# 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

#### Answer:

**Zeros are**

#### Explanation:

Zeros are

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

[Ans]

#### 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

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

#5-11+7-1 = 0#

Hence

#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

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

The remaining linear factor

So the zeros of

#1" "# with multiplicity#2#

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

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