x^3+5x^2-2x-15=0. Use the Rational Root Theorem to list all possible rational roots and find any actual rational roots please.

Question

Having trouble with this one :/. 

Answer 1

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) = x³ + 5x² -2x -15 = 0

 

all 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) = 4x⁵ - 2x⁴ + 30x³ -15x² +50x -25

 

all 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) = 6x⁴ - 11x³ + 8x² -11x -30

 

all 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.

 

Answer 2

OK. The Rational Root Theorem, very roughly, says that factors of the constant term in your polynomial, divided by factors of the first term, make good first guesses (though no guarantees) for your actual roots.

 

In this case, your first term is 1. So the "good first guesses are 1, 3, 5, 15, -1, -3, -5, -15. Plug some of those into f(x) = x³ + 5x² - 2x - 15 and see if they make f(x) go to 0. Neither 1 nor -1 works. Neither 3 nor -3 works. You go on, and you find none of the integers above work.

 

BUT. A couple of those integers are CLOSE. Especially -5. You can nose around near there and find one solution. Then you can simplify the function to a quadratic, and find the other two solutions with the Quadratic Equation.

 

Cheers!

Answer 3

Descartes' Rule of signs shows that this polynomial has 1 positive real root and 2 or 0 negative real roots

(f(x) has 1 change of sign, and f(-x) has 2 changes of sign)/

 

The possible rational roots will be +/- factors of the constant term:  +/- 1, +/- 3, +/- 5, and +/-15.

 

Since there is one positive real root, try to find it first:  f(1) = -11; f(3) = 32, no need to check 5 or 15 because the one root, which is not rational, lies between 1 and 3 (f(x) changed from negative to positive, so the intermediate value theorem says a zero must lie between the two x values).

 

Now test the negative possibilities:  f(-1) = -9; f(-3) = 28 (note that an irrational root lies between x = -1 and x = -3), f(-5) = -5, so the other negative root, which is also irrational, lies between x = -3 and x = -5.

 

There are no rational roots for this polynomial.