lokiando one year ago 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?

1. misty1212

HI!!

2. misty1212

is this $6x^4-8x^2+3x^2+4$?

3. lokiando

yes

4. misty1212

it is the same as $6x^4-5x^2+4$ when you combine like terms

5. lokiando

yup

6. misty1212

but it does not factor, so the question "what should Faelyn do next" has not real answer other than "go have a beer"

7. misty1212

not sure what the options are looks like you were given choices

8. lokiando

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

9. misty1212

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

10. misty1212

b) Faelyn should go back and regroup the terms in Step 1 as (6x^4 + 3x^2) – (8x^2 + 4). this does not help

11. misty1212

c) In Step 2, Faelyn should factor only 2x out of the first expression. this does not help either

12. misty1212

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

13. lokiando

this is the problem

14. misty1212

it still does not factor

15. misty1212

no matter what you do might as well throw in the towel

16. lokiando

oh i got it, tnx

17. lokiando

its prime

18. UsukiDoll

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.

19. UsukiDoll

sorry only saw earlier post when the equation wasn't combined for the first line of my comment