What are all the possible rational zeros for #f(x)=5x^3-11x^2+7x-1# and how do you find all zeros?

2 Answers
Sep 3, 2017

Answer:

Zeros are #x=0.2 and x=1#

Explanation:

# f(x) = 5x^3 -11x^2 +7x-1 = 5x^3 -5x^2-6x^2+6x +x-1 #

#= 5x^2 (x-1)- 6x (x-1) +1 (x-1) #

#= (x-1)(5x^2-6x+1) = (x-1)(5x^2-5x-x+1)#

#= (x-1){5x(x-1) -1(x-1} = (x-1)(x-1)( 5x-1) #

#= (x-1)^2(5x-1)#

Zeros are

#(x-1=0 or x=1 and 5x-1=0 or 5x=1 or x=0.2#

graph{5x^3-11x^2+7x-1 [-10, 10, -5, 5]}
[Ans]

Sep 3, 2017

Answer:

The "possible" rational zeros are:

#+-1/5, +-1#

The actual zeros are:

#1" "# with multiplicity #2#

#1/5" "# with multiplicity #1#

Explanation:

Given:

#f(x) = 5x^3-11x^2+7x-1#

By the rational root theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-1# and #q# a divisor of the coefficient #5# of the leading term.

That means that the only possible rational zeros are:

#+-1/5, +-1#

To find the actual zeros, we could just try each of these in turn, but there is a shortcut:

Note that the sum of the coefficients of #f(x)# is zero. That is:

#5-11+7-1 = 0#

Hence #x=1# is a zero and #(x-1)# a factor:

#5x^3-11x^2+7x-1 = (x-1)(5x^2-6x+1)#

The same is true of the remaining quadratic:

#5-6+1 = 0#

Hence #x=1# is a zero again and #(x-1)# a factor:

#5x^2-6x+1 = (x-1)(5x-1)#

The remaining linear factor #5x-1# gives us a zero #x=1/5#

So the zeros of #f(x)# are:

#1" "# with multiplicity #2#

#1/5" "# with multiplicity #1#

graph{5x^3-11x^2+7x-1 [-2.107, 2.89, -1.14, 1.36]}