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Simplify (2z – 3)^2
a)4z^2 – 9
b)4z^2 – 9z
c)4z^2 – 12z – 9
d)4z^2 – 12z + 9
 2 years ago
 2 years ago
Simplify (2z – 3)^2 a)4z^2 – 9 b)4z^2 – 9z c)4z^2 – 12z – 9 d)4z^2 – 12z + 9
 2 years ago
 2 years ago

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AccessDeniedBest ResponseYou've already chosen the best response.3
(2z  3)^2 = (2z  3)(2z  3) do you know how to use FOIL on the binomials?
 2 years ago

AccessDeniedBest ResponseYou've already chosen the best response.3
hmm.. I'll stick with using distributive property twice then (2z  3)(2z  3) (2z)(2z  3) + (3)(2z  3) 4z^2  6z + (6z) + 9 4z^2  12z + 9 ^ this is sufficient for the answer, im just going to go over FOIL a bit. maybe it will help you out. :)  FOIL is just another way to multiply them together, and somewhat more convenient to do than distributive twice. "F irst O uter I nner L ast" I'll box the terms I am referring to in each step of it: ([2z]  3)([2z]  3) The 2z 's are first, so we multiply these first (2z)(2z) = 4z^2 ([2z]  3)(2z [ 3]) These are the outer terms. next we multiply these and add it onto the previous expression 4z^2 + 2z(3) = 4z^2  6z (2z [ 3)([2z]  3) These are the inner terms. we next multiply these and add it to the last again 4z^2  6z + (3)(2z) = 4z^2  6z  6z = 4z^2  12z (2z [ 3])(2z [ 3]) These are the 'last' terms. Let's just multiply these together and we'll add it to the last and be finished! 4z^2  12z + (3)(3) = 4z^2  12z + 9
 2 years ago

mitosukiBest ResponseYou've already chosen the best response.1
(2z  3)^2 (2z  3)(2z  3) FOIL: Multiply the FRONT two (for F in FOIL) 2z * 2z 4z^2 Multiply the OUTER terms (O in FOIL) 2z * (3) 6z Multiply the INNER terms (I in FOIL) 3 * 2z 6z Multiply LAST terms (L in FOIL) 3 * (3) 9 Therefore, 4z^2  6z  6z + 9 Combine like terms, 4z^2  12z + 9
 2 years ago
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