Factoring polynomials of degree three. Includes finding a greatest common factor that includes variables, prime polynomials and what it means to factor completely, factoring a four-term polynomial by grouping, the sum and difference of two cubes patterns, and reading the dimensions of a box from a factored volume.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 10 — Polynomials and Factoring
Factoring Cubic Polynomials
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 616-622 — the lesson these objectives are drawn from
Warm-up
The last four lessons factored quadratics. Cubics are handled by removing a factor until a quadratic is left, then using what you already know.
Discussion prompt
Factor nine x squared plus fifteen. Then say what would be different if the expression were nine x cubed plus fifteen x squared.
Hint: Look for what both terms share.
Answer:
\[ 9x^2 + 15 = 3(3x^2 + 5) \]
The first has a common factor of three. The second also has a common factor of x squared, since both terms contain at least two x's — so three x squared comes out, leaving three x plus five. Variables can be part of a common factor, and that is what makes cubics manageable.
Concept
When factoring a cubic polynomial, factor out the greatest common factor first, including any variable factors, and then look for the patterns of the earlier lessons in what remains.
prime polynomial — A polynomial that cannot be factored using integer coefficients. To factor a polynomial completely is to write it as a product of monomial and prime factors.
The greatest common factor may contain variables as well as numbers.
Figure (svg): The greatest common factor found from prime factorisations
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 616-617
Section
Section 1
Concept
The greatest common factor of two terms includes every prime factor they share, and every variable factor they share. Writing each term out as a product makes both visible.
\[ 14x^3 + 21x^2 = 7x^2(2x + 3) \]
The variable part takes the smaller of the two exponents.
Figure (svg): The greatest common factor found from prime factorisations
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 616-616 — Example 1, Find the Greatest Common Factor
Picture it
Shared factors are then visible.
Figure (svg): The greatest common factor found from prime factorisations
The two terms share one seven and two x's, so the factor is seven x squared. Writing them out is slower than guessing and considerably more reliable.
Worked example
This is Example 1 from the textbook.
\[ \text{Factor the greatest common factor out of } 14x^3 + 21x^2. \]
Write the first term out
Why: As a product of primes and variables.
\[ 2 \cdot 7 \cdot x \cdot x \cdot x \]
Write the second term out
Why: The same way.
\[ 3 \cdot 7 \cdot x \cdot x \]
Take what is shared
Why: One seven and two x's.
\[ 7 x ^{2} \]
Divide it out
Why: What is left in each term.
\[ 7 x ^{2}(2 x + 3) \]
Figure (svg): The greatest common factor found from prime factorisations
\[ 14x^3 + 21x^2 = 7x^2(2x + 3) \]
Verify: expand the answer
Why: Seven x squared times two x is fourteen x cubed, and seven x squared times three is twenty-one x squared. Both terms are recovered, so nothing was dropped and nothing extra was taken out.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 616-616
Faded example
Numbers and variables both.
Fill in the blanks
14x^3 = 2 \cdot 7 \cdot x^3, \; 21x^2 = 3 \cdot 7 \cdot x^2 \;\Longrightarrow\; \text7 = 2x^___}
Why: The variable part takes the smaller of the two exponents, since that is how many x's both terms are guaranteed to have. Taking the larger would leave a negative exponent in one term.
Worked example
Guided Practice 3, 5 and 6.
\[ \text{Factor out the greatest common factor of } 8x^3 - 16x, \; 4y^3 + 10y^2 \text{ and } 9x^3 + 6x^2 - 18x. \]
Take the first
Why: Eight and sixteen share eight; both have an x.
\[ 8 x(x ^{2} - 2) \]
Take the second
Why: Four and ten share two; both have y squared.
\[ 2 y ^{2}(2 y + 5) \]
Take the third
Why: All share three and one x.
\[ 3 x(3 x ^{2} + 2 x - 6) \]
Check each remainder
Why: Nothing more comes out.
Figure (svg): The greatest common factor found from prime factorisations
\[ 8x(x^2 - 2), \; 2y^2(2y + 5), \; 3x(3x^2 + 2x - 6) \]
Verify: check the second one carefully
Why: Four and ten share only a two, not a four, so taking four out would leave a fraction in the second term. Taking the greatest common factor means the largest one that divides every term exactly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 616-616
Trap
\[ 14x^3 + 21x^2 = 7x(2x^2 + 3x) \]
Take out seven and one x
Why: Both terms do contain a seven and an x.
They also both contain a second x, so seven x squared comes out rather than seven x. The bracket still has a common factor of x inside it, which means the expression is not fully factored.
\[ 14x^3 + 21x^2 = 7x^2(2x + 3) \]
Take every shared factor, then check the bracket has none left
Why: Greatest, not merely common.
A quick look inside the bracket for a remaining common factor catches this every time.
Matching
Shared numbers and shared variables.
Match the pairs
Why: The third takes only a two rather than a four, because ten is not divisible by four. The greatest common factor of the coefficients is a separate question from that of the variables, and both have to be answered.
Elimination
For 14x cubed + 21x squared.
Eliminate the wrong options
Which has taken out the greatest common factor?
Survives elimination: A
Why: Only one leaves a bracket with nothing left to remove. The three wrong answers each took a genuine common factor, just not the greatest, which is why checking inside the bracket is the reliable test.
Socratic
It could seem natural to take the larger.
Discussion prompt
Explain why the common variable factor uses the smaller of the two exponents. Then say what would go wrong if the larger were used.
Hint: How many x's does each term actually contain?
Answer:
A common factor has to divide every term, and the term with fewer x's limits how many can be taken. Fourteen x cubed has three x's and twenty-one x squared has two, so only two are available in both — taking a third would take from a term that does not have it.
Using the larger exponent would leave the second term with a negative exponent, which means a variable in a denominator and is not a polynomial at all. That is the same restriction that keeps monomials to whole number exponents in Lesson 10.1, and it is why the smaller exponent is forced rather than merely conventional.
Section
Section 2
Concept
A polynomial is prime if it cannot be factored using integer coefficients. To factor completely is to write the polynomial as a product of a monomial factor and prime factors.
Stopping after the common factor is a partial answer.
Figure (svg): A cubic factored into a monomial and two prime factors
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 617-617 — the definition of a prime polynomial and Example 2, Factor Completely
Picture it
Three factors, none reducible.
Figure (svg): A cubic factored into a monomial and two prime factors
Removing the four x left a trinomial whose leading coefficient is one, which is exactly the shape Lesson 10.5 handles. Cubics become manageable the moment a factor comes out.
Worked example
This is Example 2 from the textbook.
\[ \text{Factor } 4x^3 + 20x^2 + 24x \text{ completely.} \]
Find the common factor
Why: All three share four and one x.
\[ 4 x \]
Remove it
Why: Divide each term.
\[ 4 x(x ^{2} + 5 x + 6) \]
Factor the trinomial
Why: Two and three.
\[ (x + 2) (x + 3) \]
Write the complete factorisation
Why: Monomial and two primes.
\[ 4 x(x + 2) (x + 3) \]
Figure (svg): A cubic factored into a monomial and two prime factors
\[ 4x^3 + 20x^2 + 24x = 4x(x + 2)(x + 3) \]
Verify: expand back in stages
Why: x plus two times x plus three is x squared plus five x plus six, and multiplying by four x gives four x cubed plus twenty x squared plus twenty-four x. Expanding in the reverse order of the factoring keeps the check short.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 617-617
Sorting
Check every bracket for further structure.
Sort into buckets
Sort each factorisation by whether it is complete.
All three incomplete ones have a quadratic bracket that fits a pattern from Lesson 10.7. That is the usual place a factorisation stops one step early.
Worked example
Guided Practice 7, 9 and 12.
\[ \text{Factor } 2n^3 + 4n^2 + 2n, \; 5m^3 - 45m \text{ and } 6p^3 - 21p^2 + 9p \text{ completely.} \]
Take the first
Why: Common factor two n; the rest is a perfect square.
\[ 2 n(n + 1) ^{2} \]
Take the second
Why: Common factor five m; the rest is a difference of squares.
\[ 5 m(m + 3) (m - 3) \]
Take the third
Why: Common factor three p.
\[ 3 p(2 p ^{2} - 7 p + 3) \]
Finish the third
Why: The trinomial factors by trial.
\[ 3 p(2 p - 1) (p - 3) \]
Figure (svg): A cubic factored into a monomial and two prime factors
\[ 2n(n+1)^2, \; 5m(m+3)(m-3), \; 3p(2p-1)(p-3) \]
Verify: notice which pattern each remainder needed
Why: The three remainders were a perfect square trinomial, a difference of squares and an ordinary trial factorisation — every method from the last three lessons, applied after the same first step. That is why the common factor comes out first regardless of what follows.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 617-617
Error analysis
The student was asked to factor a cubic completely.
Annotate
On: \( \begin{aligned} 3x^3 - 12x &= 3x(x^2 - 4) \\ \text{so the answer is } &3x(x^2 - 4) \end{aligned} \)
Nothing here is wrong, only unfinished, which makes it easy to overlook. After removing a common factor it is worth looking at the bracket specifically for the patterns of Lesson 10.7, since a difference of squares is exactly what tends to survive that first step.
Faded example
Common factor, then a pattern.
Fill in the blanks
3x^3 - 12x = 3x(x^2 - 4) = 3x(x + 2)(x - 2)
Why: The bracket left after the common factor is a difference of two squares, so it factors again. Looking at the bracket for a pattern is the habit that turns a partial answer into a complete one.
Elimination
For a polynomial rather than a number.
Eliminate the wrong options
Which polynomial is prime over the integers?
Survives elimination: A
Why: A sum of a square and a positive constant has a negative discriminant, so it has no real roots and no linear factors. Note that a prime polynomial is one with no integer factorisation, which is a different idea from a prime number but plays the same role.
Socratic
The order could be reversed.
Discussion prompt
Give two reasons for taking out the greatest common factor before anything else. Then say what happens to the degree of what remains.
Hint: What kind of expression is left?
Answer:
First, it reduces a cubic to a quadratic, which is the only kind of polynomial the last four lessons can factor. Second, it shrinks every coefficient, which makes any subsequent search shorter and any pattern easier to recognise — as with fifty minus ninety-eight x squared in Lesson 10.7.
Removing a factor of x lowers the degree by one, so a cubic becomes a quadratic and everything already learnt applies. That is the whole strategy of this lesson: cubics are not factored by a new method but reduced until an old method fits.
Section
Section 3
Concept
A polynomial with four terms can sometimes be factored by grouping the terms into two pairs, factoring each pair, and then removing the bracket both pairs share.
The method stalls if the two brackets differ.
Figure (svg): A four-term polynomial factored by grouping
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 617-617 — Example 3, Factor by Grouping
Picture it
Group, factor, extract, finish.
Figure (svg): A four-term polynomial factored by grouping
The final step used a difference of squares, so grouping produced something the earlier lessons could finish. That combination is typical of complete factorisations.
Worked example
This is Example 3 from the textbook.
\[ \text{Factor } x^3 + 2x^2 - 9x - 18 \text{ completely.} \]
Group in pairs
Why: First two, then last two.
\[ (x ^{3} + 2 x ^{2}) + (-9 x - 18) \]
Factor each pair
Why: x squared, then negative nine.
\[ x ^{2}(x + 2) + (-9) (x + 2) \]
Remove the shared bracket
Why: It appears in both.
\[ (x + 2) (x ^{2} - 9) \]
Finish with a pattern
Why: A difference of two squares.
\[ (x + 2) (x + 3) (x - 3) \]
Figure (svg): A four-term polynomial factored by grouping
\[ x^3 + 2x^2 - 9x - 18 = (x + 2)(x + 3)(x - 3) \]
Verify: substitute a value
Why: At x equal to one the original gives one plus two minus nine minus eighteen, which is negative twenty-four, and the factors give three times four times negative two, also negative twenty-four. Three linear factors means three solutions, which only a cubic can have.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 617-617
Faded example
The sign of the second factor is a choice.
Fill in the blanks
-9x - 18 = (-9)(x + 2)
Why: Removing negative nine leaves x plus two, matching the first pair's bracket. Removing positive nine would leave negative x minus two, and the method would appear to fail.
Worked example
Guided Practice 14 and 15.
\[ \text{Factor } x^3 + 5x^2 - 4x - 20 \text{ and } x^3 + 4x^2 - 9x - 36 \text{ completely.} \]
Group the first
Why: Factor each pair.
\[ x ^{2}(x + 5) - 4(x + 5) \]
Finish the first
Why: The bracket is a difference of squares.
\[ (x + 5) (x + 2) (x - 2) \]
Group the second
Why: Factor each pair.
\[ x ^{2}(x + 4) - 9(x + 4) \]
Finish the second
Why: Again a difference of squares.
\[ (x + 4) (x + 3) (x - 3) \]
Figure (svg): A four-term polynomial factored by grouping
\[ (x+5)(x+2)(x-2), \qquad (x+4)(x+3)(x-3) \]
Verify: notice the sign that had to be handled
Why: In each case the second pair was factored with a negative in front — negative four rather than four — so that the bracket would match the first pair's. Choosing that sign deliberately is what makes the two brackets agree.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 617-617
Trap
\[ x^3 + 2x^2 - 9x - 18 = x^2(x + 2) + 9(-x - 2) \]
Take out a positive nine from the last two terms
Why: Nine divides both, so nine was removed.
The bracket then reads negative x minus two, which is not the same as x plus two, and the shared bracket disappears. Taking out negative nine instead gives x plus two and the method continues.
\[ = x^2(x + 2) + (-9)(x + 2) \]
Choose the sign of the second factor so that the brackets match
Why: That is what the step is for.
If no choice of sign makes them match, the grouping was wrong and the pairs should be rearranged.
Sorting
Count the terms.
Sort into buckets
Sort each polynomial by the first method to try.
The four-term polynomials here have no overall common factor, which is exactly when grouping earns its place. Checking for a common factor first still costs nothing.
Prediction
After factoring each pair.
Predict first
What should you do?
Correct: Try grouping the terms differently, or conclude grouping does not work.
A cubic that resists grouping may still factor after a common factor is removed, or by finding one root and dividing.
Why: Matching brackets are what the method depends on, and if the natural pairing does not produce them, a different pairing sometimes will — the first and third terms with the second and fourth, for instance. Changing a sign arbitrarily would change the polynomial, and adding the brackets has no justification. Grouping is one tool among several, and its failure does not mean the polynomial is prime.
Socratic
It looks like a trick.
Discussion prompt
Explain what grouping is really doing in terms of the distributive property. Then say why four terms are the natural case for it.
Hint: What is being treated as a single object?
Answer:
Once both pairs have been factored, the expression is x squared times a bracket plus something else times the same bracket — which is exactly the shape a common factor comes out of, with the bracket playing the role of the shared factor. The distributive property does not care whether the shared factor is a number, a variable or a whole expression.
Four terms are natural because two pairs each need two terms, and each pair must leave a single bracket behind. With six terms one can group in three pairs and look for the same shared bracket, so the idea extends — but four is the smallest case where the method has anything to do.
Section
Section 4
Concept
A sum of two cubes and a difference of two cubes each factor into a binomial and a trinomial. Unlike squares, a sum of two cubes does factor.
\[ a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2) \]
The trinomial factor is always prime over the integers.
Figure (svg): The sum and difference of two cubes patterns
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 618-618 — the Sum and Difference of Two Cubes Patterns and Examples 4 and 5
Picture it
Only the first varies.
Figure (svg): The three signs in a cube factorisation
The binomial takes the original sign, the trinomial's middle term takes the opposite, and the last term is always positive. One decision fixes all three.
Worked example
This is Example 4 from the textbook.
\[ \text{Factor } x^3 + 27. \]
Write as cubes
Why: Twenty-seven is three cubed.
\[ x ^{3} + 3 ^{3} \]
Write the binomial
Why: The same sign as the original.
\[ (x + 3) \]
Write the trinomial
Why: Squares at the ends, opposite middle sign.
\[ (x ^{2} - 3 x + 9) \]
Check the trinomial
Why: It does not factor further.
Figure (svg): The sum and difference of two cubes patterns
\[ x^3 + 27 = (x + 3)(x^2 - 3x + 9) \]
Verify: expand the product
Why: x times the trinomial gives x cubed minus three x squared plus nine x, and three times it gives three x squared minus nine x plus twenty-seven. Everything cancels except x cubed and twenty-seven, which is what makes the pattern work.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 618-618
Faded example
One decision fixes all of them.
Fill in the blanks
n^3 - 64 = (n - 4)(n^2 + 4n + 16)
Why: The binomial takes the original sign and the trinomial's middle term takes the opposite. The final term is positive in every case, because it is a square.
Worked example
This is Example 5 from the textbook, with a guided item added.
\[ \text{Factor } n^3 - 64 \text{ and } 2n^3 - 250. \]
Write the first as cubes
Why: Sixty-four is four cubed.
\[ n ^{3} - 4 ^{3} \]
Apply the pattern
Why: Middle sign is now positive.
\[ (n - 4) (n ^{2} + 4 n + 16) \]
Take the second
Why: Remove the common factor first.
\[ 2(n ^{3} - 125) \]
Apply the pattern
Why: A hundred and twenty-five is five cubed.
\[ 2(n - 5) (n ^{2} + 5 n + 25) \]
Figure (svg): The three signs in a cube factorisation
\[ (n-4)(n^2 + 4n + 16), \qquad 2(n-5)(n^2 + 5n + 25) \]
Verify: compare the two middle signs
Why: The first came from a difference, so its trinomial has a plus in the middle; a sum would have given a minus there. The last term is positive in both, since it is a square, and that is the check that the pattern was copied correctly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 618-618
Trap
A sum of two squares does not factor, so a sum of two cubes cannot either.
Generalise from the squares case
Why: The two look analogous.
A sum of two cubes does factor: x cubed plus twenty-seven is x plus three times x squared minus three x plus nine. The analogy fails, and it fails because a cube keeps the sign of its base while a square does not.
\[ x^3 + 27 = (x + 3)(x^2 - 3x + 9) \]
Use the cube pattern for cubes and the square pattern for squares
Why: They are different rules.
The reason is visible in the roots: negative three cubed is negative twenty-seven, so x equal to negative three really is a root.
Translation
Identify the two cube roots.
Match the pairs
Why: In every case the trinomial's last term is the square of the cube root, not the cube. Twenty-seven has cube root three, and nine rather than twenty-seven appears at the end.
Hypothesis
The two patterns disagree.
Predict first
What explains the difference?
Correct: A negative number has a real cube root, so a sum of cubes has a real root.
\[ (-3)^3 = -27 \qquad \text{but no real } x \text{ has } x^2 = -9 \]
Why: Setting x cubed plus twenty-seven equal to nought gives x cubed equal to negative twenty-seven, and negative three cubed is negative twenty-seven — so a real root exists and a real linear factor must too. Setting x squared plus nine equal to nought would need a real number squaring to negative nine, which does not exist, so there is no factor. The difference comes from odd powers preserving sign where even powers destroy it, which is the same fact behind Lesson 9.1's count of square roots.
Socratic
It is asserted in both examples.
Discussion prompt
Use the discriminant to explain why the trinomial from a cube pattern never factors further. Then say what that means in practice.
Hint: Compute b squared minus four a c for x squared minus three x plus nine.
Answer:
For x squared minus three x plus nine the discriminant is nine minus thirty-six, which is negative twenty-seven. In general the trinomial is x squared minus a b x plus b squared, whose discriminant is b squared minus four b squared, or negative three b squared — always negative for non-zero b.
In practice it means the factorisation stops there, so no time need be spent hunting for factors of the trinomial. Recognising that the second factor is automatically prime turns the cube patterns into a one-step complete factorisation rather than the start of another search.
Section
Section 5
Concept
A volume is a product of three lengths, so a cubic that factors into three linear pieces gives a possible set of dimensions directly. The factorisation must be complete for the reading to work.
A monomial factor is a dimension too.
Figure (svg): A box whose volume is a cubic polynomial
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 616-622 — the lesson opener on the dimensions of a terrarium
Picture it
Each factor is an edge.
Figure (svg): A box whose volume is a cubic polynomial
The factorisation does not say which factor is which edge — any assignment gives the same volume. What it gives is a set of three lengths that multiply correctly.
Worked example
The lesson's situation, with a volume stated.
\[ \text{A terrarium has volume } x^3 + 5x^2 + 6x. \text{ Find possible dimensions.} \]
Remove the common factor
Why: Every term has an x.
\[ x(x ^{2} + 5 x + 6) \]
Factor the trinomial
Why: Two and three.
\[ x(x + 2) (x + 3) \]
Read the three factors
Why: Each is a length.
\[ x, \; x + 2, \; x + 3 \]
Check positivity
Why: All three need x positive.
\[ x > 0 \]
Figure (svg): A box whose volume is a cubic polynomial
\[ V = x(x + 2)(x + 3) \]
Verify: test with a number
Why: Take x equal to four: the dimensions are four, six and seven, a volume of a hundred and sixty-eight. The polynomial gives sixty-four plus eighty plus twenty-four, which is also a hundred and sixty-eight.
Faded example
Common factor, then the trinomial.
Fill in the blanks
x^3 + 5x^2 + 6x = x(x^2 + 5x + 6) = x(x + 2)(x + 3)
Why: The x that came out is itself one of the three dimensions, so the monomial factor is not merely a tidying step. All three factors are lengths.
Worked example
Working backwards from a required volume.
\[ \text{For what } x \text{ does the terrarium hold } 60 \text{ cubic units?} \]
Set the volume
Why: Sixty cubic units.
\[ x(x + 2) (x + 3) = 60 \]
Try small whole numbers
Why: x equal to two gives twenty-four.
Try the next
Why: x equal to three gives ninety.
Interpret
Why: The answer lies between them.
\[ \text{about } 2.6 \]
Figure (svg): A box whose volume is a cubic polynomial
\[ x(x+2)(x+3) = 60 \;\Longrightarrow\; x \approx 2.6 \]
Verify: check the estimate
Why: At x equal to 2.6 the dimensions are 2.6, 4.6 and 5.6, giving about sixty-seven — a little high, so the true value is slightly below 2.6. Note that the zero-product property does not apply here, because the right side is sixty rather than nought.
Trap
\[ x(x + 2)(x + 3) = 60 \;\Longrightarrow\; x = 60 \text{ or } x + 2 = 60 \text{ or } x + 3 = 60 \]
Set each factor equal to the volume
Why: The equation is factored, so each factor was set to the right-hand side.
The zero-product property applies only when the product is nought. Sixty can be reached by countless triples of numbers, so no single factor is forced to any value — and x equal to sixty would give a volume of over two hundred thousand.
Estimate numerically, or expand and solve the cubic.
Check that the right side is nought before splitting factors
Why: Only nought has that property.
This is the same restriction that Lesson 10.4 established, and it does not weaken for cubics.
Elimination
For a volume of x cubed + 5x squared + 6x.
Eliminate the wrong options
Which set of three dimensions is correct?
Survives elimination: A
Why: The absence of a constant term means one dimension must be x itself, which rules out option C immediately. Reading the constant term is a quick check on any proposed set of dimensions.
Prediction
For a volume polynomial.
Predict first
If a volume polynomial has no constant term, what follows?
Correct: One of the dimensions is a multiple of the variable.
\[ x(x+2)(x+3): \; \text{constant } 0 \qquad (x+1)(x+2)(x+3): \; \text{constant } 6 \]
Why: The constant term of a product is the product of the constants, so it is nought exactly when one factor has no constant — that is, when one dimension is x or a multiple of it. That also means the volume is nought when x is nought, which makes physical sense: a box with one edge of length nought holds nothing. Reading the constant term is a fast structural check before any factoring begins.
Socratic
The expanded form is the same polynomial.
Discussion prompt
Say what the factored form of a volume tells a designer that the expanded form hides. Then say what it does not determine.
Hint: Think about what a manufacturer needs to know.
Answer:
It gives three lengths that can actually be cut and assembled, which is what someone building the box needs. The expanded form gives the same volume for any value of x but says nothing about the shape, and recovering the dimensions from it would mean factoring anyway.
It does not determine which factor is the length, the width or the height, since multiplication can be reordered freely — any assignment gives the same volume. Nor does it rule out other factorisations over the reals: a box of the same volume could have quite different proportions, and the polynomial only describes this particular family of shapes.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Situation | Method | What is left |
|---|---|---|
| Every term shares a factor | remove the GCF | a quadratic to factor by earlier methods |
| Four terms, no overall factor | group in pairs | a shared bracket times a quadratic |
| Two terms, both cubes | use a cube pattern | a binomial times a prime trinomial |
None of these factors a cubic directly. Each one reduces it to something the earlier lessons already handle, which is the whole strategy of the chapter's last lesson.
Pattern
To factor any cubic polynomial completely, these five moves cover it.
Step one comes first every time, because it is the only step that always applies and it makes every later step shorter.
OpenStax Elementary Algebra 2e, §7.1 Greatest Common Factor and Factor by Grouping §7.1
Check
Greatest, not merely common.
Check your understanding
Factor the greatest common factor out of 14x cubed + 21x squared.
Answer: A
Why: Both terms share a seven and two x's, so the greatest common factor is seven x squared and the bracket has nothing left to remove.
Check
Match the brackets.
Check your understanding
Factor x cubed + 2x squared - 9x - 18 completely.
Answer: A
Why: Grouping gives x plus two times x squared minus nine, and the second bracket is a difference of two squares.
Check
Three signs, one decision.
Check your understanding
Factor x cubed + 27.
Answer: A
Why: The binomial takes the original plus sign, the trinomial's middle term takes the opposite, and its last term is a square and so positive.
Real world
This is the terrarium question from the lesson opener. A terrarium is a closed box, so its volume is the product of its three dimensions.
Discussion prompt
A terrarium's volume is given by x cubed plus five x squared plus six x cubic inches. Find possible dimensions, say what the missing constant term tells you, and find roughly what x makes the volume 60 cubic inches.
Hint: Factor completely; each factor is a length.
Answer:
\[ x^3 + 5x^2 + 6x = x(x + 2)(x + 3) \]
The three dimensions are x, x plus two and x plus three, so the terrarium is two inches longer in one direction than another and three inches longer in the third.
The missing constant term means one dimension is x itself, so the volume falls to nought when x does — which is what a box with an edge of length nought should do. For a volume of sixty, trying whole numbers brackets the answer between two and three, since two gives twenty-four and three gives ninety, and about 2.6 is close. The zero-product property cannot be used here, because sixty is not nought, and that restriction is exactly the one from Lesson 10.4.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Is 3x(x squared - 4) a complete factorisation of 3x cubed - 12x?
Correct: No, because x squared - 4 is a difference of two squares.
\[ 3x^3 - 12x = 3x(x^2 - 4) = 3x(x + 2)(x - 2) \]
Why: The common factor of three x was found correctly, so the first step is right and the expression does expand back to the original. But factoring completely requires every remaining factor to be prime, and x squared minus four is not — it factors as x plus two times x minus two, giving three x times x plus two times x minus two. The common factor of three alone would be an incomplete removal rather than a wrong one, and cubics factor perfectly well once their degree is reduced. The lesson is that after removing a common factor it is always worth looking at the bracket specifically for the patterns of Lesson 10.7, since a difference of squares is exactly what tends to survive that first step.
Explain it
They grouped a four-term polynomial and said the method failed because the brackets did not match.
Discussion prompt
In no more than four sentences, explain the sign choice they probably missed. Then tell them what to try if it genuinely does not work.
Hint: What sign came out of the second pair?
Answer:
A usable answer: when you factor the second pair, you can take out either a positive or a negative factor, and the choice decides the sign inside the bracket. Taking out negative nine from negative nine x minus eighteen leaves x plus two, which matches the first bracket, while taking out positive nine leaves negative x minus two, which does not.
If no sign choice makes them match, try pairing the terms differently — the first with the third, and the second with the fourth. If that also fails, grouping is not the right method for that polynomial.
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: The greatest common factor is fixed by writing both terms out as products and taking the smaller exponent. Completeness is fixed by looking inside every bracket for a remaining pattern. Grouping is fixed by choosing the sign of the second factor deliberately so the brackets agree. Cube signs are fixed by remembering that the trinomial's middle sign is the opposite of the binomial's and its last term is always positive. 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 two terms out as full products of primes and variables, ring the shared factors, and read off the greatest common factor, then use it to factor the expression. Underneath, factor a cubic completely in two clearly separated stages, labelling the monomial factor and the prime factors, and beside it write an incomplete version and say what is still missing from it. In the middle, factor a four-term polynomial by grouping, writing all five lines, and circle the sign you chose in the second pair so that the brackets would match. Beneath that, write both cube patterns and label the three signs in each with arrows saying which is fixed and which follows, then work one example of each and compute the discriminant of the trinomial factor to show it is prime. In the lower half, draw a box, label its three edges with the linear factors of a volume you have factored, and test the whole thing with a numerical value of the variable. Finally, in the margin, write the order of the steps and why the greatest common factor always comes first.
Every discriminant you compute for a cube pattern's trinomial should be negative. If one comes out as a perfect square, the pattern was copied wrongly, most likely with the middle sign the wrong way round.
Recap
Five things, and the first one always comes first.
| If the question says | Your first move is |
|---|---|
| Factor a cubic | Take out the greatest common factor |
| Four terms and no common factor | Group them in pairs |
| Two terms, both perfect cubes | Use a cube pattern |
| Factor completely | Check every bracket for further structure |
| Find dimensions from a volume | Factor completely and read the linear factors |
That completes Chapter 10. Chapter 11 turns to expressions with variables in a denominator — rational expressions — where factoring becomes the tool for simplifying rather than for solving, and where the restrictions on the variable finally start to matter.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.8 Factoring Cubic Polynomials §10.8, pp. 616-622 — everything on these slides traces back here
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