Algebra 1

Multiplying two binomials

Two sums are multiplied together. Which products does that make, and how can you be sure none is missing?

You need

The distributive property with variablesMath 7 Area of rectangles and squaresGrade 3

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Degree, terms and standard formAlgebra 1 Special productsAlgebra 1 Factoring when the leading coefficient is oneAlgebra 1 CountingGrades 7 to 11 The binomial theoremAlgebra 2 +2 more

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Algebra9

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▴ 5 above Variables5 Distribute, then combine

Multiplying two binomials

Rewriting an expression to show what it means Equivalent expressions: the distributive property and like terms ▾ Show all 12 (2 more)

When you reach for it

You already know how to multiply a number into a sum: 3(x + 5) = 3x + 15, by the distributive property. Now both factors are sums. The product (x + 3)(x + 5) is, for example, the area of a rectangle with sides x + 3 and x + 5, and you want that area as a single sum of terms with no brackets.

A sum of two terms is a binomial, and rewriting a product of binomials as a sum is called expanding it. Reach for it when brackets stand in the way: comparing two areas, collecting terms in an equation, reading off the constant or the highest power, or checking a factorization.

The procedure

The distributive property, used twice

For any value of x, the bracket x + 5 is just a number, and the distributive property holds for every number. So distribute the whole bracket x + 5 over the sum x + 3:

(x + 3)(x + 5) = x(x + 5) + 3(x + 5)

This is (a + b)c = ac + bc, the distributive property with the common factor on the right (the commutative property of multiplication turns it into the familiar c(a + b)). Each piece is now a single term times a binomial, so distribute again:

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

where x \cdot x = x^2 by the exponent rules. Last, 5x and 3x are like terms, so they combine:

(x + 3)(x + 5) = x^2 + 8x + 15

Why there are four products

The two distributions made x \cdot x, x \cdot 5, 3 \cdot x and 3 \cdot 5. Each of the 2 terms of the first bracket met each of the 2 terms of the second exactly once, so there are 2 \times 2 = 4 products, the same count as choosing one of 2 shirts and one of 2 hats. That is the whole rule: every term of the first times every term of the second, then collect like terms.

Some books name an order for the four products (first, outer, inner, last). A name only helps you check that four were made; the reason they are the right four is the distributive property.

The area model

Figure 1The rectangle (x + 3) by (x + 5) cut where each x ends. Its four pieces are the four products.

Cut each side of the rectangle where its x ends. The rectangle falls into four smaller ones, one for each product, and since area is length times width and the pieces fill the whole without overlapping, their areas add to the whole area. This is the area model you met with numbers, now with a letter on the sides.

The picture also shows why the middle collects. The pieces 5x and 3x each have one side x and one side a plain number, so they are like terms. The x^2 piece has two sides x and the 15 piece has none, so neither combines with anything.

Negative terms and the grid

A subtraction is adding the opposite, x - 4 = x + (-4), and the sign travels with its term (combining like terms). The four products then carry their signs by the rules for multiplying negatives. A rectangle cannot have a side of length -4, so the picture stops being a literal area. Keep the same 2 by 2 layout as a grid: the terms of one factor across the top, the other down the side, each cell their product (Example 1).

Checking by substitution

(x + 3)(x + 5) and x^2 + 8x + 15 are equivalent: equal for every value of x. So substitute a number into both. At x = 2: 5 \cdot 7 = 35 and 4 + 16 + 15 = 35.

A mismatch at any value proves the expansion wrong. A match at one value is evidence, not proof. Avoid x = 0, which tests only the constant, and x = 1, where x^2 and x are both 1 and a wrong power goes unseen.

Worked examples

Example 1

A subtraction in one bracket

Expand (x - 4)(x + 7) and check the result.

Solution

Figure 2The grid for (x - 4)(x + 7). With a subtraction the cells are no longer areas, only a table of the four products.

Write x - 4 as x + (-4) and fill the grid. Top row, x times each term: x \cdot x = x^2 and x \cdot 7 = 7x. Bottom row, -4 times each term: (-4) \cdot x = -4x and (-4) \cdot 7 = -28. Add the four cells and collect the like terms 7x and -4x:

\begin{aligned} (x - 4)(x + 7) &= x^2 + 7x - 4x - 28\\ &= x^2 + 3x - 28 \end{aligned}

Check at x = 2: (2 - 4)(2 + 7) = (-2)(9) = -18, and 4 + 6 - 28 = -18. They agree.

Example 2

Coefficients and two negatives

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

Solution

Every term of the first times every term of the second, signs attached:

\begin{aligned} (2x)(3x) &= 6x^2 & (2x)(-1) &= -2x\\ (-3)(3x) &= -9x & (-3)(-1) &= 3 \end{aligned}

(2x)(3x) = 6x^2 multiplies the numbers and the letters separately, using the commutative and associative properties. The last product is positive because a negative times a negative is positive. Collect -2x - 9x = -11x:

(2x - 3)(3x - 1) = 6x^2 - 11x + 3

Check at x = 2: (1)(5) = 5, and 24 - 22 + 3 = 5.

Example 3

Two variables

Expand (a + 2b)(3a - b).

Solution

The four products are a \cdot 3a = 3a^2, a \cdot (-b) = -ab, 2b \cdot 3a = 6ab and 2b \cdot (-b) = -2b^2. Here 6ab came out as 6ba first; the commutative property makes ba and ab the same, so -ab and 6ab are like terms, and -ab + 6ab = 5ab. The terms a^2 and b^2 have different letters and do not combine:

(a + 2b)(3a - b) = 3a^2 + 5ab - 2b^2

Check at a = 2, b = 1: (4)(5) = 20, and 12 + 10 - 2 = 20.

Example 4

A vegetable bed made bigger

Idris has a rectangular vegetable bed 6 m long and 3 m wide. He extends it by x meters in both directions, so it becomes x meters longer and x meters wider. Write the new area as a sum of terms, and find the area he added.

Solution

Figure 3Idris’s bed, extended by x meters both ways. The added area is the two strips and the corner square.

The new bed is (x + 6) by (x + 3) meters. The four pieces of the figure are the four products: the old bed 6 \cdot 3 = 18, a strip 3x across the short end, a strip 6x along the long side, and the corner x \cdot x = x^2 where the two extensions meet. Collect 3x + 6x:

(x + 6)(x + 3) = x^2 + 9x + 18

square meters. The added area is everything except the old 18, which is x^2 + 9x square meters. It is not 9x alone: the corner square belongs to both extensions and is counted once, as x^2. Check at x = 1: the bed is 7 by 4, so 28 - 18 = 10 was added, and 1 + 9 = 10.

Example 5

A tray of brownies, one side longer and one shorter

Lorelei bakes brownies in a tray 30 cm long and 20 cm wide. A new tray is x cm longer and x cm narrower, for 0 < x < 20. Expand its area and compare it with the old tray.

Solution

The new tray is (30 + x) by (20 - x) cm. Every term times every term:

\begin{aligned} (30 + x)(20 - x) &= 600 - 30x + 20x - x^2\\ &= -x^2 - 10x + 600 \end{aligned}

written with the highest power first. The old area is 600 square centimeters, so the change is -x^2 - 10x. For every allowed x both of those terms are negative, so this new tray is smaller than the old one, even though its perimeter is still 100 cm. Check at x = 5: the tray is 35 by 15, area 525, and -25 - 50 + 600 = 525.

The -10x came from lengthening the side that was already longer. Lengthen the 20 cm side and shorten the 30 cm side instead, and (20 + x)(30 - x) = -x^2 + 10x + 600, which is more than 600 for 0 < x < 10. The expansion, not the perimeter, says which way the area moves.

Example 6

A square, and a product whose middle vanishes

Expand (x + 9)^2 and (x + 9)(x - 9).

Solution

A square is a product of two equal factors, so (x + 9)^2 = (x + 9)(x + 9):

\begin{aligned} (x + 9)(x + 9) &= x^2 + 9x + 9x + 81\\ &= x^2 + 18x + 81 \end{aligned}

The two middle products are equal, so the middle term is twice 9x. In the second product they are opposites:

\begin{aligned} (x + 9)(x - 9) &= x^2 - 9x + 9x - 81\\ &= x^2 - 81 \end{aligned}

and -9x + 9x = 0 leaves no middle term. Both are ordinary products here; their shapes are studied on sight in special products.

Boundaries

Four products belong to two binomials, 2 \times 2. A factor with three terms gives 2 \times 3 = 6 products by the same rule, every term of the first times every term of the second.
The area model is a literal area only when every term is a positive length. With a subtracted term it is a grid of products, and a cell such as -28 is not an area.
Only like terms combine: the same letters to the same powers. x^2, x and a constant never combine with each other, and neither do a^2 and b^2.
A substitution check can prove an expansion wrong. A match at one value does not prove it right.

Where this goes next

Factoring a trinomial runs this backwards, from x^2 + 8x + 15 back to (x + 3)(x + 5), which is why the middle term being the sum 3 + 5 and the constant the product 3 \cdot 5 will matter. Special products name the square and the sum times difference from Example 6. Raising a binomial to higher powers leads to the binomial theorem, whose coefficients come from the same counting of choices, one term from each bracket, that gave 2 \times 2 here (Counting).

Exercises

Positive terms and the area model

Expand (x + 2)(x + 6).
Expand (y + 7)(y + 1).
Expand (3n + 2)(n + 4).
A rectangle has sides 2x + 1 and x + 4. Give the area of each of its four pieces, then the total area as a sum of terms.

Negative terms

Expand (x - 2)(x + 9).
Expand (m - 6)(m - 2).
Expand (4x + 1)(x - 5).
Expand (5 - x)(2x + 3) and write the result with the highest power first.
Expand \displaystyle \left(x + \frac{1}{2}\right)\left(x - \frac{1}{4}\right).

Two variables and higher powers

Expand (p + q)(p + 3q).
Expand (2a - b)(a - 4b).
Expand (x^2 + 3)(x - 2). Are there any like terms to collect?

Checking

Is x^2 - 3x + 18 an expansion of (x - 6)(x + 3)? Substitute x = 2 into both forms to test it, and give the expansion.
Find the number k for which (x + k)(x + 4) = x^2 + 11x + 28.

Problems

Philippa’s photo print is x cm wide and 11 cm taller than it is wide. A second print is 4 cm wider and 5 cm shorter than the first. Expand the area of the second print, and find for which widths it is larger than the first.
A rectangle has sides 2x + 5 and x + 3 cm. Expand its area, then find the area at x = 4 both from your expansion and from the side lengths.

Reasoning

Use the area model to explain why (x + a)(x + b) = x^2 + (a + b)x + ab for any positive numbers a and b.
Suppose every expansion is checked only by substituting x = 0. Explain what that value tests and what it cannot, using (x + 2)(x + 7) and the form x^2 + 14.

Answers

x^2 + 8x + 12. The products are x^2, 6x, 2x, 12; collect 6x + 2x.
y^2 + 8y + 7. The products are y^2, y, 7y, 7.
3n^2 + 14n + 8. The products are 3n^2, 12n, 2n, 8.
Pieces 2x^2, 8x, x and 4, from 2x \cdot x, 2x \cdot 4, 1 \cdot x, 1 \cdot 4. Total 2x^2 + 9x + 4.
x^2 + 7x - 18. The products are x^2, 9x, -2x, -18.
m^2 - 8m + 12. The products are m^2, -2m, -6m and (-6)(-2) = 12.
4x^2 - 19x - 5. The products are 4x^2, -20x, x, -5.
-2x^2 + 7x + 15. The products are 10x, 15, -2x^2, -3x; collect 10x - 3x.
\displaystyle x^2 + \frac{1}{4}x - \frac{1}{8}. The products are x^2, \displaystyle -\frac{1}{4}x, \displaystyle \frac{1}{2}x and \displaystyle -\frac{1}{8}, and \displaystyle \frac{1}{2} - \frac{1}{4} = \frac{1}{4}.
p^2 + 4pq + 3q^2. The products are p^2, 3pq, pq, 3q^2.
2a^2 - 9ab + 4b^2. The products are 2a^2, -8ab, -ab and (-b)(-4b) = 4b^2.
x^3 - 2x^2 + 3x - 6. No: the four products x^3, -2x^2, 3x, -6 all have different powers, so all four stay.
At x = 2 the product is (-4)(5) = -20 but x^2 - 3x + 18 gives 4 - 6 + 18 = 16, so it is not an expansion. The last product is (-6)(3) = -18: the correct expansion is x^2 - 3x - 18, which gives 4 - 6 - 18 = -20.
k = 7. Expanding gives x^2 + (k + 4)x + 4k, so 4k = 28, and then k + 4 = 11 agrees.
The second print is (x + 4) by (x + 6), area x^2 + 10x + 24. The first is x(x + 11) = x^2 + 11x. The difference is 24 - x, so the second is larger when x < 24, the same at x = 24, and smaller above. At x = 4: 80 against 60.
2x^2 + 11x + 15 square centimeters. At x = 4: 32 + 44 + 15 = 91, and the sides are 13 and 7, with 13 \cdot 7 = 91.
Cut the rectangle x + a by x + b where each x ends. The pieces are x^2, bx, ax and ab. The two strips each have one side x, so they combine to (a + b)x, and the corner piece is the plain number ab.
At x = 0 every term containing x is 0, so only the constant is tested. Both forms give 14, yet x^2 + 14 is wrong: at x = 1 the product is 3 \cdot 8 = 24 and x^2 + 14 is 15. A match at one value cannot prove an expansion; the missing 9x vanished at x = 0.