Question

What are all the possible rational zeros for `y=30x^3-x^2-6x+1` and how do you find all zeros?

Answer 1

Use the Rational Zero Theorem, synthetic division, and factoring.

Explanation:

According to the Rational Zero Theorem, the list of all possible rational zeros is obtained by dividing all the factors of the constant term 1 by all the factors of the leading coefficient term 30.

The possible factors of 1 are `+-1`

The possible factors of 30 are `+-1, 2, 3, 5, 6,10,15,30`

The possible zeros are:
`+-1/1, +-1/2,+-1/3,+-1/5,+-1/6,+-1/10,+-1/15,+-1/30`

To find all the zeros, use synthetic division. Pick one of the possible zeros as a divisor. If the remainder is zero, the divisor is a zero. If the remainder is not zero, choose another possible zero and try again. I chose 1/3 because I "cheated" and first checked the zeros using a graphing utility.

`1/3~|30color(white)(aa)-1color(white)(aaa)-6color(white)(aaaa)1`
`color(white)(aaaaaaaaa)10color(white)(aaaaa)3color(white)(1a)-1`
`color(white)(aaa)`_________

`color(white)(aaa)30color(white)(aaaaa)9color(white)(aaa)-3color(white)(aaaa)0`

`1/3` is a zero because the remainder is zero.

Write the quotient using the coefficients found in synthetic division and set it equal to zero.
`30x^2+9x-3=0`

Factor and solve to find the remaining zeros:
`3(10x^2+3x-1)=0`
`3(5x-1)(2x+1)=0`
`x=1/5` and `x=-1/2`

The three zeros are `x =-1/2, x=1/5, x=1/3`