List all the possible rational zeros of p(x) = 3x^3 - 3x^2 + 4x - 12

Answer 1

3x^3 -3x^2 + 4x - 12

has at least one real zero. A 3rd degree polynomial crosses the x axis at least once, and possibly 3 times.

The real root(s) are positive, given the change in signs among terms. 3 changes in sign, potentially 3 zeros. Following DeCartes

f(1) = -8

f(0) = -12

f(2) = +8, so one real zero is between 1 and 2

others potentially greater than 2

check the turning points, there are potentially 2 for a cubic polynomial.

take the derivative and set equal to zero

6x^2 - 6x + 4 = 0

3x^2 -3x + 2 = 0

check the discriminant, b^2- 4ac = 9 - 4(3)(2) <0, so the other two zeros are imaginary. the polynomial graph only intersects the x axis between x=1 and x = 2

That's the one and only real root. 1<x<2

look at the constant term -12 and the coefficient of the cubic term 3. zeros will be factors of one over the other. 3 has factors 1 and 3. 12 has factors 1, 2, 3, 4, 6 and 12. Divide one by the other and get a number between 1 and 2: 3/2, or 4/3,

plug them in

3(3/2)^3 -3((3/2)^2 +4(3/2) - 12 = 81/8 - 27/4 + 12/2 - 12 = (81-54 + 48 - 96)/8 does not equal zero

try 4/3

3(4/3)^3 - 3(4/3)^2 + 4(4/3) - 12 = 192/27 - 48/9 + 16/3 -12 = (192 - 144 + 144 - 174) does not equal zero

the real root looks irrational, meaning zero rational roots

Answer 2

The list of possible rational zeros (sometimes called P/Q) is determined by the coefficient of the leading term and the constant.

3x^3 - 3x^2 + 4x - 12

The leading term, containing the highest exponent, is 3x^3.

We will use the coefficient of 3.

The constant, the number alone, is -12.

Let’s find the factors of 12 and the factors of 3

12:

1, 2, 3, 4, 6, 12

The factors of the constant are the P’s

3:

1, 3

The factors of the leading coefficient are the Q’s

The list of possible rational zeroes is all the combinations of +or- P/Q —> the possible fractions of the factors of 12 divided by the factors of 3.

We will make every combination of +or-P/Q, then reduce and eliminate any equivalent fractions

+ or -

1/1, 1/3, 2/1, 2/3, 3/1, 3/3, 4/1, 4/3, 6/1, 6/3, 12/1, 12/3

Simplify/reduce fractions if possible

+ or -

1 1/3 2 2/3 3 1 4 4/3 6 2 12 4

eliminate any doubles

+ or -

1 1/3 2 2/3 3 4 4/3 6 12

It is optional but it is common to arrange the numbers from smallest to largest

+ or - 1/3 2/3 1 4/3 2 3 4 6 12

These are the possible rational zeroes. All other zeroes either contain square roots or imaginary numbers.

Answer 3

For a polynomial of the form p(x) = axⁿ + bx^n-1+cx^n-2...+qx⁰, we find the possible rational roots by finding all factors of q and dividing by all factors of a. Remember, if the polynomial is factorable then at least some of the factors of the first term and last term will show up in the expression. We don't care about which ones show up for now, we just want to know which ones may show up.

In the example provided, a is 3 and corresponds to the coefficient of the "3x³" and q is -12 and corresponds to the last coefficient. So, all the factors of 3 are ±3 and ±1 while all the factors of -12 are ±1, ±2, ±3, ±4, ±6, and ±12. We divide all possible factors of q by all possible factors of a and remove any duplicates. Going in the order listed, the possible rational roots are ±1/3, ±1, ±2/3, ±2, ±3, ±4/3, ±4, ±6 and ±12. Note, ±1, ±2 and ±4 are duplicates which I did not count twice.

The problem doesn't ask for this, but knowing the possible rational zeros can help you check your work if you factor the polynomial to graph it by hand. If your graph crosses the x axis at points that are not possible rational roots, then you know you need to check your work. You can also guess and check the actual zeros if factoring the polynomial is complicated by plugging the possible rational roots in p(x). If none of them amount to 0, then the polynomial has no actual rational roots.