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How do you multiply two binomials? - Answers

Apply the Distrubutive Property twice as you do when multiplying two-digit numbers Use the FOIL methed First Inner Outter Last. In the equation (3x + 4)(9x + 1) you get. (3x)*(9x) + (3x)*1 + (9x)*4 + 4*1 Now you multiply. 3x*9x = 27x^2 3x*1 + 9x*4 = 3x + 36x = 39x 1*4 = 4 So your entire answer is (3x)*(9x) + (3x)*1 + (9x)*4 + 4*1 = 27x^2 + 39x + 4



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How do you multiply two binomials? - Answers

https://math.answers.com/algebra/How_do_you_multiply_two_binomials

Apply the Distrubutive Property twice as you do when multiplying two-digit numbers Use the FOIL methed First Inner Outter Last. In the equation (3x + 4)(9x + 1) you get. (3x)*(9x) + (3x)*1 + (9x)*4 + 4*1 Now you multiply. 3x*9x = 27x^2 3x*1 + 9x*4 = 3x + 36x = 39x 1*4 = 4 So your entire answer is (3x)*(9x) + (3x)*1 + (9x)*4 + 4*1 = 27x^2 + 39x + 4



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https://math.answers.com/algebra/How_do_you_multiply_two_binomials

How do you multiply two binomials? - Answers

Apply the Distrubutive Property twice as you do when multiplying two-digit numbers Use the FOIL methed First Inner Outter Last. In the equation (3x + 4)(9x + 1) you get. (3x)*(9x) + (3x)*1 + (9x)*4 + 4*1 Now you multiply. 3x*9x = 27x^2 3x*1 + 9x*4 = 3x + 36x = 39x 1*4 = 4 So your entire answer is (3x)*(9x) + (3x)*1 + (9x)*4 + 4*1 = 27x^2 + 39x + 4

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      Apply the Distrubutive Property twice as you do when multiplying two-digit numbers Use the FOIL methed First Inner Outter Last. In the equation (3x + 4)(9x + 1) you get. (3x)*(9x) + (3x)*1 + (9x)*4 + 4*1 Now you multiply. 3x*9x = 27x^2 3x*1 + 9x*4 = 3x + 36x = 39x 1*4 = 4 So your entire answer is (3x)*(9x) + (3x)*1 + (9x)*4 + 4*1 = 27x^2 + 39x + 4

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