Use the Rational Zeros Theorem to find the possible zeros of the following polynomial function: #f(x)=33x^3-245x^2+407x-35#?
1 Answer
The possible rational zeros are:
#+-1/33, +-1/11, +-5/33, +-7/33, +-5/11, +-7/11, +-1/3, +-1, +-35/33, +-5/3, +-7/3, +-35/11, +-5, +-7, +-35/3, +-35#
Explanation:
Given:
#f(x) = 33x^3-245x^2+407x-35#
By the rational zeros theorem, any rational zeros of
The divisors of
#+-1, +-5, +-7, +-35#
The divisors of
#+-1, +-3, +-11, +-33#
So the possible rational zeros are:
#+-1, +-5, +-7, +-35#
#+-1/3, +-5/3, +-7/3, +-35/3#
#+-1/11, +-5/11, +-7/11, +-35/11#
#+-1/33, +-5/33, +-7/33, +-35/33#
or in increasing order of size:
#+-1/33, +-1/11, +-5/33, +-7/33, +-5/11, +-7/11, +-1/3, +-1, +-35/33, +-5/3, +-7/3, +-35/11, +-5, +-7, +-35/3, +-35#
Note that these are only the rational possibilities. The rational zeros theorem does not tell us about possible irrational or complex zeros.
Using Descartes' Rule of Signs, we can determine that this cubic has no negative zeros and
So the only possible rational zeros are:
#1/33, 1/11, 5/33, 7/33, 5/11, 7/11, 1/3, 1, 35/33, 5/3, 7/3, 35/11, 5, 7, 35/3, 35#
Trying each in turn, we find:
#f(1/11) = 33(color(blue)(1/11))^3-245(color(blue)(1/11))^2+407(color(blue)(1/11))-35#
#color(white)(f(1/11)) = (3-245+4477-4235)/121#
#color(white)(f(1/11)) = 0#
So
#33x^3-245x^2+407x-35 = (11x-1)(3x^2-22x+35)#
To factor the remaining quadratic we can use an AC method:
Find a pair of factors of
The pair
Use this pair to split the middle term then factor by grouping:
#3x^2-22x+35 = (3x^2-15x)-(7x-35)#
#color(white)(3x^2-22x+35) = 3x(x-5)-7(x-5)#
#color(white)(3x^2-22x+35) = (3x-7)(x-5)#
So the other two zeros are:
#x=7/3" "# and#" "x = 5#