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Asked by Hailey

Name 2 values of that make the expression factorable

6x^2+bx-11
4 years ago

Answers

Answered by oobleck
the discriminant is b^2 + 264
so you need that to be a perfect square
25+264 = 289 = 17^2
6x^2+5x-11 = (6x+11)(x-1)
6x^2-5x-11 = (6x-11)(x+1)
4 years ago
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