pls help! Faelyn grouped the terms and factored the GCF out of the groups of the polynomial 6x4 – 8x2 + 3x2 + 4. Her work is shown. Step 1: (6x^4 – 8x^2) + (3x^2 + 4) Step 2: 2x^2(3x^2 – 4) + 1(3x^2 + 4) Faelyn noticed that she does not have a common factor. Which accurately describes what Faelyn should do next?

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pls help! Faelyn grouped the terms and factored the GCF out of the groups of the polynomial 6x4 – 8x2 + 3x2 + 4. Her work is shown. Step 1: (6x^4 – 8x^2) + (3x^2 + 4) Step 2: 2x^2(3x^2 – 4) + 1(3x^2 + 4) Faelyn noticed that she does not have a common factor. Which accurately describes what Faelyn should do next?

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HI!!
is this \[6x^4-8x^2+3x^2+4\]?
yes

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it is the same as \[6x^4-5x^2+4\] when you combine like terms
yup
but it does not factor, so the question "what should Faelyn do next" has not real answer other than "go have a beer"
not sure what the options are looks like you were given choices
a) Faelyn should realize that her work shows that the polynomial is prime. b) Faelyn should go back and regroup the terms in Step 1 as (6x^4 + 3x^2) – (8x^2 + 4). c) In Step 2, Faelyn should factor only 2x out of the first expression. d) Faelyn should factor out a negative from one of the groups so the binomials will be the same
hmm not really clear probably the first one, because it is prime but that is not a proof that it is prime we can check the other ones
b) Faelyn should go back and regroup the terms in Step 1 as (6x^4 + 3x^2) – (8x^2 + 4). this does not help
c) In Step 2, Faelyn should factor only 2x out of the first expression. this does not help either
d) Faelyn should factor out a negative from one of the groups so the binomials will be the same this is wrong as well, not really clear it is prime, so no matter what you do you will not be able to factor it
this is the problem
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it still does not factor
no matter what you do might as well throw in the towel
oh i got it, tnx
its prime
actually we can combine like terms for the x^2 part making the equation \[6x^4-5x^2+4\] now we can factor if and only if we have a perfect square result from the discriminant formula which is \[b^2-4ac\] so letting b = -5,a = 6, and c = 4 \[(-5)^2-4(6)(4)\] \[25-96 = -71\] not a perfect square so this quadratic equation can't be factored. The quadratic formula needs to be used, but it's not asking to solve further in this question. So, this is a prime polynomial because in the discriminant we had -71. the number 71 is indeed a prime number. However, taking the square root of -71 we have \[\sqrt{71}i\]. since negative numbers aren't allowed in the radical the result is imaginary or i. all the roots for this equation is complex.
sorry only saw earlier post when the equation wasn't combined for the first line of my comment

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