Algebra 1 · A.APR.A.1

Multiplying polynomials

One term multiplies a whole sum, or a sum multiplies a sum. How do you multiply out every piece and know that none was missed?

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Degree, terms and standard formAlgebra 1

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Adding, subtracting and multiplying functionsAlgebra 2 Polynomial long divisionAlgebra 2 Factoring trinomialsGrade 9

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Multiplying polynomials

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When you reach for it

A product written as factors, such as 3x^2(2x^3 - 5x + 4) or (x + 1)(x + 2)(x + 3), hides what kind of polynomial it is. Multiplied out and collected, it shows its degree, its constant term and every coefficient, and it can be added to another polynomial term by term. Areas and volumes arrive as products, a length times a length, and multiplying out is how they become one polynomial.

The only tool is the distributive property, used as often as needed, with the exponent rules doing each small product. The lesson runs over two days. Day 1 multiplies a polynomial by a single term, a monomial. Day 2 multiplies a polynomial by a polynomial, starting from the two-binomial case of Multiplying two binomials and carrying the same idea to longer factors and to three factors in a row.

The procedure

Day 1: a monomial times a polynomial

The distributive property, a(b + c) = ab + ac, works for any number of terms in the bracket. For three, group the first two (the associative property of addition) and use it twice:

\begin{aligned} a(b + c + d) &= a\big((b + c) + d\big) \\ &= a(b + c) + ad \\ &= ab + ac + ad \end{aligned}

A fourth term is covered by the same step again. So a monomial times a polynomial is the monomial times each term, added up.

Each small product is a product of two monomials. Reorder and regroup its factors (the commutative and associative properties of multiplication) so the numbers sit together and each letter sits with its own powers. Then multiply the numbers and add the exponents on each base, since x^m \cdot x^n = x^{m+n}.

A subtraction inside the bracket is the addition of a negative term, so the sign travels with its term. In 3x^2(2x^3 - 5x + 4) the terms are 2x^3, -5x and 4:

\begin{aligned} 3x^2 \cdot 2x^3 &= 6x^5 \\ 3x^2 \cdot (-5x) &= -15x^3 \\ 3x^2 \cdot 4 &= 12x^2 \end{aligned}

Added up, 3x^2(2x^3 - 5x + 4) is 6x^5 - 15x^3 + 12x^2. Nothing needs collecting. Multiplying by 3x^2 raises every exponent by the same 2, so terms with different powers before still have different powers after. A monomial times a polynomial whose like terms are already collected never creates new like terms.

The area picture. When every term is positive, the product is an area. 4x(x + 2y + 3) is a strip 4x tall and x + 2y + 3 long. Cut it where each term of the length ends, and it falls into three rectangles:

Figure 1One strip. 4x(x + 2y + 3) is the whole area, and the three pieces are 4x^2, 8xy and 12x.

The whole area is the sum of the pieces, 4x^2 + 8xy + 12x: the distributive property, drawn. A negative term has no rectangle of negative width, but the algebra never needed one.

The degree adds. A term of degree m times a term of degree n has degree m + n, because the exponents add. So the highest-degree term of the polynomial gives the highest-degree term of the product, and its coefficient cannot be 0, being a product of two non-zero numbers. The degree of the product is the sum of the two degrees. Above, 3x^2 has degree 2, the bracket has degree 3, and the product’s leading term 6x^5 has degree 5.

Day 2: a polynomial times a polynomial

Multiplying two binomials treated the second bracket as one quantity and distributed it over the first: (x + 3)(x + 5) = x(x + 5) + 3(x + 5). Nothing in that move needed the brackets to be short. For a binomial times a trinomial,

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

and each piece is a Day 1 product. Followed through, every term of the first factor multiplies every term of the second, and the results are added.

That also says how many products there are before anything is collected. Each of the 2 terms of the first factor meets each of the 3 terms of the second, so there are 2 \times 3 = 6 products, just as 2 shirts and 3 hats make 6 outfits. In general the count is the number of terms in one factor times the number in the other, and counting is a check that none was missed.

A grid keeps the products in order. One factor’s terms go across the top, the other’s down the side, and each cell holds the product of its row and its column:

\times	x^2	4x	-5
2x	2x^3	8x^2	-10x
-3	-3x^2	-12x	15

Six cells, six products. Now collect like terms: 8x^2 - 3x^2 = 5x^2 and -10x - 12x = -22x. With both factors written highest power first, like terms sit on the diagonals running down to the left.

\begin{aligned} &(2x - 3)(x^2 + 4x - 5) \\ &= 2x^3 + 5x^2 - 22x + 15 \end{aligned}

Six products became four terms. The degree is 1 + 2 = 3 for the Day 1 reason: the two leading terms give 2x \cdot x^2 = 2x^3, the only product of that power.

Three factors. Multiplication is associative, so three polynomials are multiplied two at a time: multiply any two, collect, then multiply the result by the third.

Checking by substitution. The factored and expanded forms are equal for every value of x, so put a number into both. At x = 2, (2 \cdot 2 - 3)(4 + 8 - 5) = 7, and 16 + 20 - 44 + 15 = 7. A mismatch at any value proves the expansion wrong; a match is good evidence, not a proof. Avoid 0 and 1: x = 0 tests only the constant term, and x = 1 turns every power into 1, so it tests only the sum of the coefficients.

Closed under multiplication. Each product of two terms is again a term with whole-number exponents, and a sum of such terms is a polynomial, so a polynomial times a polynomial is always a polynomial, as an integer times an integer is always an integer.

Worked examples

Example 1

Day 1: negative terms and a fraction

Multiply, and give the degree of each result.

(a) -4m(m^3 - 2m^2 + 6) \qquad (b) \displaystyle (6y^2 - 9y + 3) \cdot \frac{2}{3}y^3

Solution

(a) Multiply -4m by each of m^3, -2m^2 and 6, keeping each sign with its term:

\begin{aligned} (-4m)(m^3) &= -4m^4 \\ (-4m)(-2m^2) &= 8m^3 \\ (-4m)(6) &= -24m \end{aligned}

So -4m(m^3 - 2m^2 + 6) is -4m^4 + 8m^3 - 24m. The middle product is positive because a negative times a negative is positive. The degree is 1 + 3 = 4.

(b) The monomial stands on the right. Multiplication is commutative, so it multiplies each term just the same, and \displaystyle \frac{2}{3} of 6, -9 and 3 is 4, -6 and 2:

\begin{aligned} 6y^2 \cdot \tfrac{2}{3}y^3 &= 4y^5 \\ (-9y) \cdot \tfrac{2}{3}y^3 &= -6y^4 \\ 3 \cdot \tfrac{2}{3}y^3 &= 2y^3 \end{aligned}

So the product is 4y^5 - 6y^4 + 2y^3, of degree 2 + 3 = 5.

Example 2

Day 1: two variables

Multiply -2ab(a^2 - 3ab + b^2).

Solution

Each letter keeps its own exponent, and only powers of the same base add:

\begin{aligned} -2ab \cdot a^2 &= -2a^3b \\ -2ab \cdot (-3ab) &= 6a^2b^2 \\ -2ab \cdot b^2 &= -2ab^3 \end{aligned}

So -2ab(a^2 - 3ab + b^2) is -2a^3b + 6a^2b^2 - 2ab^3. Every term in the bracket has degree 2 (in several letters, a term’s degree is the sum of its exponents) and -2ab has degree 2, so every term of the product has degree 4.

Check at a = 1, b = 2: the left side is -4(1 - 6 + 4) = 4, and the right side is -4 + 24 - 16 = 4.

Example 3

Day 1: two products, then collect

Simplify 2x(x^2 - 3x + 1) - x^2(4x - 5).

Solution

Read the minus sign as belonging to the second monomial, so the second product is -x^2 times 4x - 5:

\begin{aligned} 2x(x^2 - 3x + 1) &= 2x^3 - 6x^2 + 2x \\ -x^2(4x - 5) &= -4x^3 + 5x^2 \end{aligned}

Neither product has like terms on its own, but the two together do. Collect 2x^3 - 4x^3 = -2x^3 and -6x^2 + 5x^2 = -x^2:

\begin{aligned} &2x(x^2 - 3x + 1) - x^2(4x - 5) \\ &= -2x^3 - x^2 + 2x \end{aligned}

Check at x = 2: 4(-1) - 4(3) = -16, and -16 - 4 + 4 = -16.

Example 4

Day 2: a trinomial times a trinomial, and a collapse

(a) Multiply (x^2 + 2x - 1)(2x^2 - x + 3). \qquad (b) Multiply (a - 2b)(a^2 + 2ab + 4b^2).

Solution

(a) Three terms times three terms make 3 \times 3 = 9 products:

\times	2x^2	-x	3
x^2	2x^4	-x^3	3x^2
2x	4x^3	-2x^2	6x
-1	-2x^2	x	-3

Collect along the diagonals: -x^3 + 4x^3 = 3x^3, then 3x^2 - 2x^2 - 2x^2 = -x^2, then 6x + x = 7x.

\begin{aligned} &(x^2 + 2x - 1)(2x^2 - x + 3) \\ &= 2x^4 + 3x^3 - x^2 + 7x - 3 \end{aligned}

Nine products, five terms, degree 2 + 2 = 4. Check at x = 2: 7 \cdot 9 = 63, and 32 + 24 - 4 + 14 - 3 = 63.

(b) Two terms times three make 6 products. Distribute the trinomial over a and over -2b:

\begin{aligned} &a(a^2 + 2ab + 4b^2) = a^3 + 2a^2b + 4ab^2 \\ &-2b(a^2 + 2ab + 4b^2) \\ &\quad = -2a^2b - 4ab^2 - 8b^3 \end{aligned}

The a^2b terms cancel and so do the ab^2 terms:

(a - 2b)(a^2 + 2ab + 4b^2) = a^3 - 8b^3

Six products, two terms. The count is a check on the work before collecting, not a forecast of how long the answer will be. Products that collapse like this are read backwards in factoring special patterns.

Example 5

Day 2: the vertical layout, and 312 \times 23

Multiply (3x^2 + x + 2)(2x + 3) in a vertical layout, then compare it with 312 \times 23.

Solution

Write the longer factor on top and multiply it by each term of the shorter one, one row each, with like terms in columns, the way long multiplication lines up place values:

\begin{array}{rrrr} & 3x^2 & +\,x & +\,2 \\ \times & & 2x & +\,3 \\ \hline & 9x^2 & +\,3x & +\,6 \\ 6x^3 & +\,2x^2 & +\,4x & \\ \hline 6x^3 & +\,11x^2 & +\,7x & +\,6 \end{array}

The first row is 3 times the top. The second is 2x times the top, and it starts one column to the left because multiplying by x raises every power by one. Adding down the columns collects the like terms, so the product is 6x^3 + 11x^2 + 7x + 6.

Now put x = 10. Then 3x^2 + x + 2 = 312 and 2x + 3 = 23, so the product must be 312 \times 23. The polynomial gives 6000 + 1100 + 70 + 6 = 7176, and indeed 312 \times 23 = 7176. Long multiplication by place value is the same calculation:

\begin{array}{rrrr} & 3 & 1 & 2 \\ \times & & 2 & 3 \\ \hline & 9 & 3 & 6 \\ 6 & 2 & 4 & \\ \hline 6 & 11 & 7 & 6 \end{array}

The column sums are the coefficients 6, 11, 7, 6. A number then carries, since 11 hundreds is 1 thousand and 1 hundred, giving 7176. A polynomial never carries: 11x^2 cannot be traded for a multiple of x^3, because x is not a fixed number like 10.

Example 6

Day 2: three factors in a row

Expand (x + 1)(x + 2)(x + 3).

Solution

Multiply two of the factors first, as two binomials:

(x + 1)(x + 2) = x^2 + 3x + 2

Then multiply that trinomial by the remaining binomial, 3 \times 2 = 6 products:

\begin{aligned} &(x^2 + 3x + 2)(x + 3) \\ &= x^3 + 3x^2 + 3x^2 + 9x + 2x + 6 \\ &= x^3 + 6x^2 + 11x + 6 \end{aligned}

The degree is 1 + 1 + 1 = 3. Check at x = 2: 3 \cdot 4 \cdot 5 = 60, and 8 + 24 + 22 + 6 = 60.

By the associative and commutative properties any pair could go first: (x + 1)(x + 3) first gives (x^2 + 4x + 3)(x + 2), which multiplies out to the same cubic. Notice too that the constant term, 6 = 1 \cdot 2 \cdot 3, can only come from the three constants, and the coefficient of x^2 is 1 + 2 + 3 = 6.

Example 7

Day 2: the volume of an open box

An open box is folded from a card 24 cm by 18 cm by cutting a square of side x cm from each corner and folding up the sides. Write the volume as a polynomial in x, and use it to find the volume when x = 3.

Figure 2Cut a square of side x from each corner and fold up the flaps. The floor is (24 - 2x) by (18 - 2x), and the box is x deep.

Solution

Each side of the card loses x at both ends, so the floor is (24 - 2x) by (18 - 2x), and the flaps make the box x deep. Volume is length times width times height:

V = x(24 - 2x)(18 - 2x)

Multiply the two binomials first, then the monomial x, a Day 1 product:

\begin{aligned} &(24 - 2x)(18 - 2x) \\ &= 432 - 48x - 36x + 4x^2 \\ &= 4x^2 - 84x + 432 \end{aligned}

Then x times each term:

V = 4x^3 - 84x^2 + 432x

At x = 3: V = 108 - 756 + 1296 = 648 cubic centimeters. Directly, the floor is 18 by 12 and the box is 3 deep, and 18 \cdot 12 \cdot 3 = 648. The polynomial describes a box only for 0 < x < 9: at x = 9 the 18 cm side is used up.

Boundaries

Every term meets every term. A product that skips a pairing is not the expansion. The count (terms in one factor) \times (terms in the other) counts the products before collecting, with each factor’s terms counted as written.

Only like terms combine. They need the same letters to the same powers: 3x^2 + 2x^3 stays as it is, and x^2 \cdot y^3 is the single term x^2y^3. Exponents add only on the same base.

The degree adds for non-zero factors. Multiplying by the zero polynomial gives 0, which has no degree.

The area picture needs positive lengths; the algebra does not. With a negative term the grid is a table of products, not a drawing of areas.

Substitution proves only a mismatch. One matching value is evidence, and x = 0 or x = 1 is weak evidence.

A word problem limits the variable. The box polynomial is defined for every x, but it is a volume only for 0 < x < 9.

Where this goes next

Some products come up so often that their results are worth knowing on sight, such as the square of a binomial, in special products. Factoring runs this lesson backwards: exercise 8 pulls a common monomial back out, as in factoring out the greatest common factor, and factoring trinomials finds the two binomials behind a quadratic. Polynomial long division undoes multiplication in the same vertical layout. Multiplying functions is this lesson with f(x) and g(x) as the factors. And multiplying (x + 1) by itself again and again gives coefficients in a famous pattern, the binomial theorem.

Exercises

Day 1: a monomial times a polynomial

5x(2x^2 - 3x + 4)
-3y^2(4y^3 + 2y - 7)
2pq(3p^2 - pq + 5q^2)
\displaystyle (8n^2 - 4n + 12) \cdot \left(-\frac{3}{4}n^2\right)
-a^2b(ab^2 - 2a + 3b)

Day 1: collecting, and working backwards

Simplify 3x(x^2 + 2x - 1) - 2x(x^2 - 4).
Simplify 4y^2(y - 3) + y(2y^2 + 5y).
Find the monomial M for which M(3x^2 - 2x + 6) equals 12x^6 - 8x^5 + 24x^4.
A rectangle is 5x wide and 2x + 3y + 1 long. Write its area as a polynomial, and say what the three pieces of its area model are.
Without multiplying out, give the degree of (a) -6x^4(3x^5 - x^2 + 2) and (b) 2a^2b(a^3b + 4).

Day 2: a polynomial times a polynomial

(x + 2)(x^2 - 3x + 5)
(3y - 1)(2y^2 + y - 4)
(x^2 - x + 2)(x^2 + 3x - 1)
(2a + b)(a^2 - 3ab + b^2)
(x^2 + 1)(x^3 - x + 4)
Multiply (2x^2 + 4x + 1)(3x + 2) in a vertical layout, then use x = 10 to find 241 \times 32.
How many products are there before collecting in (x^3 + x^2 + x + 1)(x^2 + x + 1)? Find the product.

Day 2: three factors, checks and problems

Expand (x - 1)(x + 2)(x + 4).
Expand 2x(x - 3)(x + 5).
A claim says that (x + 3)(x^2 - 2x + 4) is x^3 + 5x^2 - 2x + 12. Test the claim at x = 2, then find the correct product.
A photo is w cm wide and w + 6 cm tall, with a mat x cm wide around it on every side. (a) Write the outer width and height. (b) Write the area of the mat alone as a polynomial. (c) Find that area when w = 20 and x = 5.

Figure 3The mat in exercise 21 is the shaded border, x wide on every side.

Reasoning

A trinomial times a trinomial gives 9 products before collecting. Explain why, and why the final answer can have fewer than 9 terms.
Explain why a polynomial of degree m times a polynomial of degree n has degree exactly m + n, and why no other product can cancel its leading term.
Rosamund tests the claim (x + 2)(x^2 + 3) = 3x^3 + 2x^2 + x + 6 at x = 1, and both sides give 12. Show that x = 2 exposes the claim as false, find the correct product, and explain why x = 1 missed the error.

Answers

10x^3 - 15x^2 + 20x. Each term times 5x, adding exponents.
-12y^5 - 6y^3 + 21y^2. The last product, (-3y^2)(-7), is positive.
6p^3q - 2p^2q^2 + 10pq^3. Each letter’s exponents add separately.
-6n^4 + 3n^3 - 9n^2. \displaystyle -\frac{3}{4} of 8, -4 and 12 is -6, 3 and -9.
-a^3b^3 + 2a^3b - 3a^2b^2. For example, (-a^2b)(-2a) = 2a^3b.
x^3 + 6x^2 + 5x. 3x^3 + 6x^2 - 3x - 2x^3 + 8x, then collect.
6y^3 - 7y^2. 4y^3 - 12y^2 + 2y^3 + 5y^2.
M = 4x^4. 12x^6 \div 3x^2 = 4x^4, and it fits the other terms: 4x^4 \cdot (-2x) = -8x^5 and 4x^4 \cdot 6 = 24x^4.
10x^2 + 15xy + 5x. The pieces are 5x by 2x, 5x by 3y, and 5x by 1.
(a) 9, since 4 + 5. (b) 7, since 2a^2b has degree 3 and a^3b + 4 has degree 4.
x^3 - x^2 - x + 10. Six products: x^3 - 3x^2 + 5x + 2x^2 - 6x + 10.
6y^3 + y^2 - 13y + 4. 6y^3 + 3y^2 - 12y - 2y^2 - y + 4.
x^4 + 2x^3 - 2x^2 + 7x - 2. Nine products; the x^2 terms are -x^2 - 3x^2 + 2x^2.
2a^3 - 5a^2b - ab^2 + b^3. Before collecting: 2a^3 - 6a^2b + 2ab^2 + a^2b - 3ab^2 + b^3.
x^5 + 4x^2 - x + 4. Six products; -x^3 and x^3 cancel.
6x^3 + 16x^2 + 11x + 2, so 241 \times 32 = 7712. The rows are 4x^2 + 8x + 2 and 6x^3 + 12x^2 + 3x; at x = 10, 6000 + 1600 + 110 + 2 = 7712.
12 products, since 4 \times 3. The product is x^5 + 2x^4 + 3x^3 + 3x^2 + 2x + 1.
x^3 + 5x^2 + 2x - 8. (x - 1)(x + 2) = x^2 + x - 2, then times x + 4.
2x^3 + 4x^2 - 30x. (x - 3)(x + 5) = x^2 + 2x - 15, then times 2x.
At x = 2 the left side is 5 \cdot 4 = 20 but the claim gives 36, so it is wrong. The correct product is x^3 + x^2 - 2x + 12: the x^2 terms are -2x^2 + 3x^2.
(a) w + 2x by w + 6 + 2x. (b) 4x^2 + 4wx + 12x. The outer area (w + 2x)(w + 2x + 6) is w^2 + 4wx + 4x^2 + 6w + 12x, and the photo takes w^2 + 6w of it. (c) 560 square centimeters: 100 + 400 + 60, and indeed 30 \cdot 36 - 20 \cdot 26 = 560.
Each of the 3 terms of the first factor multiplies each of the 3 terms of the second, and every pairing happens once, so there are 3 \times 3 = 9 products. Several products can have the same power, and collecting merges them into one term, or cancels them.
Write the leading terms as ax^m and bx^n with a and b non-zero. Every other product has a smaller power in at least one factor, so its degree is below m + n. The only product of degree m + n is abx^{m+n}, and ab \neq 0, so nothing cancels it.
At x = 2: (4)(7) = 28, but the claim gives 24 + 8 + 2 + 6 = 40. The correct product is x^3 + 2x^2 + 3x + 6. At x = 1 every power of x is 1, so each side is just the sum of its coefficients, and 3 + 2 + 1 + 6 and 1 + 2 + 3 + 6 are both 12.