Question

If `f(x)=ax^6 + bx^4 + cx^3`, where `a`, `b` and `c` are integers, how many distinct rational zeros could `f(x)` have? a. 1 or 3 b. 2 or 4 c. 1, 3, or 5 d. 2, 4, or 6

Answer 1

None of those options is complete and correct...

Explanation:

`f(x) = ax^6+bx^4+cx^3 = (ax^3+bx+c)x^3`

This has one zero of multiplicity `3`, namely `x=0`.

The remaining cubic has at least one real zero, but such a zero is not necessarily rational. Any zeros of the cubic are necessarily non-zero.

• `x^3+x+1` has no rational zeros

• `x^3+2x-3` has one rational zero, namely `1`

• `x^3-3x+2` has two distinct rational zeros, namely `1` and `-2`

• `x^3-7x+6` has three distinct rational zeros, namely `1`, `2` and `-3`

`f(x)` has one more distinct rational zero than the cubic, so `1`, `2`, `3` or `4`.

In addition, if `a, b, c` are allowed to be `0` then:

`0x^3+0x+0` has infinitely many distinct rational zeros, namely all the rational numbers.