How to Multiply With Three Factors in Algebra

FOIL can only take you so far with polynomials.

You may have learned to multiply binomials using FOIL: First, Inner, Outer, Last. While FOIL works for multiplying two terms, e.g. (x+1)(x+3), it tells you nothing about multiplying more than two terms, as in (x+1)(x+2)(x+3). FOIL's weakness is that it doesn't explain the concept behind multiplying binomials. There is no specific order in which terms "must" be multiplied: all that matters is that you multiply every term within a quantity by every term outside of that quantity.

Instructions

1. Review the distributive property by solving this problem: x(y+3). Multiplication is distributive over addition, so x(y+3) = xy + 3y. All you did was multiply each term inside the parentheses by each term outside of the parentheses: y and 3 are inside the parentheses and x is outside, so you multiplied y by x and 3 by x. Evaluate the expression, 2 * (3 * 5) = (2 * 3) * 5 to see the distributive property at work on integers.

2. Evaluate this expression: (2 * 3 ) * 4 = (2 * 3 * 4) = 2 * (3 * 4). This reduces to 24 = 24 = 24, so the expression is true. This is the associative property: (AB)C = (ABC) = A(BC). Remember, all parentheses mean is that you solve the expression inside of the parentheses first; but when only multiplication is involved, this usually doesn't make a difference. That's why we can move the parentheses around.

3. Combine like terms by looking at exponents and variables. Two terms can be combined if they have the same variable and the same exponents. x + 5x + x^2 = 6x + x^2 because x^2 and x are not like terms, but 5x and x are. Similarly, x^3 + x^2 + 4x^3 = 5x^3 + x^2, because 4x^3 and x^3 have the same exponent.

4. Write the following expression on a piece of paper: (x+1)(y+3). Solve it using FOIL: multiply the first term of each binomial, followed by the outer terms, followed by the inner terms, followed by the last terms, then add them together. x*y + 1y + 3x + 3 = xy + y + 3x + 3. Retrace your steps and notice what you did: you multiplied each term inside of each quantity by the terms outside. This problem was the same as Step 1, except x was changed to (x + 1).

5. Write the expression, (x+1)(x+2)(x+3). Recalling the associative property from Step 2, you can deduce that (AB)(CD)(EF) = ((AB)(CD))(EF). So you can add parentheses in order to slice the problem up into smaller pieces and solve them one at a time, because multiplication is associative.

6. Add parentheses around the first two terms of (x+1)(x+2)(x+3). This yields ( (x+1)(x+2) )(x+3). Simplify (x+1)(x+2) using FOIL to get x^2 + 3x + 2. Replace ((x+1)(x+2)) with (x^2 + 3x + 2), yielding (x^2 + 3x + 2)(x + 3). Recalling the distributive property, A(B + C) = (AB + AC), so: ( ( x^2 + 3x + 2)(x + 3) ) = ( x(x^2 + 3x + 2) + 3(x^2 + 3x + 2) ). This looks confusing, but all you did was replace A with ( x^2 + 3x + 2) and (B+C) with (x + 3).

7. Simplify x(x^2 + 3x + 2) + 3(x^2 + 3x + 2) to (x^3 + 3x^2 + 2x) + (3x^2 + 6x + 6). Combine like terms as you did in Step 3, yielding x^3 + 6x^2 + 8x + 6. You have completely simplified the polynomial.

References

• West Texas University: Intermediate Algebra Tutorial on Multiplying Polynomials
• Regentsprep.org: Multiplying Binomials
• Basic-Mathematics.com: Multiplying Binomials