\[ \text{Solve } 21n^2 - 14n - 7 = 6n - 11. \]
Write in standard form
Why: Move the right side across.
\[ 21 n ^{2} - 20 n + 4 = 0 \]
Factor
Why: Both constants negative.
\[ (3 n - 2) (7 n - 2) = 0 \]
Set each factor to nought
Why: Two equations.
\[ 3n - 2 = 0 \text{ or } 7n - 2 = 0 \]
Solve
Why: Divide out each coefficient.
\[ n = \tfrac{2}{3}, \; \tfrac{2}{7} \]
Figure (svg): An equation rearranged, factored and solved
\[ n = \tfrac{2}{3} \text{ and } n = \tfrac{2}{7} \]
Verify: expand the factored form
Why: Three n minus two times seven n minus two is twenty-one n squared minus six n minus fourteen n plus four, which is twenty-one n squared minus twenty n plus four. That matches the standard form, so both the rearrangement and the factoring are confirmed.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 605-605
Faded example
Two steps, not one.
Fill in the blanks
3n - 2 = 0 \;\to\; 3n = 2 \;\to\; n = 2/3
Why: The addition and the division are separate steps, and skipping the second gives two rather than two thirds. A factor with a coefficient almost never yields a whole-number solution.
Worked example
Guided Practice 10 to 12.
\[ \text{Solve } 2x^2 + 7x + 3 = 0, \; 2x^2 + x - 3 = 0 \text{ and } 4x^2 - 16x + 15 = 0. \]
Take the first
Why: Factor, then split.
\[ (2 x + 1) (x + 3) = 0 \]
Solve it
Why: One fraction, one integer.
\[ x = -\tfrac{1}{2}, \; -3 \]
Take the second
Why: Opposite signs, since c is negative.
\[ (2 x + 3) (x - 1) = 0 \]
Take the third
Why: Both constants negative.
\[ (2 x - 3) (2 x - 5) = 0 \]
Figure (svg): An equation rearranged, factored and solved
\[ -\tfrac{1}{2}, -3; \quad -\tfrac{3}{2}, 1; \quad \tfrac{3}{2}, \tfrac{5}{2} \]
Verify: check one solution in its equation
Why: For the third, substituting three halves gives four times nine quarters, which is nine, minus twenty-four, plus fifteen — that is nought. Fractional solutions check just as cleanly as whole ones, provided the squaring is done before the multiplication.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 605-605
Trap
\[ (3n - 2)(7n - 2) = 0 \;\Longrightarrow\; n = 2 \text{ or } n = 2 \]
Take the constant from each bracket
Why: Both brackets contain a two.
Each factor gives an equation to solve, not a number to copy. Three n equals two gives two thirds, and seven n equals two gives two sevenths — the coefficients have to be divided out.
\[ 3n = 2 \;\Longrightarrow\; n = \tfrac{2}{3}; \qquad 7n = 2 \;\Longrightarrow\; n = \tfrac{2}{7} \]
Write and solve each factor's equation on its own line
Why: Two steps each when a coefficient is present.
The two brackets contained the same constant and gave different solutions, which alone shows the coefficient matters.
Translation
One equation per factor.
Match the pairs
Why: Every factor carrying a coefficient produced a fraction and every factor without one produced an integer. That correspondence is worth expecting rather than being surprised by.
Hypothesis
With a leading coefficient other than one.
Predict first
What makes fractional solutions the normal case here?
Correct: Each factor's coefficient has to be divided out at the last step.
\[ (3n - 2)(7n - 2) = 0 \;\Longrightarrow\; n = \tfrac{2}{3}, \tfrac{2}{7} \]
Why: A factor like three n minus two gives n equal to two over three, and the denominator is exactly the leading coefficient of that factor. Since the two leading coefficients multiply to a, a leading coefficient other than one guarantees at least one factor carries a coefficient, and so at least one solution is a fraction unless the numerator happens to divide evenly. Fractions here are a consequence of the structure rather than a symptom of a mistake.
Socratic
Both give exact answers.
Discussion prompt
Say when factoring beats the quadratic formula for an equation like this one, and when it does not. Then describe a quick way to decide.
Hint: Think about the discriminant and the size of the coefficients.
Answer:
Factoring wins when the coefficients are small and the discriminant is a perfect square, since the trial table is then short and the answers come out without any radical arithmetic. It loses when a and c have many factor pairs, because the table grows quickly, and it fails outright when the discriminant is not a perfect square.
Computing the discriminant first settles it in one subtraction: a perfect square means the trinomial factors over the integers and the search will succeed, and anything else means it will not. For twenty-one n squared minus twenty n plus four the discriminant is four hundred minus three hundred and thirty-six, which is sixty-four — a perfect square, so factoring was worth attempting.
Section
Section 5
Concept
A vertical motion model set equal to nought is a quadratic equation. Removing a common factor first usually makes it factorable, and the positive solution is the physical answer.
Thrown upwards makes the velocity positive.
Figure (svg): A diver's height above the water against time
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 605-605 — Example 5, Write a Quadratic Model, on a cliff diver
Picture it
Up briefly, then down.
Figure (svg): A diver's height above the water against time
The upward velocity is small compared with the drop, so the rise is barely visible. It still matters: without it the fall would take slightly less time.
Worked example
This is Example 5 from the textbook.
\[ \text{From a } 48 \text{ foot ledge with upward velocity } 8 \text{ ft/s, when does the diver reach the water?} \]
Write the model
Why: Upwards makes v positive.
\[ h = -16 t ^{2} + 8 t + 48 \]
Set the height to nought
Why: Entering the water.
\[ -16 t ^{2} + 8 t + 48 = 0 \]
Divide out the common factor
Why: Every coefficient is a multiple of eight.
\[ 2 t ^{2} - t - 6 = 0 \]
Factor and solve
Why: Keep the positive time.
\[ (2 t + 3) (t - 2) = 0 \]
Figure (svg): A diver's height above the water against time
\[ t = 2 \text{ seconds} \]
Verify: substitute the time back
Why: Sixteen times four is sixty-four, and eight times two is sixteen, so the height is negative sixty-four plus sixteen plus forty-eight, which is nought. The other solution, negative three halves, is a time before the jump and is discarded.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 605-605
Faded example
Divide through by negative eight.
Fill in the blanks
-16t^2 + 8t + 48 = 0 \;\to\; 2t^2 - t - 6 = 0 \;\to\; (2t + 3)(t - 2) = 0
Why: Dividing by negative eight flips all three signs and shrinks all three coefficients, which turns a long trial table into a short one. The solutions are unaffected, since dividing an equation by a non-zero number preserves them.
Worked example
Comparing the two routes.
\[ \text{How much does dividing by } -8 \text{ shorten the factoring of } -16t^2 + 8t + 48 = 0? \]
Look at the original coefficients
Why: Sixteen and forty-eight.
Divide through by negative eight
Why: Signs flip as well.
\[ 2 t ^{2} - t - 6 = 0 \]
Look at the new coefficients
Why: Two and six.
Factor the simplified version
Why: One short table.
\[ (2 t + 3) (t - 2) \]
Figure (svg): A common factor removed before the trinomial is factored
\[ -16t^2 + 8t + 48 = -8(2t^2 - t - 6) \]
Verify: confirm the division was exact
Why: Negative eight times two t squared is negative sixteen t squared, times negative t is positive eight t, and times negative six is positive forty-eight — all three match. Dividing an equation by a non-zero number never changes its solutions, which is why this simplification is free.
Trap
\[ -16t^2 + 8t + 48 = 0 \;\Longrightarrow\; 2t^2 + t + 6 = 0 \]
Divide every coefficient by negative eight but keep the signs
Why: Only the sizes seemed to be changing.
Dividing by a negative flips every sign, so the middle term becomes negative t and the constant negative six. This version has a negative discriminant and appears to have no solutions at all.
\[ 2t^2 - t - 6 = 0 \]
Apply the sign change to every term when dividing by a negative
Why: All three, not some.
Multiplying back to check takes one line and catches this immediately.
Elimination
The factored equation gives t = -3/2 or t = 2.
Eliminate the wrong options
Which time should be reported?
Survives elimination: A
Why: Time is measured from the jump, so only a positive value describes something that happens. Saying explicitly why the negative root was rejected is part of a complete answer.
Prediction
The same ledge, but with no upward velocity.
Predict first
How would the entry time change?
Correct: It would be slightly less than 2 seconds.
\[ -16t^2 + 48 = 0 \;\Longrightarrow\; t = \sqrt{3} \approx 1.73 \]
Why: Without the upward push the model becomes negative sixteen t squared plus forty-eight, giving t squared equal to three and a time of about 1.73 seconds. The jump adds a small rise before the fall begins, so it buys about a quarter of a second in the air. That is the same dropped-object model from Lesson 9.2, which is this model with the middle term removed.
Socratic
The quadratic formula would also work.
Discussion prompt
Say what makes the motion model a good candidate for factoring. Then say what would make you switch to the formula.
Hint: Look at the coefficients after simplifying.
Answer:
The coefficients share a large common factor, so dividing through leaves small numbers with few factor pairs — two and six here, each with one or two pairs. A short trial table and a perfect-square discriminant make factoring both quick and exact.
A model with an awkward initial height or velocity would give coefficients with no common factor and a discriminant that is not a perfect square, and then no amount of searching would find integer factors. Computing the discriminant first tells you which situation you are in, and in this case sixty-four plus four hundred and eighty is five hundred and forty-four for the original — but for the simplified equation it is one plus forty-eight, which is forty-nine, a perfect square.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Lesson 10.5 | Lesson 10.6 | |
|---|---|---|
| Numbers to choose | two | four |
| What gets split | the constant only | both a and c |
| Arrangements per choice | one | two |
Every row in the right-hand column is the left-hand one made larger. The method is unchanged; only the size of the search grows.
Pattern
To factor any trinomial with a leading coefficient other than one, these five moves cover it.
Steps one and two are both free and both shorten the table dramatically, which matters far more here than it did with a leading coefficient of one.
OpenStax Elementary Algebra 2e, §7.3 Factor Trinomials of the Form ax2+bx+c §7.3
Check
The arrangement matters.
Check your understanding
Factor 2x squared + 11x + 5.
Answer: A
Why: The Outer and Inner products are ten x and x, adding to eleven x. The other arrangements of the same numbers give seven x.
Check
Common factor first.
Check your understanding
Factor 6x squared + 2x - 4 completely.
Answer: A
Why: All three coefficients are even, so the two comes out first, leaving three x squared plus x minus two to factor.
Check
Divide out the coefficient.
Check your understanding
Solve (3n - 2)(7n - 2) = 0.
Answer: A
Why: Three n equals two gives two thirds, and seven n equals two gives two sevenths. Each factor's coefficient becomes the denominator.
Real world
This is the cliff diver from the lesson opener. A diver leaves a ledge 48 feet above the ocean with an initial upward velocity of 8 feet per second.
Discussion prompt
Write the vertical motion model, find when the diver enters the water, and say how much difference the upward jump made compared with simply stepping off.
Hint: Divide out the common factor before factoring.
Answer:
\[ -16t^2 + 8t + 48 = 0 \;\to\; 2t^2 - t - 6 = 0 \;\to\; (2t + 3)(t - 2) = 0 \]
The solutions are negative three halves and two, and only the positive one is a time after the jump, so the diver enters the water two seconds after leaving the ledge.
Stepping off with no upward velocity gives negative sixteen t squared plus forty-eight equals nought, so t is the root of three, about 1.73 seconds. The small jump therefore bought roughly a quarter of a second of extra airtime — which is exactly what the linear term in the model contributes, and why divers who want time for a somersault jump upwards rather than simply falling.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Factor 6x squared - 19x + 15.
Correct: (2x - 3)(3x - 5).
\[ (2x - 3)(3x - 5) = 6x^2 - 10x - 9x + 15 \]
Why: The constant is positive, so both constants in the brackets must share a sign, and the middle coefficient is negative, so both must be negative — which rules out the first option immediately without computing anything. Among the remaining candidates, the Outer and Inner products decide: two x times negative five is negative ten x and negative three times three x is negative nine x, adding to negative nineteen x as required. The third option gives negative fifteen x minus six x, or negative twenty-one x, and the fourth expands to a middle term of negative thirty-three x and also leaves a common factor of three inside its first bracket. Checking the sign pattern first is what makes this a two-candidate problem rather than a six-candidate one.
Explain it
They found the right four numbers but keep getting the wrong middle term.
Discussion prompt
In no more than four sentences, explain what they are missing. Then tell them the fastest way to test a trial.
Hint: The numbers are right; where are they?
Answer:
A usable answer: with a leading coefficient other than one, the same four numbers can be arranged two ways and the two arrangements give different middle terms. Swapping which constant sits with which leading coefficient changes the Outer and Inner products, so both arrangements have to be tested.
To test one quickly, compute only the Outer and Inner products and add them. The first and last terms are guaranteed by how you chose the numbers, so there is no need to expand the whole thing.
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: Arrangements are fixed by writing a two-column table of trial factors and middle terms rather than working in your head. Common factors are fixed by checking all three coefficients before anything else. Signs are fixed by asking about c first and b second. Factors with coefficients are fixed by writing each as its own two-step equation. 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 the general factorisation with m, n, p and q, and label the three conditions with an arrow to each place they appear in the expanded form. Underneath, factor a trinomial whose leading coefficient is prime by drawing a two-column table of trial factors and middle terms, filling in every arrangement so the difference between them is visible. In the middle, factor one where neither coefficient is prime: write the sign decisions first as two short sentences, then list the factor pairs, then build the table and strike through the trials the sign rules had already excluded. Beneath that, take a trinomial with a common factor, remove it as its own line, factor what remains, and write the complete answer with the factor restored, then expand the whole thing to check. In the lower half, solve a motion model: substitute the velocity and height, set the height to nought, divide out the common factor showing every sign flip, factor, and mark which solution the situation rejects. Finally, in the margin, write how many arrangements a given pair of factor pairs produces and why.
Your trial tables should show the middle terms growing when extreme factors are paired together. If they do not, check whether both the Outer and the Inner product were computed rather than just one.
Recap
Five things, and the first two are the ones that keep the search short.
| If the question says | Your first move is |
|---|---|
| Factor a trinomial with a coefficient | Look for a common factor first |
| The constant is positive | Both constants share a sign |
| Test a trial factorisation | Compute only the Outer and Inner products |
| Solve by factoring | Standard form, then factor, then zero-product |
| A motion model reaches the ground | Set h to nought and divide out the common factor |
Lesson 10.7 returns to the special products of Lesson 10.3 and runs them backwards. A trinomial that is a perfect square, or a binomial that is a difference of two squares, can be factored on sight with no trial table at all.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.6 Factoring ax^2 + bx + c §10.6, pp. 603-608 — everything on these slides traces back here
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