Algebra for calculus:

[unknownfly]

Hello all! I need some help with this question: (x^4) - (5x^3) + (7x^2) - (5x) + 6 = 0
This is the last question on a homework sheet that I am trying to complete on solving general polynomial equations with Descartes law of signs & Rational Root Theorem, but I am having a hard time with it so any help is appreciated!

 

[Deleted member 4993]

Hello all! I need some help with this question: (x^4) - (5x^3) + (7x^2) - (5x) + 6 = 0
This is the last question on a homework sheet that I am trying to complete on solving general polynomial equations with Descartes law of signs & Rational Root Theorem, but I am having a hard time with it so any help is appreciated!

Did you apply "rational root theorem" to the given function? Did it provide any root/s for the polynomial? If it did - what were those?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.

Also, please tell us what the exact question was. Was it "Solve for 'x'" or something else? You did not tell us!

 

[Steven G]

I just tried x=2 and it worked. No pen and paper, nothing. Why did you not try x=1, x=-1, x=2. So what do you do now that you know that x=2 is a solution?

 

[topsquark]

Hello all! I need some help with this question: (x^4) - (5x^3) + (7x^2) - (5x) + 6 = 0
This is the last question on a homework sheet that I am trying to complete on solving general polynomial equations with Descartes law of signs & Rational Root Theorem, but I am having a hard time with it so any help is appreciated!

Using the rational root theorem the possible rational roots are [math]\pm 1, ~ \pm 2, ~\pm 3, ~ \pm 6[/math]. By substitution we see that x = 2 and x = 3 are solutions. Do synthetic division by these roots and you have a quadratic equation which you can solve.

-Dan