How do you use the rational roots theorem to find all possible zeros of # p(x)= 2x^5+7x^3+6x^2-15#?

1 Answer
Mar 1, 2016

Answer:

#p/q: ⇒ ±1/2, ±1, ±2#

Explanation:

If #f(x)=a_nx^n+a_(n−1)x^(n−1)+…+a_1x+a_0# has integer coefficients and p/q (where p/q is reduced) is a rational zero, then #p# is a factor of the constant term #a_0# and #q# is a factor of the leading coefficient #a_n#.
Given this all what we need to do here is identify the factors of #a_0# and #a_n# and build the #p/q# ratios that have the potential to be roots.
We have #2x^5 + 7x^3+6x^2-15#

so p⇒ ±1,±2 and q⇒±1,±2
and the ratio:
#p/q: ⇒ ±1/2, ±1, ±2#

This are the potential roots, but remember they are not guaranteed... In this case you can try (x-1), use long division and see if you can divide without a remainder...
You were not asked to do that but you can divide #(x-1)# and check:
#2x^4+2x^3+9x^2+15x+15#