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1963 TMTA High School Algebra I Contest Problem 6

Revision as of 21:33, 1 February 2021 by Coolmath34 (talk | contribs) (Created page with "== Problem == <math>(3x+2)(4x-5)</math> is equal to: <math>\text{(A)} \quad 12x^2-10 \quad \text{(B)} \quad 12x^2-23x-10 \quad \text{(C)} \quad 12x^3+7x-10</math> <math>\tex...")
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Problem

$(3x+2)(4x-5)$ is equal to:

$\text{(A)} \quad 12x^2-10 \quad \text{(B)} \quad 12x^2-23x-10 \quad \text{(C)} \quad 12x^3+7x-10$

$\text{(D)} \quad 12x^2-7x-10 \quad \text{(E) NOTA}$

Solution

Use the FOIL method to expand: \[(3x+2)(4x-5) = 12x^2 + 8x- 15x - 10 = \boxed{12x^2- 7x-10}\] The answer is $\boxed{\text{(D)}}.$

See Also

1963 TMTA High School Mathematics Contests (Problems)
Preceded by
Problem 5 		TMTA High School Mathematics Contest Past Problems/Solutions Followed by
Problem 7
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