Multiplying two binomials gave . Given only that answer, how do you find the two binomials?

Algebra 1 · A.SSE.A.2

Factoring when the leading coefficient is one

Multiplying two binomials gave x^2 + bx + c. Given only that answer, how do you find the two binomials?

You need

Factoring out the greatest common factorMath 7 Multiplying two binomialsAlgebra 1

Unlocks

Factoring when the leading coefficient is not oneAlgebra 1 Solving quadratics by factoringAlgebra 1

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

Multiplying two binomials turns a product into a sum:

(x + 2)(x + 11) = x^2 + 13x + 22

Factoring runs the same road the other way. You are handed x^2 + 13x + 22 and asked for the two binomials that multiply to it. The answer is an equivalent expression: it has the same value as the trinomial for every x, written as a product instead of a sum.

Reach for this move when a trinomial x^2 + bx + c, with leading coefficient 1 and integer b and c, would be more useful as a product: to see an area as a length times a width, to find where an expression equals zero, or to cancel a factor shared by the top and bottom of a fraction. Factoring out a common factor pulled one shared factor to the front. Here the terms share no factor, and the trinomial splits into two binomials instead.

This lesson runs over two days. Day 1 builds the idea and handles a positive constant c. Day 2 handles a negative c, trinomials that do not factor, two variables, and terms written out of order.

The procedure

Where the two numbers come from

Multiply two binomials that both begin with x, where p and q stand for any two numbers:

\begin{aligned} (x + p)(x + q) &= x^2 + qx + px + pq \\ &= x^2 + (p + q)x + pq \end{aligned}

The first line is the distributive property, used twice. The second gathers the two x terms, which is the distributive property read backwards: qx + px = (p + q)x. So the middle coefficient of the product is the sum of the two numbers, and the constant is their product.

Read right to left, this says how to factor x^2 + bx + c: look for two numbers p and q with

p + q = b \qquad\text{and}\qquad pq = c

and then x^2 + bx + c = (x + p)(x + q).

Two facts make this search the whole job. First, any pair that works gives a factorization, by the identity above. Second, no factorization of another shape is being missed. If x^2 + bx + c is a product of two binomials with integer coefficients, the coefficients of x in them multiply to 1, so both are 1 or both are -1; and (-x - p)(-x - q) is the same product as (x + p)(x + q). The order of the two factors does not matter either, since (x + p)(x + q) = (x + q)(x + p) by the commutative property of multiplication.

The picture: tiles into a rectangle

A collage has one square tile x by x, seven strips x long and 1 wide, and six unit tiles 1 by 1. Their total area is x^2 + 7x + 6. Can they be laid edge to edge as one rectangle?

Figure 1One x^2 square, seven x-strips and six unit tiles fit together as a rectangle x + 6 wide and x + 1 tall, so x^2 + 7x + 6 = (x + 6)(x + 1).

The square sits in a corner, and the strips line up along two of its sides, some to the right and some below. That leaves a gap in the far corner, and the unit tiles must fill it exactly. With p strips standing to the right and q lying below, the gap is p tiles wide and q tiles tall, so it holds pq tiles, and the strips number p + q. The tile puzzle is the number puzzle: p + q = 7 and pq = 6.

Six tiles can fill a gap 6 by 1, which needs 6 + 1 = 7 strips. They can also fill a gap 3 by 2, but that uses only 5 strips and leaves two over. So the rectangle is x + 6 by x + 1, and

x^2 + 7x + 6 = (x + 6)(x + 1)

The picture works only when p and q are positive, since no strip has a negative width. It shows why the method works; the algebra above covers every sign.

Day 1: finding the pair when c is positive

Search the products, not the sums. Infinitely many pairs of integers add to b, but only a few multiply to c, since each number in such a pair divides c. So list the factor pairs of c, in order: start with 1 and c, then try 2, 3, 4 and so on as the smaller number, keeping each one that divides c. Stop once the smaller number passes \displaystyle \sqrt{c}, because after that the pairs repeat in reverse. For 24 the list is 1 \cdot 24, 2 \cdot 12, 3 \cdot 8, 4 \cdot 6, and it stops there since \displaystyle 5 > \sqrt{24} \approx 4.9.

Read the signs first. When c is positive, p and q have the same sign, because a positive times a negative is negative. Two positive numbers add to a positive number and two negative numbers add to a negative one, so both numbers take the sign of b. That halves the search: list the pairs of c with the sign of b on both.

Check by multiplying back. A factorization claims two expressions are equal, and expanding the product tests the claim in one line.

Day 2: a negative c, and when no pair works

Opposite signs. When c is negative, one of p and q is positive and the other negative. The sum of a positive and a negative number is the difference of their sizes, carrying the sign of the one with the larger size. So look for a factor pair of c whose sizes differ by the size of b, and give the larger one the sign of b. If b = 0, the two sizes are equal.

When no pair works. If the complete list of factor pairs holds no pair with the right sum, the trinomial cannot be written as (x + p)(x + q) with integers p and q. It is then called prime over the integers. The full list is the reason, not a failure to spot the pair: any pair of integers with product c is on it, because each number divides c.

Two variables. The same identity holds with py and qy in place of p and q:

\begin{aligned} (x + py)(x + qy) &= x^2 + qxy + pxy + pqy^2 \\ &= x^2 + (p + q)xy + pqy^2 \end{aligned}

so x^2 + bxy + cy^2 is factored with the same two numbers, each followed by y.

Order and missing terms. Read b and c only after the terms are in the order x^2, then x, then the constant. Reordering a sum is the commutative property of addition, so the value does not change. If there is no x term at all, then b = 0.

Worked examples

Example 1

Day 1: both numbers positive

Factor x^2 + 11x + 18.

Solution

Here b = 11 and c = 18. The constant is positive, so the two numbers share a sign, and the sign of b makes them both positive. List the factor pairs of 18:

Factors of 18	Sum
1,\ 18	19
2,\ 9	11
3,\ 6	9

The list is complete: 4 does not divide 18, and 5 is past \displaystyle \sqrt{18} \approx 4.2. The pair 2 and 9 multiplies to 18 and adds to 11, so

x^2 + 11x + 18 = (x + 2)(x + 9)

Check. (x + 2)(x + 9) = x^2 + 9x + 2x + 18, which is x^2 + 11x + 18.

You may stop as soon as a pair works. The whole table is written out here to show that no other pair works, so this factorization is the only one apart from the order of the factors.

Example 2

Day 1: both numbers negative

Factor x^2 - 11x + 24.

Solution

The constant 24 is positive, so the numbers share a sign, and b = -11 is negative, so both are negative. List the pairs of 24 with a minus sign on each:

Factors of 24	Sum
-1,\ -24	-25
-2,\ -12	-14
-3,\ -8	-11
-4,\ -6	-10

The pair -3 and -8 has product 24 and sum -11, so

x^2 - 11x + 24 = (x - 3)(x - 8)

Check. (x - 3)(x - 8) = x^2 - 8x - 3x + 24, which is x^2 - 11x + 24.

The same 3 and 8 factor x^2 + 11x + 24 as (x + 3)(x + 8). Changing the sign of b alone, with c positive, changes only the signs inside the brackets.

Example 3

Day 1: the sides of a stained-glass panel

A rectangular stained-glass panel has a square pane of clear glass in one corner, x centimeters on each side, and colored glass filling the rest. The whole panel has area x^2 + 20x + 96 square centimeters, and each side is x plus a whole number of centimeters. Find the two sides.

Figure 2The stained-glass panel of Example 3. Its four panes have areas x^2, 12x, 8x and 96, and its sides are x + 12 and x + 8.

Solution

Call the sides x + p and x + q, with p and q whole numbers. Area is length times width, so

\begin{aligned} (x + p)(x + q) &= x^2 + (p + q)x + pq \\ &= x^2 + 20x + 96 \end{aligned}

and we need p + q = 20 and pq = 96. Since \displaystyle \sqrt{96} \approx 9.8, list the pairs up to 9 (of the numbers up to 9, only 5, 7 and 9 do not divide 96):

Factors of 96	Sum
1,\ 96	97
2,\ 48	50
3,\ 32	35
4,\ 24	28
6,\ 16	22
8,\ 12	20

The sides are x + 8 and x + 12 centimeters. The figure shows the four panes: the clear square x^2, the strips 12x and 8x, and the corner 8 \cdot 12 = 96.

Check. (x + 8)(x + 12) = x^2 + 12x + 8x + 96, which is x^2 + 20x + 96. With a number as well: if x = 10, the panel is 18 by 22, which is 396 square centimeters, and 10^2 + 20 \cdot 10 + 96 = 396.

Only one pair in the table works, so no other whole-number sides give this area formula.

Example 4

Day 2: a negative constant

Factor (a) x^2 + 5x - 50 \qquad (b) x^2 - 5x - 50

Solution

In both, c = -50 is negative, so the two numbers have opposite signs, and their sum is the difference of their sizes. Look for a pair of 50 whose sizes differ by 5:

Factors of 50	Difference
1,\ 50	49
2,\ 25	23
5,\ 10	5

(a) Here b = 5 is positive, so the larger size, 10, is positive: the numbers are 10 and -5, with product -50 and sum 5.

x^2 + 5x - 50 = (x + 10)(x - 5)

Check. x^2 - 5x + 10x - 50 = x^2 + 5x - 50.

(b) Now b = -5, so the larger size is negative: the numbers are -10 and 5.

x^2 - 5x - 50 = (x - 10)(x + 5)

Check. x^2 + 5x - 10x - 50 = x^2 - 5x - 50.

One table served both. The sign of b decided only which number carries the minus.

Example 5

Day 2: when no pair works

Factor, or show that the trinomial is prime over the integers.

(a) x^2 + 6x - 20 \qquad (b) x^2 - 9x + 16

Solution

(a) The constant is negative, so the signs are opposite and the sizes must differ by 6.

Factors of 20	Difference
1,\ 20	19
2,\ 10	8
4,\ 5	1

No pair differs by 6. Every pair of integers with product -20 is one of these three with a minus sign on one number, so none has been missed. x^2 + 6x - 20 is prime over the integers.

(b) The constant is positive and b is negative, so both numbers are negative and their sizes must add to 9.

Factors of 16	Sum
-1,\ -16	-17
-2,\ -8	-10
-4,\ -4	-8

No pair adds to -9, so x^2 - 9x + 16 is prime over the integers.

A prime trinomial is not broken. It still has a value for every x; it just has no factorization of the form (x + p)(x + q) with integers p and q.

Example 6

Day 2: two variables

Factor (a) x^2 + 7xy + 12y^2 \qquad (b) a^2 - ab - 42b^2

Solution

(a) Look for (x + py)(x + qy), which expands to x^2 + (p + q)xy + pqy^2. So p + q = 7 and pq = 12, both positive. The pairs of 12 are 1, 12 (sum 13), 2, 6 (sum 8) and 3, 4 (sum 7).

x^2 + 7xy + 12y^2 = (x + 3y)(x + 4y)

Check. Expanding gives x^2 + 4xy + 3xy + 12y^2, which is x^2 + 7xy + 12y^2.

(b) The coefficient of ab is -1 and the coefficient of b^2 is -42. The numbers have opposite signs, their sizes differ by 1, and the larger one is negative. Among the pairs of 42 (1, 42; 2, 21; 3, 14; 6, 7), only 6 and 7 differ by 1, so the numbers are -7 and 6.

a^2 - ab - 42b^2 = (a - 7b)(a + 6b)

Check. Expanding gives a^2 + 6ab - 7ab - 42b^2, which is a^2 - ab - 42b^2.

Example 7

Day 2: terms out of order, and a missing term

Factor (a) 30 + x^2 - 11x \qquad (b) x^2 - 121

Solution

(a) Put the terms in order by the commutative property of addition: x^2 - 11x + 30. Now b = -11 and c = 30 can be read off. Both numbers are negative and their sizes add to 11. The pairs of 30 are 1, 30; 2, 15; 3, 10; 5, 6, and only 5 + 6 = 11.

30 + x^2 - 11x = (x - 5)(x - 6)

Check. x^2 - 6x - 5x + 30 = x^2 - 11x + 30, which is the original expression reordered.

(b) There is no x term, so this is x^2 + 0x - 121 with b = 0. The numbers need sum 0 and product -121: opposite signs and equal sizes, so 11 and -11.

x^2 - 121 = (x + 11)(x - 11)

Check. x^2 - 11x + 11x - 121 = x^2 - 121. The two middle terms cancel, which is exactly why the expression had no x term. This is a difference of two squares, a pattern with its own lesson.

Boundaries

The leading coefficient must be 1. The search for p + q = b and pq = c comes from (x + p)(x + q), whose x^2 term is exactly x^2. When every term shares a factor, take it out first, as in 2x^2 + 26x + 72 = 2(x^2 + 13x + 36), which is 2(x + 4)(x + 9); that order of work is factoring completely. When the leading coefficient is not 1 and nothing divides out, as in 3x^2 + 8x + 5, the move changes, and its own lesson handles it.

Prime means prime over the integers. The list of factor pairs covers integer p and q only. Some trinomials that are prime here can be split using square roots, but that needs other tools and is not a factorization over the integers.

The tile picture needs positive numbers. A strip of width -3 cannot be laid down. The identity (x + p)(x + q) = x^2 + (p + q)x + pq holds for every p and q, and it is what the method rests on.

A missing x term with a positive constant never factors. For x^2 + c with c > 0, the sum p + q = 0 forces q = -p, and then pq = -p^2 is never positive.

Where this goes next

Factoring gives an equivalent expression; it does not solve anything by itself. Setting x^2 - 11x + 24 = 0 and using its factored form (x - 3)(x - 8) is the work of solving by factoring. Read as a function, the factored form shows where a graph meets the x-axis, which is the subject of the factored form of a quadratic and roots and x-intercepts.

When the leading coefficient is not 1, the same search for two numbers returns with the product ac in place of c, and the middle term is split into two pieces that are then factored by grouping. The special patterns, such as the difference of squares in Example 7, are faster routes for particular shapes, and factoring completely puts every move in order.

Exercises

Day 1: both numbers positive

Factor.

x^2 + 10x + 9
x^2 + 13x + 42
n^2 + 16n + 48
x^2 + 15x + 56

Day 1: both numbers negative

Factor.

x^2 - 9x + 14
y^2 - 14y + 45
x^2 - 17x + 72

Day 1: tiles, panels and missing numbers

Fiametta’s collage has one x^2 square, nine x-strips and eight unit tiles. She lays them edge to edge as one rectangle. What are its sides?
A stained-glass panel has area x^2 + 14x + 48 square centimeters, and each side is x plus a whole number of centimeters. Find the sides.
Find every integer b for which x^2 + bx + 12 factors as (x + p)(x + q) with integers p and q.

Day 1: reasoning

In x^2 + bx + c, the constant c is positive and b is negative. Explain why p and q must both be negative.
Explain why it is quicker to list the pairs of integers that multiply to c than the pairs that add to b.

Day 2: a negative constant

Factor.

x^2 + 3x - 40
x^2 - 4x - 32
t^2 - t - 56
x^2 + 10x - 39

Day 2: factor, or show it is prime

x^2 + 5x + 9
x^2 + 2x - 63
Gunnar has one x^2 square, seven x-strips and eight unit tiles. Can he lay all of them edge to edge as one rectangle?

Day 2: two variables, order and missing terms

Factor.

m^2 - 12mn + 27n^2
a^2 + 4ab - 45b^2
6x + x^2 - 27

Day 2: reasoning

When c is negative, explain why the number with the larger size takes the sign of b.
Show that x^2 + 144 is prime over the integers, and that x^2 - 144 is not.

Answers

(x + 1)(x + 9). Both positive: 1 \cdot 9 = 9 and 1 + 9 = 10; the other pair, 3, 3, adds to 6.
(x + 6)(x + 7). Pairs of 42: 1, 42; 2, 21; 3, 14; 6, 7, and only 6 + 7 = 13.
(n + 4)(n + 12). Pairs of 48: 1, 48; 2, 24; 3, 16; 4, 12; 6, 8, and only 4 + 12 = 16.
(x + 7)(x + 8). 7 \cdot 8 = 56 and 7 + 8 = 15.
(x - 2)(x - 7). Both negative: (-2)(-7) = 14 and -2 + (-7) = -9.
(y - 5)(y - 9). (-5)(-9) = 45 and -5 + (-9) = -14.
(x - 8)(x - 9). Pairs of 72: 1, 72; 2, 36; 3, 24; 4, 18; 6, 12; 8, 9, and only 8 + 9 = 17.
x + 8 by x + 1. The area is x^2 + 9x + 8, so p + q = 9 and pq = 8: 1 and 8 (the pair 2, 4 adds to 6). Eight strips stand beside the square, one lies below, and the unit tiles fill an 8 by 1 gap.
x + 6 and x + 8 centimeters. Pairs of 48 with their sums: 1, 48 (49); 2, 24 (26); 3, 16 (19); 4, 12 (16); 6, 8 (14).
b = 13, 8, 7, -7, -8, -13. The pairs with product 12 are 1, 12; 2, 6; 3, 4 and the same with both negative, and b is the sum of a pair.
c = pq is positive, so p and q have the same sign. If both were positive, b = p + q would be positive. Since b is negative, both are negative.
When c \neq 0, each number in a pair with product c divides c, so there are only a few such pairs. Pairs that add to b never run out: 0 and b, 1 and b - 1, 2 and b - 2, and so on for ever.
(x + 8)(x - 5). Opposite signs, sizes differing by 3, the larger positive. Pairs of 40: 1, 40; 2, 20; 4, 10; 5, 8, and 8 - 5 = 3.
(x - 8)(x + 4). Sizes differing by 4, the larger negative: 8 - 4 = 4.
(t - 8)(t + 7). Here b = -1: sizes 7 and 8 differ by 1, and the larger is negative.
(x + 13)(x - 3). Pairs of 39: 1, 39 (difference 38) and 3, 13 (difference 10).
Prime over the integers. Both numbers positive with sum 5, but the pairs of 9 give 1 + 9 = 10 and 3 + 3 = 6.
(x + 9)(x - 7). Pairs of 63: 1, 63; 3, 21; 7, 9, and 9 - 7 = 2.
No. A rectangle would mean x^2 + 7x + 8 = (x + p)(x + q) with p + q = 7 and pq = 8, but the pairs of 8 add to 1 + 8 = 9 and 2 + 4 = 6. The trinomial is prime, so no rectangle exists.
(m - 3n)(m - 9n). Both numbers negative with product 27 and sum -12: -3 and -9.
(a + 9b)(a - 5b). Opposite signs, product -45, sizes differing by 4: 9 and -5.
(x + 9)(x - 3). In order it is x^2 + 6x - 27. Pairs of 27: 1, 27 (difference 26) and 3, 9 (difference 6).
One number is positive and one negative. Adding them takes the smaller size away from the larger and keeps the sign of the larger, so p + q has the sign of the number with the larger size. Since p + q = b, that sign is the sign of b. If b = 0 the sizes are equal.
x^2 + 144 = x^2 + 0x + 144 needs p + q = 0, so q = -p, and then pq = -p^2, which is never positive and so never 144. For x^2 - 144, the numbers 12 and -12 have sum 0 and product -144: x^2 - 144 = (x + 12)(x - 12).