Common monomial factors, the sum and difference of two cubes, factoring by grouping, recognising quadratic form, and solving higher-degree polynomial equations with the zero product property including a basin volume model.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 5 — Polynomials and Polynomial Functions
Factor and Solve Polynomial Equations
Objectives
Five outcomes. The first four are ways of seeing; the fifth is what they are for.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 353-357 — the lesson these objectives are drawn from
Warm-up
Chapter 4 factored quadratics. Nothing there had degree above 2, and this lesson removes that limit.
Discussion prompt
Factor x cubed plus 2x squared minus 15x completely. What is the first move, and what does it turn the problem into?
Hint: Every term contains an x.
Answer:
\[ x^3 + 2x^2 - 15x = x(x^2 + 2x - 15) = x(x+5)(x-3) \]
Taking out the common monomial dropped the degree from 3 to 2, and what was left was an ordinary Lesson 4.3 trinomial. That is the shape of most of this lesson: reduce the problem to one Chapter 4 can already do.
Concept
A polynomial of degree 3 or more is rarely factored directly. Instead a first move — pulling out a monomial, grouping the terms, or substituting for a repeated expression — reduces it to a quadratic or a known pattern, and Chapter 4's methods finish the job.
factored completely — A factorable polynomial with integer coefficients is factored completely when it is written as a product of unfactorable polynomials with integer coefficients.
\[ 4x^5 - 60x^3 + 216x = 4x(x+3)(x-3)(x^2-6) \]
Completely is the word that does the work. After every step, look again at each factor and ask whether it can be factored further — a difference of squares hiding inside a bracket is the usual culprit.
Figure (svg): Two columns separating complete factorisations from ones that can still be taken further
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 353-355
Section
Section 1
Concept
Before anything else, check whether every term shares a common monomial factor, including the highest power of the variable present in all of them. Taking it out lowers the degree and usually reveals a Chapter 4 pattern.
\[ 2y^5 - 18y^3 = 2y^3(y^2 - 9) = 2y^3(y+3)(y-3) \]
The monomial includes both the numerical factor and the variable factor: in 4z to the fourth minus 16z cubed plus 16z squared, the common factor is 4z squared, not just 4 or just z squared.
Figure (svg): Three polynomials each factored first by a common monomial and then by a Chapter 4 pattern
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 353-353 — Find a common monomial factor
Picture it
Example 1: a trinomial, a difference of squares and a perfect square.
Figure (svg): Three polynomials each factored first by a common monomial and then by a Chapter 4 pattern
In every case the monomial came out first and the pattern only became visible afterwards. Trying to spot the pattern before taking the monomial out is much harder and often impossible.
Worked example
Example 1, all three parts.
\[ \text{Factor } x^3 + 2x^2 - 15x, \; 2y^5 - 18y^3, \; 4z^4 - 16z^3 + 16z^2. \]
First: take out the common x
Why: Every term has at least one x, and no numerical factor is shared.
\[ x(x ^{2} + 2 x - 15) \]
First: factor the trinomial
Why: A product of negative 15 and a sum of 2 give the pair 5 and negative 3.
\[ x(x + 5) (x - 3) \]
Second: take out 2y cubed
Why: Both terms are even and both contain y cubed.
\[ 2 y ^{3}(y ^{2} - 9) \]
Second: recognise the difference of squares
Why: Nine is 3 squared.
\[ 2 y ^{3}(y + 3) (y - 3) \]
Third: take out 4z squared, then recognise the perfect square
Why: The bracket z squared minus 4z plus 4 has middle term twice z times 2.
\[ 4 z ^{2}(z - 2) ^{2} \]
Figure (svg): The solution to Worked example factor completely shown as a ladder of expressions, one row per algebraic move
\[ x(x+5)(x-3), \; 2y^3(y+3)(y-3), \; 4z^2(z-2)^2 \]
Verify: multiply one back out
Why: For the second: 2y cubed times y plus 3 times y minus 3 is 2y cubed times y squared minus 9, which is 2y to the fifth minus 18y cubed — the original. Note that the answer is not complete until the bracket left behind is tested; stopping at 2y cubed times the quantity y squared minus 9 would be a legal but unfinished factorisation.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 353-353
Matching
Take the greatest factor shared by every term.
Match the pairs
Why: The monomial takes the greatest common numerical factor and the lowest power of the variable that appears in every term. In the third, 4 divides 4, 16 and 16, and z squared is the lowest power present — taking 4z cubed would leave a z in the denominator.
Worked example
Guided Practice 1 and 2.
\[ \text{Factor } x^3 - 7x^2 + 10x \text{ and } 3y^5 - 75y^3. \]
First: take out x
Why: No numerical factor is common to 1, negative 7 and 10.
\[ x(x ^{2} - 7 x + 10) \]
First: factor the trinomial
Why: A product of 10 and a sum of negative 7 give the pair negative 2 and negative 5.
\[ x(x - 2) (x - 5) \]
Second: take out 3y cubed
Why: Both 3 and 75 are divisible by 3, and both terms contain y cubed.
\[ 3 y ^{3}(y ^{2} - 25) \]
Second: finish with the pattern
Why: Twenty-five is 5 squared.
\[ 3 y ^{3}(y + 5) (y - 5) \]
Figure (svg): The solution to Worked example two more monomial factorisations shown as a ladder of expressions, one row per algebraic move
\[ x(x-2)(x-5), \qquad 3y^3(y+5)(y-5) \]
Verify: check the degrees add up
Why: The first is a product of three linear factors, which is degree 3 — matching the original. The second is a monomial of degree 3 times two linear factors, which is degree 5. Adding the degrees of the factors is a fast check that no factor has been lost or invented.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 354-354
Trap
\[ 3x^3 - 12x \]
Take out the common monomial and stop
Why: The monomial 3x comes out and the job looks finished.
\[ 3x(x^2 - 4) \quad \text{(not factored completely)} \]
The bracket x squared minus 4 is a difference of squares and factors further, so the answer is incomplete.
\[ 3x^3 - 12x = 3x(x^2 - 4) = 3x(x+2)(x-2) \]
Re-test every factor after every step
Why: A factorisation is complete only when no factor other than a number can be factored again.
\[ 3x(x+2)(x-2) \]
The textbook makes this the definition of factored completely and flags it in an Avoid Errors note. The usual culprit is a difference of squares left inside a bracket.
Sorting
Test every factor for one more factorisation.
Sort into buckets
Sort each factorisation.
The second item is the important contrast: x squared minus 3 looks like x squared minus 4 but is not a difference of squares over the integers, so it is genuinely finished.
Fill the middle
Example 1c.
Fill in the blanks
4z^4 - 16z^3 + 16z^2 = 4z^2(z^2 - 4z + 4)
Why: Four divides all three coefficients and z squared is the lowest power present, so 4z squared comes out. Dividing each term: 4z to the fourth over 4z squared is z squared, negative 16z cubed over 4z squared is negative 4z, and 16z squared over 4z squared is 4.
Prediction
Commit before reasoning.
Predict first
What does taking out the common monomial buy you?
Correct: It lowers the degree and shrinks the coefficients, revealing a pattern.
\[ 2y^5 - 18y^3 = 2y^3(y^2-9): \; \text{degree } 5 \to \text{ a visible degree-2 pattern} \]
Why: Two y to the fifth minus 18y cubed shows no obvious pattern; 2y cubed times the quantity y squared minus 9 immediately does. The monomial does not change what the polynomial equals, and it does add a zero at x equal to 0 when the monomial contains x — which matters for solving, as the fifth idea shows. What it buys is visibility, and that is exactly why it goes first.
Section
Section 2
Concept
A sum of two cubes factors as a binomial times a trinomial, and so does a difference. The binomial copies the sign between the cubes; the trinomial's middle term takes the opposite sign, and its outer terms are the squares.
\[ a^3 + b^3 = (a+b)(a^2 - ab + b^2), \quad a^3 - b^3 = (a-b)(a^2 + ab + b^2) \]
Unlike a sum of squares, a sum of cubes does factor. That asymmetry is worth noticing, because it is easy to assume that sums never factor after Lesson 4.3.
Figure (svg): The sum and difference of two cubes patterns, each with a worked instance
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 354-354 — Special Factoring Patterns
Picture it
The sum and the difference, each with an instance.
Figure (svg): The sum and difference of two cubes patterns, each with a worked instance
The signs are the whole difficulty. The binomial keeps the original sign; the trinomial's middle term flips it, and the trinomial's last term is always positive.
Worked example
Example 2, both parts. The second needs a monomial first.
\[ \text{Factor } x^3 + 64 \text{ and } 16z^5 - 250z^2. \]
First: write both terms as cubes
Why: Sixty-four is 4 cubed, so a is x and b is 4.
\[ x ^{3} + 4 ^{3} \]
First: apply the sum pattern
Why: The binomial is x plus 4; the trinomial is x squared minus 4x plus 16.
\[ (x + 4) (x ^{2} - 4 x + 16) \]
Second: take out the common monomial
Why: Both terms are divisible by 2 and both contain z squared.
\[ 2 z ^{2}(8 z ^{3} - 125) \]
Second: write the bracket as cubes
Why: Eight z cubed is the cube of 2z, and 125 is 5 cubed.
\[ 2 z ^{2} [(2 z) ^{3} - 5 ^{3}] \]
Second: apply the difference pattern
Why: The binomial is 2z minus 5; the trinomial is 4z squared plus 10z plus 25.
\[ 2 z ^{2}(2 z - 5) (4 z ^{2} + 10 z + 25) \]
Figure (svg): The solution to Worked example factor cubes shown as a ladder of expressions, one row per algebraic move
\[ (x+4)(x^2-4x+16), \; 2z^2(2z-5)(4z^2+10z+25) \]
Verify: multiply the first back out
Why: The quantity x plus 4 times x squared minus 4x plus 16 gives x cubed minus 4x squared plus 16x plus 4x squared minus 16x plus 64. Every middle term cancels, leaving x cubed plus 64. That cancellation is exactly why the pattern's middle sign has to be the opposite of the binomial's.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 354-354
Matching
Identify a and b as cube roots first.
Match the pairs
Why: In the last two, a is 2x and 4x rather than x, so a squared is 4x squared and 16x squared. Taking the cube root of the whole term, coefficient included, is the same discipline as taking the square root of the whole term in Lesson 4.4.
Worked example
Guided Practice 3 and 4.
\[ \text{Factor } 16b^5 + 686b^2 \text{ and } w^3 - 27. \]
First: take out the monomial
Why: Sixteen and 686 are both even, and both terms contain b squared.
\[ 2 b ^{2}(8 b ^{3} + 343) \]
First: identify the cubes
Why: Eight b cubed is the cube of 2b, and 343 is 7 cubed.
\[ a = 2 b, b = 7 \]
First: apply the sum pattern
Why: The trinomial's terms are 4b squared, negative 14b and 49.
\[ 2 b ^{2}(2 b + 7) (4 b ^{2} - 14 b + 49) \]
Second: apply the difference pattern directly
Why: Twenty-seven is 3 cubed, so the binomial is w minus 3.
\[ (w - 3) (w ^{2} + 3 w + 9) \]
Figure (svg): The solution to Worked example two more cubes shown as a ladder of expressions, one row per algebraic move
\[ 2b^2(2b+7)(4b^2-14b+49), \; (w-3)(w^2+3w+9) \]
Verify: check that the trinomials do not factor further
Why: For w squared plus 3w plus 9, the discriminant is 9 minus 36, or negative 27, so it has no real zeros and certainly no integer factors. The trinomial from a cube pattern never factors over the integers, which means the pattern always produces a complete factorisation in one step.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 354-354
Error analysis
A student factors a sum of two cubes.
Annotate
On: \( x^3 + 64 = (x + 4)(x^2 + 4x + 16) \)
The opposite sign is what makes the middle terms cancel. Multiplying back out is the fastest way to confirm you have the signs the right way round.
Fill the middle
Example 2a.
Fill in the blanks
x^3 + 64 = (x + 4)(x^2 - 4x + 16)
Why: The middle term is negative ab, which is negative 4x, and its sign is the opposite of the binomial's plus. If the signs matched, the middle terms would not cancel when multiplying back and the extra terms would survive.
Two truths and a lie
All three are about sums and differences.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it is the asymmetry worth remembering. A sum of squares such as x squared plus 49 has no real factorisation, as Lesson 4.3 established, but a sum of cubes does. The difference is that the cube pattern supplies a trinomial factor with no real zeros, which a two-factor split of squares cannot do.
Prediction
Commit before reasoning.
Predict first
Multiply the quantity a plus b by a squared minus ab plus b squared. Why does everything except two terms cancel?
Correct: The six products pair off, leaving only a cubed and b cubed.
\[ (a+b)(a^2-ab+b^2) = a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 = a^3 + b^3 \]
Why: The products are a cubed, negative a squared b, ab squared, a squared b, negative ab squared and b cubed. The two a squared b terms are opposites and cancel, and so are the two ab squared terms, leaving a cubed plus b cubed. The pattern is designed so that every middle term has a partner, which is exactly why the middle sign has to be the opposite of the binomial's.
Section
Section 3
Concept
A polynomial with four terms and no common monomial can often be factored by grouping: factor each pair separately, and if both pairs leave the same bracket, that bracket comes out as a common factor.
factor by grouping — A method for a four-term polynomial: split it into two pairs, factor each pair, and take out the bracket both pairs share.
\[ ra + rb + sa + sb = r(a+b) + s(a+b) = (r+s)(a+b) \]
If the two brackets do not match, try pairing the terms differently before concluding that grouping fails. The order of the terms is often what decides it.
Figure (svg): The factoring-by-grouping pattern shown in general and applied to a cubic
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 354-354 — Factor by grouping
Picture it
Example 3: a cubic with four terms.
Figure (svg): The factoring-by-grouping pattern shown in general and applied to a cubic
The bracket x minus 3 appeared twice, which is what let it come out. The remaining factor, x squared minus 16, then factored again — a reminder to re-test every factor.
Worked example
Example 3. Two pairs, one shared bracket.
\[ \text{Factor } x^3 - 3x^2 - 16x + 48 \text{ completely.} \]
Check for a common monomial first
Why: One, negative 3, negative 16 and 48 share no factor, and the last term has no x.
Split into pairs and factor each
Why: From the first pair, x squared comes out; from the second, negative 16 comes out.
\[ x ^{2}(x - 3) - 16(x - 3) \]
Take out the shared bracket
Why: Both pairs left x minus 3, so it is a common factor.
\[ (x ^{2} - 16) (x - 3) \]
Re-test every factor
Why: X squared minus 16 is a difference of squares.
\[ (x + 4) (x - 4) (x - 3) \]
Figure (svg): The solution to Worked example factor by grouping shown as a ladder of expressions, one row per algebraic move
\[ (x+4)(x-4)(x-3) \]
Verify: check the constant term
Why: The product of the three constants is 4 times negative 4 times negative 3, which is 48 — matching the original's constant term. And substituting x equal to 3 into the original gives 27 minus 27 minus 48 plus 48, which is zero, confirming that x minus 3 really is a factor.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 354-354
Fill the middle
Example 3, at the grouping step.
Fill in the blanks
x^3 - 3x^2 - 16x + 48 = x^2(x - 3) - 16(x - 3)
Why: Taking out negative 16 leaves x minus 3, matching the first bracket. Taking out positive 16 would leave negative x plus 3, which is the opposite bracket and blocks the method. Deciding the sign by looking at what bracket you want is the practical way to do it.
Worked example
Guided Practice 5, plus one more of the same shape.
\[ \text{Factor } x^3 + 7x^2 - 9x - 63 \text{ and } x^3 + 2x^2 - 9x - 18. \]
First: factor each pair
Why: From x cubed plus 7x squared, x squared comes out; from negative 9x minus 63, negative 9 comes out.
\[ x ^{2}(x + 7) - 9(x + 7) \]
First: take out the bracket and re-test
Why: Both pairs left x plus 7, and x squared minus 9 is a difference of squares.
\[ (x + 3) (x - 3) (x + 7) \]
Second: factor each pair
Why: From x cubed plus 2x squared, x squared comes out; from negative 9x minus 18, negative 9 comes out.
\[ x ^{2}(x + 2) - 9(x + 2) \]
Second: finish
Why: The shared bracket is x plus 2, and again x squared minus 9 splits.
\[ (x + 3) (x - 3) (x + 2) \]
Figure (svg): The solution to Worked example two more groupings shown as a ladder of expressions, one row per algebraic move
\[ (x+3)(x-3)(x+7), \; (x+3)(x-3)(x+2) \]
Verify: notice the shared structure
Why: Both polynomials contain x squared minus 9 as a hidden factor, differing only in the third bracket, which is why the two answers look so similar. The sign in front of the 9 in the second pair had to be pulled out as negative 9 rather than positive 9, or the brackets would not have matched — that sign is where grouping usually goes wrong.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 355-355
Trap
\[ x^3 - 3x^2 - 16x + 48 \]
Factor 16 out of the second pair
Why: The 16 is taken as positive, leaving what is inside unchanged in sign.
\[ x^2(x-3) + 16(-x + 3) \quad \text{(brackets do not match)} \]
The brackets x minus 3 and negative x plus 3 look different, so the method appears to fail — but they are opposites of each other.
\[ x^2(x-3) - 16(x-3) = (x^2-16)(x-3) \]
Take out the negative so that both brackets match
Why: When the second pair begins with a minus, factor out a negative monomial.
\[ +16(-x+3) = -16(x-3) \]
Grouping fails far less often than it appears to. Before abandoning it, try factoring out a negative from the second pair, and try pairing the terms in a different order.
Sorting
Factor each pair and compare the brackets.
Sort into buckets
Sort each polynomial.
Four terms is the signal for grouping. Three terms points at a monomial factor or a quadratic-form substitution instead.
Ranking
Factoring a four-term polynomial.
Put in order
Why: The monomial check comes first here as everywhere. The last step is not optional: in all three worked examples the remaining quadratic was a difference of squares, so stopping after step four would leave the answer incomplete every time.
Prediction
Commit before reasoning.
Predict first
Would grouping x cubed minus 16x with negative 3x squared plus 48 also work?
Correct: Yes — that pairing gives the same answer.
\[ x(x^2-16) - 3(x^2-16) = (x-3)(x^2-16) \]
Why: Pairing the first and third terms gives x times x squared minus 16, and the second and fourth give negative 3 times x squared minus 16, so x squared minus 16 comes out and the other factor is x minus 3 — the same factorisation reached differently. Any pairing that leaves matching brackets works, which is why a failed first attempt is a reason to regroup rather than to abandon the method.
Section
Section 4
Concept
An expression is in quadratic form when it can be written as a u squared plus b u plus c, where u is some expression in x. Substituting u makes it an ordinary quadratic, and the factorisation is then written back in the original variable.
quadratic form — An expression a u squared plus b u plus c, where u is any expression in the variable. Chapter 4's factoring methods apply to it directly.
\[ 16x^4 - 81 = u^2 - 81 \text{ with } u = 4x^2 \]
The signal is one exponent being exactly twice another: x to the fourth and x squared, or p to the sixth and p cubed. Once spotted, no new technique is needed.
Figure (svg): Two expressions rewritten as quadratics in a substituted variable and then factored
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 355-355 — Factor polynomials in quadratic form
Picture it
Example 4: one difference of squares and one trinomial.
Figure (svg): Two expressions rewritten as quadratics in a substituted variable and then factored
In the first, one of the two resulting brackets factors again; in the second, neither does. Re-testing after the substitution is undone is the step that catches the difference.
Worked example
Example 4, both parts.
\[ \text{Factor } 16x^4 - 81 \text{ and } 2p^8 + 10p^5 + 12p^2. \]
First: recognise the form
Why: Sixteen x to the fourth is the square of 4x squared, so with u equal to 4x squared the expression is u squared minus 81.
\[ u ^{2} - 81 \]
First: factor and substitute back
Why: The difference of squares gives u plus 9 times u minus 9, that is 4x squared plus 9 times 4x squared minus 9.
\[ (4 x ^{2} + 9) (4 x ^{2} - 9) \]
First: re-test
Why: Four x squared minus 9 is itself a difference of squares; 4x squared plus 9 is a sum and is not.
\[ (4 x ^{2} + 9) (2 x + 3) (2 x - 3) \]
Second: take out the monomial first
Why: Two p squared comes out, leaving p to the sixth plus 5p cubed plus 6.
\[ 2 p ^{2}(p ^{6} + 5 p ^{3} + 6) \]
Second: substitute and factor
Why: With u equal to p cubed the bracket is u squared plus 5u plus 6, which factors as u plus 3 times u plus 2.
\[ 2 p ^{2}(p ^{3} + 3) (p ^{3} + 2) \]
Figure (svg): The solution to Worked example factor in quadratic form shown as a ladder of expressions, one row per algebraic move
\[ (4x^2+9)(2x+3)(2x-3), \; 2p^2(p^3+3)(p^3+2) \]
Verify: check whether the cube factors split
Why: Three and 2 are not perfect cubes, so p cubed plus 3 and p cubed plus 2 do not factor over the integers, and the second answer is complete. Had the constants been 8 and 27 the sum-of-cubes pattern would have applied, so the re-test genuinely matters rather than being a formality.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 355-355
Fill the middle
Example 4b.
Fill in the blanks
p^6 + 5p^3 + 6 = u^2 + 5u + 6 \text3 u = p^___}
Why: The two exponents are 6 and 3, and 6 is exactly twice 3, so u equal to p cubed turns the expression into a quadratic in u. Looking for an exponent that is exactly double another is the whole of recognising quadratic form.
Worked example
Guided Practice 6 and 7.
\[ \text{Factor } 16g^4 - 625 \text{ and } 4t^6 - 20t^4 + 24t^2. \]
First: recognise the squares
Why: Sixteen g to the fourth is the square of 4g squared and 625 is 25 squared.
\[ (4 g ^{2} + 25) (4 g ^{2} - 25) \]
First: re-test
Why: Four g squared minus 25 is a difference of squares; the sum is not.
\[ (4 g ^{2} + 25) (2 g + 5) (2 g - 5) \]
Second: take out 4t squared
Why: All three coefficients are divisible by 4 and every term has t squared.
\[ 4 t ^{2}(t ^{4} - 5 t ^{2} + 6) \]
Second: substitute u equal to t squared
Why: The bracket becomes u squared minus 5u plus 6, which factors as u minus 3 times u minus 2.
\[ 4 t ^{2}(t ^{2} - 3) (t ^{2} - 2) \]
Figure (svg): The solution to Worked example two more shown as a ladder of expressions, one row per algebraic move
\[ (4g^2+25)(2g+5)(2g-5), \; 4t^2(t^2-3)(t^2-2) \]
Verify: confirm the second is complete
Why: Neither 3 nor 2 is a perfect square, so t squared minus 3 and t squared minus 2 do not factor over the integers. They do have real zeros, at plus or minus root 3 and plus or minus root 2, which matters when solving but not when factoring over the integers — the two questions have different answers here.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 355-355
Error analysis
A student finishes a quadratic-form factorisation.
Annotate
On: \( 16x^4 - 81 = (4x^2 + 9)(4x^2 - 9) = (2x+3)(2x-3)(2x+3)(2x-3) \)
Differences of squares split; sums do not. Multiplying back out catches this in one line and is worth the ten seconds on any factorisation with four factors.
Matching
Find the exponent that is half the largest.
Match the pairs
Why: The first and last absorb a coefficient into u as well as a power, which is what makes the leading term a perfect square: 16x to the fourth is 4x squared, squared, and 8x to the sixth is 2 times 2x cubed, squared. Exercise 1 of the lesson asks about exactly the last of these.
Prediction
Commit before reasoning.
Predict first
Is naming u actually necessary, or can you factor t to the fourth minus 5t squared plus 6 directly?
Correct: It is optional — the same pair search works directly.
\[ t^4 - 5t^2 + 6 = (t^2 - 3)(t^2 - 2) \]
Why: Looking for two numbers with product 6 and sum negative 5 gives negative 3 and negative 2, and writing t squared minus 3 times t squared minus 2 works without ever naming u. The substitution is a way of making the structure obvious, especially when the expressions are more complicated, and many people stop writing it once the pattern is familiar. Nothing is lost either way.
Comparison
Fill the blanks. Each first move suits a different shape.
Comparison matrix
| Signal | First move | Example |
|---|---|---|
| Every term shares a factor | take out the monomial | 2y^5 - 18y^3 |
| Four terms, no common factor | group in pairs | x^3 - 3x^2 - 16x + 48 |
| One exponent twice another | substitute for quadratic form | p^6 + 5p^3 + 6 |
| Two terms, both cubes | use the cube pattern | x^3 + 64 |
Reading the shape before choosing a move is most of the skill. Every route ends in Chapter 4's methods.
Section
Section 5
Concept
The zero product property extends to any number of factors. Write the equation in standard form, factor the polynomial completely, and set each factor equal to zero, keeping the real solutions.
\[ 4x(x+3)(x-3)(x^2-6) = 0 \;\Longrightarrow\; x = 0, \pm 3, \pm\sqrt{6} \]
Never divide both sides by a variable factor. Dividing by 4x would discard the solution x equal to zero, and the book flags this in an Avoid Errors note.
Figure (svg): A rectangular basin with one-foot walls, showing outer and interior dimensions
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 355-356 — Solve a polynomial equation
Picture it
Example 6: the volume model that becomes a cubic.
Figure (svg): A rectangular basin with one-foot walls, showing outer and interior dimensions
Grouping factored the cubic, and only x equal to 4 is real, because 2x squared plus 10 is positive for every x. The basin is 8 feet by 4 feet by 4 feet.
Worked example
Example 5. Standard form, factor completely, split.
\[ \text{Solve } 4x^5 + 216x = 60x^3. \]
Write in standard form
Why: Move every term to one side.
\[ 4 x ^{5} - 60 x ^{3} + 216 x = 0 \]
Take out the common monomial
Why: Four x comes out of all three terms.
\[ 4 x(x ^{4} - 15 x ^{2} + 54) = 0 \]
Factor the quadratic form
Why: With u equal to x squared, u squared minus 15u plus 54 factors as u minus 9 times u minus 6.
\[ 4 x(x ^{2} - 9) (x ^{2} - 6) = 0 \]
Re-test each factor
Why: X squared minus 9 is a difference of squares; x squared minus 6 is not, over the integers.
\[ 4 x(x + 3) (x - 3) (x ^{2} - 6) = 0 \]
Set each factor to zero
Why: The factors vanish at 0, negative 3, 3, and plus or minus root 6.
Figure (svg): The solution to Worked example solve a quintic shown as a ladder of expressions, one row per algebraic move
\[ x = 0, \; \pm 3, \; \pm\sqrt{6} \]
Verify: count the solutions against the degree
Why: The equation has degree 5 and produced five real solutions, which is the maximum a quintic allows. Notice also that dividing both sides by 4x at the start would have lost x equal to zero entirely — the solution that comes from the monomial factor is the one most often thrown away.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 355-355
Ranking
Solving a higher-degree polynomial equation.
Put in order
Why: Step two says keep, not cancel: the monomial factor supplies genuine solutions. Step four is where most incomplete answers come from, since a difference of squares left inside a bracket hides two solutions that never get found.
Worked example
Example 6. A volume model that factors by grouping.
\[ \text{A basin with } 1 \text{ ft walls has outer dimensions } 2x, x, x \text{ and holds } 36 \text{ ft}^3. \text{ Find } x. \]
Write the interior dimensions
Why: Length and width lose 2 feet each, for two walls; the height loses only 1, because there is a bottom but no lid.
\[ (2 x - 2) (x - 2) (x - 1) \]
Write the equation and expand
Why: The product equals 36; multiplying out gives 2x cubed minus 8x squared plus 10x minus 4.
\[ 36 = 2 x ^{3} - 8 x ^{2} + 10 x - 4 \]
Write in standard form
Why: Subtracting 36 gives a constant of negative 40.
\[ 0 = 2 x ^{3} - 8 x ^{2} + 10 x - 40 \]
Factor by grouping
Why: Two x squared comes out of the first pair and 10 out of the second, both leaving x minus 4.
\[ 0 = (2 x ^{2} + 10) (x - 4) \]
Solve and reject
Why: Two x squared plus 10 is positive for every real x, so only x minus 4 can vanish.
\[ x = 4 \]
Figure (svg): The solution to Worked example the basin shown as a ladder of expressions, one row per algebraic move
\[ x = 4 \;\Longrightarrow\; 8 \times 4 \times 4 \text{ ft} \]
Verify: compute the interior volume
Why: The interior is 6 by 2 by 3 feet, whose product is 36 cubic feet — the required capacity. The rejected factor is worth a moment: 2x squared plus 10 has no real zeros, so it contributes nothing, and there was no need to reject a root on physical grounds this time.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 356-356
Trap
\[ 4x^5 - 60x^3 + 216x = 0 \]
Divide both sides by 4x to simplify
Why: The common factor is cancelled rather than kept as a factor.
\[ x^4 - 15x^2 + 54 = 0 \;\Longrightarrow\; x = \pm 3, \pm\sqrt{6} \quad \text{(one solution lost)} \]
Substituting x equal to 0 into the original gives 0 equals 0, so it is a solution — and dividing by 4x threw it away silently.
\[ 4x(x^4 - 15x^2 + 54) = 0 \]
Keep the monomial as a factor and set it to zero too
Why: Every factor gets its own equation, including the one containing the variable.
\[ 4x = 0 \;\Longrightarrow\; x = 0 \]
Dividing by a constant is safe; dividing by anything containing the variable is not, because that expression might itself be zero. This is the same warning as in Lesson 4.9's inequalities, and the book flags it here too.
Fill the middle
Example 5, at the last step.
Fill in the blanks
4x(x+3)(x-3)(x^2-6) = 0 \;\Longrightarrow\; x = 0, \pm 3, \pmsqrt(6)
Why: The last factor vanishes when x squared equals 6, so x is plus or minus the square root of 6 — an irrational solution reached by Lesson 4.5's method rather than by factoring. A factor that does not split over the integers can still contribute real solutions, which is why the two questions must be kept apart.
Sorting
Read the shape of the polynomial.
Sort into buckets
Sort each equation by the first move after standard form.
The two groups also differ in their answers: the monomial group always includes x equal to zero as a solution, and the grouping examples here do not.
Prediction
Commit before reasoning.
Predict first
A polynomial equation of degree 5 is solved by factoring. What is the most real solutions it can have?
Correct: Five, one for each degree.
\[ \text{degree } n \;\Longrightarrow\; \text{at most } n \text{ real solutions} \]
Why: Each linear factor contributes one solution, and a degree-5 polynomial splits into at most five linear factors. Example 5 achieves the maximum with five distinct real solutions; the basin equation, also of odd degree, has only one, because its quadratic factor has no real zeros. Lesson 5.7's fundamental theorem sharpens this into an exact count once complex solutions are allowed.
Comparison
Fill the blanks. Read the polynomial before choosing.
Comparison matrix
| What you see | First move | Then |
|---|---|---|
| A factor in every term | take out the monomial | apply a Chapter 4 pattern |
| Two terms, both cubes | use the sum or difference of cubes | the trinomial never factors further |
| Four terms | group in pairs | take out the shared bracket |
| One exponent twice another | substitute for quadratic form | factor as a quadratic, then substitute back |
Every route ends in Chapter 4. What is new in this lesson is the recognising, not the factoring.
Pattern
One routine for factoring anything, and for solving.
Never divide both sides of an equation by a factor containing the variable. Set it to zero instead; that is where a solution lives.
OpenStax Algebra and Trigonometry 2e, §1.5 Factoring Polynomials §1.5
Check
Factoring completely. Test every factor.
Check your understanding
Factor 2y^5 - 18y^3 completely.
Answer: A
Why: Taking out 2y^3 leaves y^2 - 9, which is a difference of squares.
Check
Cubes. Watch the middle sign.
Check your understanding
Factor x^3 + 64 completely.
Answer: A
Why: With a = x and b = 4, the sum-of-cubes pattern gives (a + b)(a^2 - ab + b^2).
Check
Solving. Keep the monomial factor.
Check your understanding
What are the real solutions of 4x^5 + 216x = 60x^3?
Answer: A
Why: Factoring gives 4x(x + 3)(x - 3)(x^2 - 6) = 0, with five real solutions.
Real world
A shipping crate is built with outer dimensions x by x plus 1 by x plus 4 metres, and its walls, floor and lid are all 0.5 metres thick.
Discussion prompt
Write the interior volume as a polynomial in standard form, then find x if the interior volume is to be 6 cubic metres.
Hint: Each pair of opposite walls removes a total of 1 metre from that dimension.
Answer:
\[ V = (x-1)(x)(x+3) = x^3 + 2x^2 - 3x \]
\[ x^3 + 2x^2 - 3x = 6 \;\Longrightarrow\; x^3 + 2x^2 - 3x - 6 = 0 \]
\[ x^2(x+2) - 3(x+2) = 0 \;\Longrightarrow\; (x^2-3)(x+2) = 0 \;\Longrightarrow\; x = \sqrt{3} \approx 1.73 \]
The crate is about 1.73 by 2.73 by 5.73 metres. The roots negative root 3 and negative 2 are rejected because a dimension cannot be negative.
Two things are worth noticing. Every dimension lost exactly 1 metre because both opposite faces are half a metre thick, which is why the interior dimensions came out as x minus 1, x and x plus 3. And the cubic factored by grouping, leaving x squared minus 3 — a factor that does not split over the integers but still supplies the answer, exactly as x squared minus 6 did in Example 5.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is 3x times the quantity x squared minus 4 a complete factorisation?
Correct: No — x squared minus 4 factors further.
\[ 3x(x^2-4) = 3x(x+2)(x-2) \qquad 5x^2(x^2-3) \text{ is already complete} \]
Why: Four is a perfect square, so x squared minus 4 splits into x plus 2 times x minus 2 and the complete factorisation is 3x times x plus 2 times x minus 2. The textbook uses this exact expression to define what factored completely means. Compare it with 5x squared times the quantity x squared minus 3, which IS complete because 3 is not a perfect square — the difference is whether the constant is a square, not whether the bracket looks like a difference.
Explain it
They can factor quadratics and have just been handed a degree-5 polynomial.
Discussion prompt
In four sentences or fewer, explain the strategy for factoring something of high degree, without naming any specific pattern.
Hint: Talk about reducing rather than about factoring.
Answer:
You almost never factor a high-degree polynomial directly. Instead you make one move that turns it into something of degree 2 that you already know how to handle: take out a common monomial, group four terms into two pairs, or notice that one exponent is exactly twice another.
After that move, everything is Chapter 4 again. The one extra habit is to look at every factor you produce and ask whether it factors too, because a difference of squares hiding inside a bracket is easy to walk past.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For choosing a move, count the terms first: two, three or four each point somewhere different. For cubes, write the pattern out with letters before substituting anything. For grouping, try factoring out a negative from the second pair, then try a different pairing. For solving, never divide by anything containing the variable — set it to zero instead. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a decision page for factoring. Down the left, list the four signals — a common factor in every term, two terms that are both cubes, four terms with no common factor, and one exponent exactly twice another — and beside each write the first move it calls for and one example of your own. In the middle, take 4x to the fifth minus 60x cubed plus 216x and factor it completely, showing every stage and circling the moment each factor is re-tested. On the right, solve the corresponding equation, listing all five solutions and marking which one would have been lost by dividing by 4x. At the bottom, write the two cube patterns with letters and note where the signs differ. In a margin, write one factorisation that looks complete but is not, and complete it.
If your middle panel has four factors rather than five including the monomial, check whether x squared minus 9 was left unsplit — that is where two of the five solutions live.
Recap
Five things, and every one reduces a hard problem to a Chapter 4 problem.
| If you see | Then |
|---|---|
| A factor common to every term | Take it out first, and keep it |
| Two cubes with any sign between | Use the cube pattern |
| Four terms | Group them in pairs |
| An exponent twice another | Substitute for quadratic form |
| A difference of squares in a bracket | Split it; you are not finished |
| A variable factor in an equation | Set it to zero; never divide by it |
Everything here needed the polynomial to factor. Lesson 5.5 introduces division, which works whether or not anything factors and turns a single known zero into a complete factorisation.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.4 Factor and Solve Polynomial Equations §5.4, pp. 353-357 — everything on these slides traces back here
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