Find all rational zeros of f(x) = 6x^4− 5x^3 − 2x^2 − 8x + 3

Answer 1

The zeros of a polynomial are the x values that make y (aka f(x)) equal to zero. To find them we set f(x) equal to zero and solve:

6x⁴ − 5x³ − 2x² − 8x + 3 = 0

One way to solve this is to graph it and look to see where the graph crosses the x-axis (y = 0) but we'll pretend like that's not an option.

To start, the Rational Roots Theorem says that if there are any rational zeros, they will be (±) the factors of the constant term divided by the factors of the leading coefficient. The factors of the constant term 3 are 1 and 3 and the factors of the leading coefficient 6 are 1, 2, 3, and 6 . That makes the POSSIBLE rational zeros: ±1/6. ±1/3, ±1/2, ±1, ±3/2, and ± 3

Let's try them using synthetic division. I'll start with 1/6

1/6 | 6 -5 -2 -8 3

| 1 -2/3 -4/9 -38/27

-----------------------------------

6 -4 -8/3 -76/9 43/27

Since we didn't get a zero remainder, 1/6 is not a zero

Now let's try 1/3::

1/3 | 6 -5 -2 -8 3

| 2 -1 -1 -3

-------------------------

6 -3 -3 -9 0

We got lucky, this has a zero remainder so 1/3 is one of the zeros. Not only did we find a zero but we can also now work with a smaller polynomial that is the quotient (6x³ - 3x² - 3x - 9). This polynomial has a common factor of 3 so it can be reduced to 2x³ - x² - x - 3.

Now let's try 1/2:

1/2 | 2 -1 -1 -3

| 1 0 -1/2

-------------------------

2 0 -1 -7/2

Doesn't work, so let's try 1:

1 | 2 -1 -1 -3

| 2 1 0

-------------------------

2 1 0 -3

Doesn't work, so let's try 3/2:

3/2 | 2 -1 -1 -3

| 3 3 3

-------------------------

2 2 2 0

So since this worked (zero remainder, we know x = 3/2 is a zero and we now have a quadratic as the quotient: 2x² + 2x + 2 = 0 allowing us to use the quadratic formula:

x = (-2 ± √(2² - 4•2•2))/(2•2) = (-2 ± √-12)/4 which is an imaginary solution (no real roots).

So the rational zeros are x = 1/3 and x = 3/2.