Yohan C. answered • 04/24/16

Tutor

4
(1)
Math Tutor (up to Calculus) (not Statistics and Finite)

Hey Malik,

Rational Root Theorem gives you all the "possible" factors for your polynomial:

**(all factors of constant) / (all factors of leading coefficient of highest power)**

For yours will be: 15 / 1 f(x) = x

^{3}+ 5x^{2}-2x -15 = 0all factors of 15 (constant) =

**(+/-) 1, (+/-) 3, (+/-) 5, (+/-) 15**all factors of 1 (leading coefficient of highest power) =

**(+/-) 1**So, all your possible rational roots are: (+/-) [1, 3, 5, 15]

**(+/-)1, (+/-)3, (+/-)5, (+/-)15**

*** total of 8 but there are no rational roots for yours.**

Good luck to you.

Here are some examples (so you can understand what just happened):

f(x) = 4x

^{5}- 2x^{4}+ 30x^{3}-15x^{2}+50x -25all factors of 25 (constant) = (+/-) 1, (+/-) 5, (+/-) 25

all factors of 4 (leading coefficient of highest power) = (+/-) 1, (+/-) 2, (+/-) 4

All possible rational roots are:

1,5,25

-------

1,2,4

(+/-) 1, (+/-) 5, (+/-) 25

(+/-) 1/2, (+/-) 5/2, (+/-) 25/2

(+/-) 1/4, (+/-) 5/4, (+/-) 25/4

(+/-) [1, 5, 25, 1/2, 5/2, 25/2, 1/4, 5/4, 25/4]

* There are 6 possible rational roots out of total of all 9 roots.

f(x) = 6x

^{4}- 11x^{3}+ 8x^{2}-11x -30all possible factors for 30 (constant): 1,2,3,5,6,10,15,30

all possible factors for 6 (leading coefficient of highest power): 1,2,3,6

1,2,3,5,6,10,15,30

-------------------

1,2,3,6

(there will be repeated factors/roots throughout the process for this particular one)

(1/1,2,3,6) (2/1,2,3,6) (3/1,2,3,6) (5/1,2,3,6)

(6/1,2,3,6) (10/1,2,3,6) (15/1,2,3,6) (30/1,2,3,6)

(+/-)[1, 1/2, 1/3, 1/6] (+/-)[2, 1, 2/3, 1/3] (+/-)[3, 3/2, 1, 1/2] (+/-)[5, 5/2, 5/3, 5/6]

(+/-)[6, 3, 2, 1] (+/-)[10, 5, 10/3, 5/3] (+/-)[15, 15/2, 5, 5/2] (+/-)[30, 15, 10, 5]

(+/-) [1, 2, 3, 5, 6, 10, 15, 30, 1/2, 3/2, 5/2, 15/2, 1/3, 2/3, 5/3, 10/3, 1/6, 5/6]

*There are 10 possible rational roots out of all 36 roots.

Ron G.

04/24/16