Multiplying Binomials
To multiply binomials, you need to distribute twice. Notice in the pattern above that first the a was distributed to everything in the 2^{nd} parenthesis (c and d). Then, the b was distributed to everything in the 2^{nd} parenthesis (c and d). One way to remember this is with the acronym FOIL F: First - Multiply the first term in each parenthesis (a times c) O: Outer - Multiply the outside terms (a times d) I: Inner - Multiply the inner terms (b times c) L: Last - Multiply the last term in each parenthesis (b times d) Examples: 1) (x+5)(x + 2) = x^{2} + 2x + 5x + 10 = x^{2} + 7x + 10 First, we distribute the x (x + 2) and get x^{2}+2x. (Or, if you're thinking of FOIL, we do the first and outer.) Next, we distribute the 5 (x + 2)and get 5x+10. (Or, if you're thinking of FOIL, we do the inner and last.) Now we have x^{2} + 2x + 5x + 10, but we are not finished because there is a set of like terms that we can add together. Add 2x and 5x to get 7x. Make sure the final answer is in standard form: x^{2}+7x+10. 2) (2y + 6)(y - 4) = 2y^{2} - 8y + 6y - 24 = 2y^{2} - 2y - 24 First, we distribute the 2y (y - 4) and get 2y^{2} - 8y. (Or, if you're thinking of FOIL, we do the first and outer.) Next, we distribute the 6 (y - 4)and get 6y - 24. (Or, if you're thinking of FOIL, we do the inner and last.) Now we have 2y^{2} - 8y + 6y - 24, but we are not finished because there is a set of like terms that we can add together. Add -8y and 6y to get -2y. Make sure the final answer is in standard form: 2y^{2}-2y - 24. Practice: 1) (y + 3)(y + 8) 2) (x - 4)(x + 2) 3) (y^{2} + 9)(y + 7) 4) (a - 3)(a - 10) 5) (x^{2} - 2x)(x + 4) Answers: 1) y^{2} + 11y + 24 2) x^{2} - 2x - 8 3) y^{3} + 7y^{2} + 9y + 63 4) a^{2} - 13a + 30 5) x^{3} + 2x^{2} - 8x |