Algebra 1 · A.SSE.A.2

Factoring completely and choosing a strategy

A polynomial has been factored once. How do you know when you are finished?

You need

Factoring special patternsAlgebra 1 Factoring when the leading coefficient is not oneAlgebra 1

Unlocks

Simplifying a rational expression and where it is undefinedAlgebra 2

Where you are 3 in Factoring trinomials

Algebra9

Polynomials10

Factoring trinomials3

Factoring when the leading coefficient is one Factoring when the leading coefficient is not one

Factoring completely and choosing a strategy

When you reach for it

You now have five ways to factor: taking out a common factor, a trinomial with leading coefficient one, a trinomial with any leading coefficient, grouping, and the special patterns. Each lesson handed you a polynomial already shaped for its method. Out in the open nobody says which method a polynomial needs, and one method is often not enough: 3x^2 - 75 needs two, and x^4 - 256 needs the same pattern twice.

This lesson is about the whole job. Reach for it whenever a polynomial must be factored and the result will be used: to cancel factors in a fraction, to read off where a product is zero, or to split a volume into its edges. Each of those needs every factor, so stopping early gives a wrong answer later.

The procedure

Day 1: what “completely” means, and the common factor first

A polynomial with integer coefficients and degree at least 1 is prime over the integers when its coefficients share no common factor other than 1 and it cannot be written as a product of two polynomials with integer coefficients, each of degree at least 1. So x + 4, 2x - 3 and x^2 + 9 are prime; 2x + 8 = 2(x + 4) is not, and nor is x^2 - 9 = (x + 3)(x - 3).

A polynomial is factored completely when it is written as a monomial in front (a number times powers of the variables, possibly just 1 or -1) times factors that are each prime over the integers. A prime factor may appear more than once, written as a power. The number in front is left as one number: 6(x + 2), not 2 \cdot 3(x + 2).

For example, 18x + 24 = 2(9x + 12) is true but not finished, because 9x + 12 still holds a 3. Finished, it is 6(3x + 4).

Take out the greatest common factor first. It has to be in the answer anyway: every term contains it, so by the distributive property it comes out whatever else happens. Taking it out first pays twice.

It leaves smaller numbers to search. In 3x^2 + 6x - 105 the trinomial search would hunt among the pairs of 3 \cdot (-105) = -315; after the 3 comes out, it hunts among the pairs of -35 (Example 1).

It uncovers patterns. Neither 3x^2 nor 75 is a perfect square, so 3x^2 - 75 does not look like a difference of squares. Take out the 3 and it is one:

\begin{aligned} 3x^2 - 75 &= 3(x^2 - 25) \\ &= 3(x + 5)(x - 5) \end{aligned}

A negative leading coefficient comes out with the common factor. The trinomial methods are set out for a positive first term, so take out -1 times the greatest common factor. Every sign inside the bracket flips, and the bracket then starts with a positive term (Example 3).

Then count the terms. Every method undoes one shape of product, and the shape of a product fixes how many terms it has. A sum times a difference loses its middle terms, so it leaves two. A binomial times a binomial with like terms to combine leaves three. Two binomials with no like terms leave four. So the number of terms in what is left points at the method:

What you see (after the common factor)	What it suggests	Why
Two terms, a^2 - b^2	(a + b)(a - b), difference of squares	(a + b)(a - b) has middle terms ab and -ab, which cancel
Two terms, a^3 + b^3 or a^3 - b^3	(a + b)(a^2 - ab + b^2) or (a - b)(a^2 + ab + b^2)	the same canceling, one degree up
Two terms, a sum of two squares of degree 2, such as x^2 + 9	prime over the integers	a linear factor would give a real number where it is zero, and a sum of two squares is never zero
Three terms, first and last perfect squares, middle twice their product	(a + b)^2 or (a - b)^2	the square of a binomial has exactly these three terms
Three terms, leading coefficient 1	two numbers with product c and sum b	(x + p)(x + q) = x^2 + (p + q)x + pq
Three terms, leading coefficient not 1	split the middle term, or trial and check	the middle term is made of two products
Four terms	grouping	a binomial times a binomial with no like terms gives four

The table points; it does not decide. If the method it suggests fails, the polynomial may be prime, or the terms may need another order.

Check, in two senses. First, is the answer equal to the polynomial? Multiply it back out, or, as a quick test, substitute a number into both. A value such as 2 or 3 catches more than 0 or 1, which make many terms vanish or all powers equal. Second, is it complete? Look at every factor again and ask whether any of them still has a common factor or fits a shape in the table.

Day 2: factoring again, and choosing

Every new factor gets the same look. A method can produce a factor that fits a pattern itself:

\begin{aligned} x^4 - 256 &= (x^2 + 16)(x^2 - 16) \\ &= (x^2 + 16)(x + 4)(x - 4) \end{aligned}

x^2 - 16 is a difference of squares, so it splits again; x^2 + 16 is a sum of two squares, so it stays. Stopping after one step is the usual way an answer ends up equal but not complete.

Cubics with a common factor. When x is in every term, it comes out with the number, and a quadratic is left:

\begin{aligned} 2x^3 - 6x^2 - 80x &= 2x(x^2 - 3x - 40) \\ &= 2x(x - 8)(x + 5) \end{aligned}

since -8 \cdot 5 = -40 and -8 + 5 = -3.

Several variables. The common factor can include variables, as in 2y or 3xy, and the patterns work with a and b standing for whole terms: 9x^2 - 4y^2 has a = 3x and b = 2y.

When nothing works. A polynomial is prime over the integers only when a method’s own test says so, for example when no pair of integers has the right product and sum. Failing to spot a pair is not the same thing; list the pairs.

Once a polynomial is completely factored, setting it equal to zero splits it into one small equation per factor, which is the zero-product property.

Worked examples

Example 1

Day 1: a common factor, then a trinomial

Factor 3x^2 + 6x - 105 completely.

Solution

All three coefficients are multiples of 3, and x is not in the last term, so the greatest common factor is 3:

3x^2 + 6x - 105 = 3(x^2 + 2x - 35)

Three terms with leading coefficient 1: two numbers with product -35 and sum 2. They are 7 and -5. So

3x^2 + 6x - 105 = 3(x + 7)(x - 5)

Each bracket has degree 1 and no common factor, so each is prime. Check at x = 2: 12 + 12 - 105 = -81, and 3 \cdot 9 \cdot (-3) = -81.

Skipping the first step still gives a true answer. Splitting 6x in 3x^2 + 6x - 105 needs a pair with product -315 and sum 6, which is 21 and -15, and grouping gives 3x(x + 7) - 15(x + 7) = (x + 7)(3x - 15). That multiplies back correctly, but it is not complete: 3x - 15 = 3(x - 5) still holds the 3 that the first step would have taken out at once.

Example 2

Day 1: patterns that appear after the common factor

Factor completely: (a) 12x^2 - 27 \qquad (b) 8y^2 - 40y + 50 \qquad (c) 2x^3 + 54

Solution

(a) The greatest common factor is 3, and the bracket is a difference of squares, since 4x^2 = (2x)^2 and 9 = 3^2:

\begin{aligned} 12x^2 - 27 &= 3(4x^2 - 9) \\ &= 3(2x + 3)(2x - 3) \end{aligned}

(b) The greatest common factor is 2. Inside, 4y^2 = (2y)^2, 25 = 5^2 and 2 \cdot 2y \cdot 5 = 20y, so the bracket is a perfect square trinomial with a minus sign in the middle:

\begin{aligned} 8y^2 - 40y + 50 &= 2(4y^2 - 20y + 25) \\ &= 2(2y - 5)^2 \end{aligned}

\begin{aligned} 2x^3 + 54 &= 2(x^3 + 27) \\ &= 2(x + 3)(x^2 - 3x + 9) \end{aligned}

The quadratic factor is prime: two integers with product 9 and sum -3 would have to be both negative, and -1, -9 and -3, -3 add to -10 and -6.

In all three the pattern was hidden. 12x^2 and 27 are not perfect squares, nor are 8y^2 and 50, and 54 is not a cube; each pattern sat behind the common factor.

Example 3

Day 1: a negative leading coefficient, and a cable channel

(a) Factor -3x^2 + 15x + 108 completely.

(b) A steel strip 28 cm wide is bent into an open channel that carries electric cables along a workshop floor, turning up x cm along each long edge. The cables lie in the channel’s cross-section. Write the area of that cross-section, factor it completely, and use the factored form to find the area when x = 5.

Figure 1Turning up x cm on each side leaves a floor 28 - 2x cm wide. The shaded cross-section has area x(28 - 2x).

Solution

(a) The common factor is 3, and the first term is negative, so take out -3. Dividing each term by -3 flips every sign:

\begin{aligned} -3x^2 + 15x + 108 &= -3(x^2 - 5x - 36) \\ &= -3(x - 9)(x + 4) \end{aligned}

since -9 \cdot 4 = -36 and -9 + 4 = -5.

(b) The two turned-up sides are walls of height x, and the floor is what is left of the width, 28 - 2x. The cross-section is a rectangle, so its area in square centimeters is

\begin{aligned} x(28 - 2x) &= 28x - 2x^2 \\ &= -2x^2 + 28x \end{aligned}

The common factor is 2x, and the leading coefficient is negative, so take out -2x:

-2x^2 + 28x = -2x(x - 14)

The same product can be written 2x(14 - x), moving the minus sign into the bracket, and this form reads the channel: height x times floor 28 - 2x = 2(14 - x). It also shows that only 0 < x < 14 makes a channel, since the floor needs a positive width. At x = 5 the area is 2 \cdot 5 \cdot 9 = 90 square centimeters; check directly, height 5 times floor 18 is 90.

Example 4

Day 1: a common factor, then grouping, in two variables

Factor 4ab + 12a - 2b - 6 completely.

Solution

All four coefficients are even, and no variable is in every term, so take out 2:

\begin{aligned} 4ab + 12a &- 2b - 6 \\ &= 2(2ab + 6a - b - 3) \end{aligned}

Four terms suggest grouping. The first pair shares 2a; take -1 out of the second pair so the brackets match:

\begin{aligned} 2(2ab &+ 6a - b - 3) \\ &= 2\big[2a(b + 3) - 1(b + 3)\big] \\ &= 2(b + 3)(2a - 1) \end{aligned}

Check at a = 2, b = 1: 8 + 24 - 2 - 6 = 24, and 2 \cdot 4 \cdot 3 = 24.

Grouping straight away works too, 4a(b + 3) - 2(b + 3) = (b + 3)(4a - 2), but 4a - 2 still holds a 2 and must be finished.

Example 5

Day 2: choosing by the shape

Factor completely, or say that the polynomial is prime over the integers.

(a) 50 - 2y^2 \qquad (b) 4x^2 + 25 \qquad (c) 6x^2 - x - 15

Solution

(a) Common factor 2, then two terms: 25 = 5^2 minus y^2, a difference of squares.

\begin{aligned} 50 - 2y^2 &= 2(25 - y^2) \\ &= 2(5 + y)(5 - y) \end{aligned}

Written with y first, the same answer is -2(y + 5)(y - 5), since 5 - y = -(y - 5). Both are complete.

(b) The coefficients 4 and 25 share no factor and there is no variable in both terms. Two terms with a plus sign between squares: a sum of two squares of degree 2, so prime over the integers. 4x^2 + 25 is positive for every real x, so no factor px + q could ever be zero.

(c) No common factor; three terms with leading coefficient 6. Split the middle term: a pair with product 6 \cdot (-15) = -90 and sum -1 is -10 and 9.

\begin{aligned} 6x^2 - x &- 15 \\ &= 6x^2 - 10x + 9x - 15 \\ &= 2x(3x - 5) + 3(3x - 5) \\ &= (2x + 3)(3x - 5) \end{aligned}

Example 6

Day 2: factoring again

Factor completely: (a) x^4 - 13x^2 + 36 \qquad (b) 2x^3 + 10x^2 - 8x - 40

Solution

(a) No common factor. The powers are x^4 = (x^2)^2, x^2 and a constant, so read it as a trinomial in the block x^2: two numbers with product 36 and sum -13, which are -4 and -9.

x^4 - 13x^2 + 36 = (x^2 - 4)(x^2 - 9)

Now look at each factor again. Both are differences of squares:

\begin{aligned} x^4 - 13x^2 &+ 36 \\ &= (x + 2)(x - 2)(x + 3)(x - 3) \end{aligned}

Four linear factors in a degree-4 polynomial: nothing is left to split.

(b) Common factor 2, then four terms, so group:

\begin{aligned} 2x^3 + 10x^2 &- 8x - 40 \\ &= 2(x^3 + 5x^2 - 4x - 20) \\ &= 2\big[x^2(x + 5) - 4(x + 5)\big] \\ &= 2(x + 5)(x^2 - 4) \\ &= 2(x + 5)(x + 2)(x - 2) \end{aligned}

Grouping left x^2 - 4, which a second look splits. Check at x = 1: 2 + 10 - 8 - 40 = -36, and 2 \cdot 6 \cdot 3 \cdot (-1) = -36.

Example 7

Day 2: the edges of a shipping crate

Carmine builds a wooden shipping crate, a box whose volume is 2x^3 + 18x^2 + 28x cubic decimeters, where x is a length in decimeters. The crate is 2x dm tall. Find the length and width of its base, and all three dimensions when x = 4.

Figure 2The crate in Example 7. Its volume 2x^3 + 18x^2 + 28x factors into its three edges.

Solution

The volume of a box is height times length times width, so a complete factoring offers the pieces. The common factor is 2x:

\begin{aligned} 2x^3 + 18x^2 + 28x &= 2x(x^2 + 9x + 14) \\ &= 2x(x + 2)(x + 7) \end{aligned}

since 2 \cdot 7 = 14 and 2 + 7 = 9. The height accounts for the factor 2x, so the base is x + 7 dm long and x + 2 dm wide.

Factoring alone could not have decided this. The same volume is x(2x + 4)(x + 7), which would be a crate x tall: the number 2 can join any factor. The complete factoring gives the prime pieces; the stated height says how they group into edges.

When x = 4 the crate is 8 dm tall, 11 dm long and 6 dm wide. Check: 8 \cdot 11 \cdot 6 = 528, and 2 \cdot 64 + 18 \cdot 16 + 28 \cdot 4 = 128 + 288 + 112 = 528.

Boundaries

“Completely” depends on which numbers are allowed. Everything here is over the integers. x^2 - 5 is prime over the integers, yet over the real numbers it is \displaystyle (x + \sqrt{5})(x - \sqrt{5}).

Apart from order and signs, the answer is unique. Every correct route ends at the same prime factors. They may be listed in any order, and a factor -1 may move between factors, as in 2(5 + y)(5 - y) = -2(y + 5)(y - 5).

The “sum of squares is prime” row is for degree 2. x^2 + 9 and 4x^2 + 25 are prime, but a higher-degree sum can factor: 4x^4 + 1 = (2x^2 - 2x + 1)(2x^2 + 2x + 1).

The table covers the shapes in this course, not every polynomial. x^3 - 3x + 2 has three terms but is not a quadratic, nor a quadratic in a block like x^2, so the trinomial rows do not fit; yet it equals (x - 1)^2(x + 2). Finding factors of such a polynomial needs methods from later courses.

A substituted number can catch a mistake but cannot prove an answer. Two different expressions can agree at one value. Only multiplying out shows the two are equal for every x.

Where this goes next

A completely factored polynomial is the form later work asks for. Set equal to zero, it gives every solution at once through the zero-product property, one small equation per factor. In a fraction of polynomials, only factors can cancel, and only complete factoring of the top and bottom shows every common factor; it also shows each value that makes the bottom zero, which simplifying a rational expression must exclude. The same habit, take out what is common and then look at each piece again, runs through all of factoring.

Exercises

Day 1: a common factor, then one more method

Factor completely.

3x^2 + 21x + 30
2y^2 - 14y + 24
6x^2 + 3x - 45
7m^2 - 28
20x^2 + 60x + 45
5a^3 - 40
18x^2y - 8y^3

Day 1: negative leading coefficients, four terms, a channel

Factor completely: -x^2 - 3x + 28
Factor completely: -4t^2 + 8t + 60
Factor completely: 3xy + 6x - 3y - 6
A copper strip 32 cm wide is bent into a cable channel the same way, turning up x cm along each long edge. Write the area of the cross-section, factor it completely, and use the factored form to find the area when x = 6.

Day 1: reasoning

Is 4x^2 - 36 = (2x + 6)(2x - 6) true? Is it complete? If not, finish it.
Factor 10x^2 + 120x + 350 completely, and explain why taking out the common factor first makes the search for two numbers shorter.

Day 2: choose a method

Factor completely, or say that the polynomial is prime over the integers.

32 - 2x^2
3x^2 + 27
12x^2 - 5x - 2
4a^2 + 28ab + 49b^2

Day 2: factor again

Factor completely.

y^3 + 4y^2 - 9y - 36
16x^4 - 1
x^4 - 10x^2 + 9
4x^4 - 4x^3 - 24x^2

Day 2: a crate

A crate has volume 3x^3 + 21x^2 + 36x cubic decimeters and is 3x dm tall. Factor the volume completely, find the length and width of the base, and give all three dimensions when x = 2.

Day 2: reasoning

The factoring 32x^4 - 162 = 2(4x^2 + 9)(4x^2 - 9) stops there. Finish it, and explain why 4x^2 + 9 stays as it is.
A proposed factoring of 3x^3 - 12x is 3x(x - 2)^2. Substituting x = 0 and then x = 2 makes both sides agree each time. Explain why both checks passed, show with a better value that the factoring is wrong, and factor correctly.

Answers

3(x + 2)(x + 5). Take out 3: x^2 + 7x + 10, and 2 \cdot 5 = 10, 2 + 5 = 7.
2(y - 3)(y - 4). Take out 2: y^2 - 7y + 12, and -3, -4 have product 12 and sum -7.
3(2x - 5)(x + 3). Take out 3: 2x^2 + x - 15; the pair 6, -5 has product -30 and sum 1, so 2x(x + 3) - 5(x + 3).
7(m + 2)(m - 2). Take out 7: m^2 - 4 is a difference of squares.
5(2x + 3)^2. Take out 5: 4x^2 + 12x + 9 has (2x)^2, 3^2 and 2 \cdot 2x \cdot 3 = 12x.
5(a - 2)(a^2 + 2a + 4). Take out 5: a^3 - 8 is a difference of cubes with b = 2.
2y(3x + 2y)(3x - 2y). Take out 2y: 9x^2 - 4y^2 = (3x)^2 - (2y)^2.
-(x + 7)(x - 4). Take out -1: x^2 + 3x - 28, and 7 \cdot (-4) = -28, 7 - 4 = 3.
-4(t - 5)(t + 3). Take out -4: t^2 - 2t - 15, and -5 \cdot 3 = -15, -5 + 3 = -2.
3(y + 2)(x - 1). Take out 3: xy + 2x - y - 2 = x(y + 2) - 1(y + 2).
x(32 - 2x) = -2x^2 + 32x = -2x(x - 16), or 2x(16 - x). At x = 6: 2 \cdot 6 \cdot 10 = 120 square centimeters, which is height 6 times floor 20.
True: (2x + 6)(2x - 6) = 4x^2 - 36. Not complete, since each bracket holds a 2: 2(x + 3) \cdot 2(x - 3) = 4(x + 3)(x - 3).
10(x + 5)(x + 7). After the 10 comes out, the search is for a pair with product 35 and sum 12, found at once as 5 and 7. Without it, splitting the middle term needs a pair with product 10 \cdot 350 = 3500 and sum 120, among the 12 positive pairs of 3500, and the result, such as (x + 5)(10x + 70), still holds the 10 to take out.
2(4 + x)(4 - x), or -2(x + 4)(x - 4). Take out 2: 16 - x^2 is a difference of squares.
3(x^2 + 9). The bracket is a sum of two squares of degree 2, prime over the integers.
(4x + 1)(3x - 2). No common factor; the pair -8, 3 has product -24 and sum -5: 4x(3x - 2) + 1(3x - 2).
(2a + 7b)^2. 4a^2 = (2a)^2, 49b^2 = (7b)^2 and 2 \cdot 2a \cdot 7b = 28ab.
(y + 4)(y + 3)(y - 3). Group: y^2(y + 4) - 9(y + 4) = (y + 4)(y^2 - 9), then split y^2 - 9.
(4x^2 + 1)(2x + 1)(2x - 1). 16x^4 - 1 = (4x^2 + 1)(4x^2 - 1), and 4x^2 - 1 splits again.
(x + 1)(x - 1)(x + 3)(x - 3). In the block x^2: (x^2 - 1)(x^2 - 9), and both split.
4x^2(x - 3)(x + 2). Take out 4x^2: x^2 - x - 6, and -3 \cdot 2 = -6, -3 + 2 = -1.
3x(x + 3)(x + 4). The height takes 3x, so the base is x + 4 dm by x + 3 dm. At x = 2: 6 dm tall, 6 dm by 5 dm, volume 180 cubic decimeters, matching 24 + 84 + 72 = 180.
2(4x^2 + 9)(2x + 3)(2x - 3), since 4x^2 - 9 = (2x)^2 - 3^2. 4x^2 + 9 is a sum of two squares of degree 2, and its coefficients 4 and 9 share no factor. It is at least 9 for every real x, so it is never zero, while a factor px + q is zero at \displaystyle x = -\frac{q}{p}. So it has no factor of degree 1 and is prime over the integers.
The correct factoring, 3x(x + 2)(x - 2), is zero at x = 0 and x = 2, and so is the proposed one, because both contain the factors x and x - 2. Values where both sides are zero cannot tell them apart. At x = 1: 3 - 12 = -9, but 3 \cdot 1 \cdot 1 = 3. The right answer is 3x(x + 2)(x - 2), from 3x(x^2 - 4).