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OregonDuck
 one year ago
@mathmate
OregonDuck
 one year ago
@mathmate

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OregonDuck
 one year ago
Best ResponseYou've already chosen the best response.0factoring by grouping 2x^+4x+1+4

OregonDuck
 one year ago
Best ResponseYou've already chosen the best response.0my answer is (2x+1)(x+2)(x+4)

OregonDuck
 one year ago
Best ResponseYou've already chosen the best response.0@mathmate @welshfella

Flvs.net
 one year ago
Best ResponseYou've already chosen the best response.1I like your answer, it looks good to me.

Flvs.net
 one year ago
Best ResponseYou've already chosen the best response.1I was in public school advanced too.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2@OregonDuck If the answer is (2x+1)(x+2)(x+4) the question should have been 2x^3+13x^2+22x+8 Do you still want to know \(how\) to solve it?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2We note that all coefficients are positives, therefore so are the factors. The leading coeff. is 2, and the constant term is 8, so the possible factors are 1/2, 1,2,4, and 8. we can try group them assuming a factor of x+2 2x^3+13x^2+22x+8 =2x^3+4x^2 + 9x^2+18x + 4x+8 =2x^2(x+2) + 9x(x+2) + 4(x+2) and it seems to work well, so =(x+2)(2x^2+9x+4) factor the quadratic the same way, if you want: =(x+2)(2x^2+8x + x+4) =(x+2)(x+4)(2x+1)

OregonDuck
 one year ago
Best ResponseYou've already chosen the best response.0so (x+2)(x+4)(2x+1) is the answer?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Yes, for the following question: 2x^3+13x^2+22x+8

OregonDuck
 one year ago
Best ResponseYou've already chosen the best response.0NO WHAT IS THE ANSWER FOR 2x^+4x+1+4

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2you cannot factor 2x^+4x+1+4 because it is not a polynomial. exponent is missing, and I suspect there are typos. Please check the question.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2It also shows that you have not made any attempt to solve the problem, when you have posted your question for over half an hour and not realize that there are typos.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Are you sure about the +1+4 part? The problem could have been written 2x^2+4x+5

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2There are no rational factors for the problem 2x^2+4x+5.

OregonDuck
 one year ago
Best ResponseYou've already chosen the best response.0k if it was written this way then the answer would be?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2There are no rational factors for the problem 2x^2+4x+5.

OregonDuck
 one year ago
Best ResponseYou've already chosen the best response.0then my answer was right then for 2x^+4x+1+4 right?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2what was your answer?

pooja195
 one year ago
Best ResponseYou've already chosen the best response.1#CAUGHT XD you got the answer and now you are presenting it here lol

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Your "answer" expands to a cubic polynomial. You're given a quadratic polynomial to factor, so your "answer" cannot be right. I already gave you the question that corresponds to your "answer".

pooja195
 one year ago
Best ResponseYou've already chosen the best response.1http://openstudy.com/study#/updates/55b25896e4b0ce10565f5149

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2There are no rational factors, as I said earlier.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Just to satisfy my curiosity, can you tell me how you obtained the answer (2x+1)(x+2)(x+4) to "factor 2x^+4x+1+4" ?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2The reason I said there are no rational factors is because the expression 2x^2+4x+5 can be factored into (x+1+sqrt(3/2)i)(x+1sqrt(3/2)i) where i is the complex number where i^2=1 by completing the square!
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