Multiplying Binomials

You know how to multiply a monomial by a polynomial using the distributive property. For instance, \(3(x + 5) = 3x + 15\). But what happens when you multiply two binomials, like \((x + 3)(x + 5)\)?

The distributive property still applies, but now you distribute twice. Each term in the first binomial gets multiplied by each term in the second binomial. The acronym FOIL is a memory aid for the four multiplications involved.

The Distributive Property Extended

When you multiply \((a + b)(c + d)\), you need to multiply every term in the first parentheses by every term in the second parentheses.

$$(a + b)(c + d) = a(c + d) + b(c + d)$$

Distribute further: $$= ac + ad + bc + bd$$

That's four multiplications total.

Example: \((x + 2)(x + 5)\)

$$= x(x + 5) + 2(x + 5)$$ $$= x^2 + 5x + 2x + 10$$ $$= x^2 + 7x + 10$$

The FOIL Method

FOIL is a mnemonic device that tells you which terms to multiply. It stands for:

First: Multiply the first terms from each binomial

Outer: Multiply the outer terms

Inner: Multiply the inner terms

Last: Multiply the last terms

Then combine any like terms.

Example: Multiply \((x + 3)(x + 4)\) using FOIL.

First: \(x \times x = x^2\)

Outer: \(x \times 4 = 4x\)

Inner: \(3 \times x = 3x\)

Last: \(3 \times 4 = 12\)

Combine: $$x^2 + 4x + 3x + 12 = x^2 + 7x + 12$$

Example: \((2y + 1)(y + 5)\)

First: \(2y \times y = 2y^2\)

Outer: \(2y \times 5 = 10y\)

Inner: \(1 \times y = y\)

Last: \(1 \times 5 = 5\)

Result: \(2y^2 + 10y + y + 5 = 2y^2 + 11y + 5\)

Quick check: use FOIL to multiply \((x + 1)(x + 6)\). Show answerF: \(x^2\)   O: \(6x\)   I: \(x\)   L: \(6\). Combine: \(x^2 + 7x + 6\)

Binomials with Subtraction

FOIL works the same way when there are negative signs. Just track the signs carefully through each step.

Example: \((x - 3)(x + 5)\)

First: \(x^2\)

Outer: \(5x\)

Inner: \(-3x\)

Last: \(-15\)

Result: \(x^2 + 5x - 3x - 15 = x^2 + 2x - 15\)

Example: \((a - 4)(a - 2)\)

First: \(a^2\)

Outer: \(-2a\)

Inner: \(-4a\)

Last: \((-4)(-2) = 8\)

Result: \(a^2 - 2a - 4a + 8 = a^2 - 6a + 8\)

Notice that when you multiply two negative numbers, you get a positive result.

Special Products

Certain binomial products come up so often that they have special patterns you can memorize.

Perfect Square Trinomials:

$$(a + b)^2 = a^2 + 2ab + b^2$$ $$(a - b)^2 = a^2 - 2ab + b^2$$

Example: \((x + 5)^2\)

Don't make the mistake of thinking this equals \(x^2 + 25\). You have to square the binomial properly.

$$= (x + 5)(x + 5)$$

Using FOIL: $$= x^2 + 5x + 5x + 25 = x^2 + 10x + 25$$

Or use the pattern: \(a^2 + 2ab + b^2\) where \(a = x\) and \(b = 5\).

$$= x^2 + 2(x)(5) + 5^2 = x^2 + 10x + 25$$

Example: \((2m - 3)^2\)

$$= (2m)^2 - 2(2m)(3) + 3^2$$ $$= 4m^2 - 12m + 9$$

Difference of Squares:

$$(a + b)(a - b) = a^2 - b^2$$

The middle terms cancel out, leaving only the difference of the squares.

Example: \((x + 7)(x - 7)\)

First: \(x^2\)

Outer: \(-7x\)

Inner: \(+7x\)

Last: \(-49\)

The middle terms cancel: \(x^2 + (-7x) + 7x - 49 = x^2 - 49\)

Example: \((3y + 2)(3y - 2)\)

$$= (3y)^2 - 2^2 = 9y^2 - 4$$

Multiplying Binomials with Coefficients

When the terms have coefficients greater than 1, FOIL still works.

Example: \((3x + 2)(2x + 5)\)

First: \(3x \times 2x = 6x^2\)

Outer: \(3x \times 5 = 15x\)

Inner: \(2 \times 2x = 4x\)

Last: \(2 \times 5 = 10\)

Result: \(6x^2 + 15x + 4x + 10 = 6x^2 + 19x + 10\)

Example: \((4a - 3)(2a + 1)\)

First: \(4a \times 2a = 8a^2\)

Outer: \(4a \times 1 = 4a\)

Inner: \(-3 \times 2a = -6a\)

Last: \(-3 \times 1 = -3\)

Result: \(8a^2 + 4a - 6a - 3 = 8a^2 - 2a - 3\)

Why FOIL Isn't Enough

FOIL is great for multiplying two binomials, but it doesn't work for larger polynomials. If you need to multiply \((x + 2)(x^2 + 3x - 1)\), FOIL won't help because the second factor has three terms.

For those situations, you go back to the basic distributive property: multiply each term in the first factor by every term in the second factor.

Example: \((x + 3)(x^2 - 2x + 5)\)

Distribute \(x\): $$x(x^2 - 2x + 5) = x^3 - 2x^2 + 5x$$

Distribute 3: $$3(x^2 - 2x + 5) = 3x^2 - 6x + 15$$

Combine: $$x^3 - 2x^2 + 5x + 3x^2 - 6x + 15$$ $$= x^3 + x^2 - x + 15$$

Give These a Shot

Multiply:

1. \((x + 2)(x + 6)\) Show answerF: \(x^2\), O: \(6x\), I: \(2x\), L: \(12\). Combine: \(x^2 + 8x + 12\)

2. \((y - 5)(y + 3)\) Show answerF: \(y^2\), O: \(3y\), I: \(-5y\), L: \(-15\). Combine: \(y^2 - 2y - 15\)

3. \((m - 4)(m - 7)\) Show answerF: \(m^2\), O: \(-7m\), I: \(-4m\), L: \(28\) (negative × negative = positive). Combine: \(m^2 - 11m + 28\)

4. \((2x + 3)(x + 1)\) Show answerF: \(2x^2\), O: \(2x\), I: \(3x\), L: \(3\). Combine: \(2x^2 + 5x + 3\)

5. \((a + 8)^2\) Show answerDon't just square each term! Use FOIL: \((a+8)(a+8)\). F: \(a^2\), O: \(8a\), I: \(8a\), L: \(64\). Result: \(a^2 + 16a + 64\)

6. \((3n - 2)(3n + 2)\) Show answerThis is a difference of squares pattern: \((3n)^2 - 2^2 = 9n^2 - 4\). The middle terms cancel out.

7. \((5p - 1)(2p + 3)\) Show answerF: \(10p^2\), O: \(15p\), I: \(-2p\), L: \(-3\). Combine: \(10p^2 + 13p - 3\)

What's Next?

A few habits to lock in. The most classic mistake is dropping the middle term when squaring a binomial: \((x + 3)^2\) is \(x^2 + 6x + 9\), not \(x^2 + 9\). When in doubt, write the square out as \((x + 3)(x + 3)\) and use FOIL. Sign errors with negatives are the next most common slip — negative times negative is positive, so \((-4)(-2) = +8\). FOIL gives you four terms; the middle two usually combine, so don't leave \(x^2 + 3x + 2x + 6\) as a final answer. And FOIL only works for two binomials. To multiply something like \((x + 2)(x^2 + 3x - 1)\), drop FOIL and distribute each term in the first factor across every term in the second.

The next step is multiplying polynomials of any size, which uses the same idea — distribute each term of the first factor across every term of the second — without the FOIL shortcut.