split - Factor Polynomial

[JSmith]

Ahhhhh... yes. thank you, I couldn't find what I missed. Thank you! I then had to factor the equation, this is what I have so far, however I can't figure out how to factor it further:

If there was a negative somewhere in the quadratic equation then it would be easy....

 

[Deleted member 4993]

Ahhhhh... yes. thank you, I couldn't find what I missed. Thank you! I then had to factor the equation, this is what I have so far, however I can't figure out how to factor it further:

If there was a negative somewhere in the quadratic equation then it would be easy....

So what was it that you needed to do?

Don't loose sight of that fact!!

 

[JSmith]

So what was it that you needed to do?

Don't loose sight of that fact!!

I just had to factor it. Can it be factored further??

 

[DrPhil]

I just had to factor it. Can it be factored further??

To find whether a quadratic and be factored, find the "discriminant":
d = b^2 - 4ac
2x^2 + 11x + 6 -->
d = 11^2 - 4(2)(6) = ...

If the discriminant has a rational square root, then the expression can be factored; otherwise the your quadratic factor is "irreducible." A special case is d=0, in which case the two roots are equal.

If d is negative, there are no real roots. Otherwise (positive but not a perfect square), the roots are irrational.

 

[Deleted member 4993]

I just had to factor it. Can it be factored further??

If you did it correctly - you cannot factorize further in rational domain.

To check your factorization - multiply the factors out and see if you get your original polynomial back. I do not think your factorization is correct.

The constant terms in the factored form are -2, -5 & 6. Multiplied together those give you +60. However, in your original equation you have -60. So.....

 

[lookagain]

Ahhhhh... yes. thank you, I couldn't find what I missed.
Thank you! I then had to factor the equation,
this is what I have so far, however I can't figure out how to factor it further:

If there was a negative somewhere in the quadratic equation then it would be easy....

JSmith,

your list of the possible rational roots is wrong. You are not to divide the constant 60 by the
leading coefficient of 2 to get 30 and then list the plus/minus divisors of 30. (Notice that the
greatest common factor for all of the coefficients and constant term for the given quartic
polynomial is 1, so a "2" cannot be factored out to leave a polynomial with all integer
coefficients and constant term.)

The list of possible rational roots consists of all of the plus/minus integers,
and/or plus/minus non-integer fractions in lowest terms, where the fractions
are initially formed with integer divisors of 60 in the numerator and integer
divisors of 2 in their denominators.


Here is the list of all of the possible rational roots:


\(\displaystyle \pm \ 1/2, \ \ \pm \ 3/2, \ \ \pm \ 5/2, \ \ \pm \ 15/2, \ \ \pm \ 1, \ \ \pm \ 2, \ \ \pm \ 3, \ \ \pm \ 4, \ \ \pm \ 5, \)

\(\displaystyle \pm \ 6, \ \ \pm \ 10, \ \ \pm \ 12, \ \ \pm \ 15, \ \ \pm \ 20, \ \ \pm \ 30, \ \ \pm \ 60\)




One of the sources:

http://en.wikipedia.org/wiki/Rational_root_theorem

 

[JSmith]

Good to know about the possible zeros, however that shouldn't be where I went wrong as I still did find factors. I can't see where my error is here, as when I multiply it all out I do in fact get +60 at the end.