OregonDuck
  • OregonDuck
@mathmate
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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OregonDuck
  • OregonDuck
factoring by grouping 2x^+4x+1+4
OregonDuck
  • OregonDuck
my answer is (2x+1)(x+2)(x+4)
OregonDuck
  • OregonDuck
@mathmate @welshfella

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More answers

OregonDuck
  • OregonDuck
@DarkMoonZ
Flvs.net
  • Flvs.net
Need help?
OregonDuck
  • OregonDuck
yes plz
Flvs.net
  • Flvs.net
I like your answer, it looks good to me.
Flvs.net
  • Flvs.net
7th advanced, you?
Flvs.net
  • Flvs.net
Oh really?
Flvs.net
  • Flvs.net
I was in public school advanced too.
mathmate
  • mathmate
@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?
OregonDuck
  • OregonDuck
yes plz
OregonDuck
  • OregonDuck
@mathmate yes plz
mathmate
  • mathmate
We 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
  • OregonDuck
so (x+2)(x+4)(2x+1) is the answer?
mathmate
  • mathmate
Yes, for the following question: 2x^3+13x^2+22x+8
OregonDuck
  • OregonDuck
NO WHAT IS THE ANSWER FOR 2x^+4x+1+4
mathmate
  • mathmate
you 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
  • mathmate
It 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.
OregonDuck
  • OregonDuck
no 2x^2+4x+1+4
mathmate
  • mathmate
Are you sure about the +1+4 part? The problem could have been written 2x^2+4x+5
OregonDuck
  • OregonDuck
yes i am sure
mathmate
  • mathmate
There are no rational factors for the problem 2x^2+4x+5.
OregonDuck
  • OregonDuck
k if it was written this way then the answer would be?
mathmate
  • mathmate
There are no rational factors for the problem 2x^2+4x+5.
OregonDuck
  • OregonDuck
then my answer was right then for 2x^+4x+1+4 right?
OregonDuck
  • OregonDuck
or prime?
mathmate
  • mathmate
what was your answer?
OregonDuck
  • OregonDuck
(2x+1)(x+2)(x+4)
pooja195
  • pooja195
#CAUGHT XD you got the answer and now you are presenting it here lol
mathmate
  • mathmate
Your "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".
OregonDuck
  • OregonDuck
so it is prime?
mathmate
  • mathmate
There are no rational factors, as I said earlier.
mathmate
  • mathmate
Just 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
  • mathmate
The 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+1-sqrt(3/2)i) where i is the complex number where i^2=-1 by completing the square!

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