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—you just have to distribute twice. Each term in the first binomial gets multiplied by each term in the second binomial. There's a handy acronym called FOIL that helps you remember which terms to multiply.

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

Binomials with Subtraction

FOIL works the same way when there are negative signs—just watch your signs carefully.

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. \((x + 2)(x + 6)\)
  2. \((y - 5)(y + 3)\)
  3. \((m - 4)(m - 7)\)
  4. \((2x + 3)(x + 1)\)
  5. \((a + 8)^2\)
  6. \((3n - 2)(3n + 2)\)
  7. \((5p - 1)(2p + 3)\)

Solutions:

  1. \(x^2 + 6x + 2x + 12 = x^2 + 8x + 12\)
  2. \(y^2 + 3y - 5y - 15 = y^2 - 2y - 15\)
  3. \(m^2 - 7m - 4m + 28 = m^2 - 11m + 28\)
  4. \(2x^2 + 2x + 3x + 3 = 2x^2 + 5x + 3\)
  5. \(a^2 + 2(a)(8) + 64 = a^2 + 16a + 64\)
  6. \((3n)^2 - 2^2 = 9n^2 - 4\)
  7. \(10p^2 + 15p - 2p - 3 = 10p^2 + 13p - 3\)

Where Things Go Wrong

Forgetting the middle term when squaring a binomial is a classic blunder. The expression \((x + 3)^2\) does NOT equal \(x^2 + 9\). You must include the middle term: \((x + 3)^2 = x^2 + 6x + 9\). Write it out using FOIL if you're not sure.

Sign errors when multiplying negative terms happen all the time. Just remember the rule: negative times negative equals positive. So \((-4)(-2) = +8\), not \(-8\).

Not combining like terms at the end leaves your answer messy. After using FOIL, you typically have four terms, but the two middle terms usually combine. Don't leave \(x^2 + 3x + 2x + 6\) as your final answer—combine those middle terms to get \(x^2 + 5x + 6\).

Trying to use FOIL for polynomials with more than two terms doesn't work. FOIL only works for binomials (two terms times two terms). If you need to multiply \((x + 2)(x^2 + 3x - 1)\), forget about FOIL and just distribute each term in the first factor to every term in the second factor.

Multiplying the coefficients incorrectly throws off the entire answer. When you multiply \((2x)(3x)\), you multiply the numbers AND the variables: \(2 \times 3 = 6\) and \(x \times x = x^2\), giving you \(6x^2\).

Related Topics