Rational Zeros of Function

Question

find all the rational zeros of these functions

x³ - 10x² + 23x - 14

x³ + 6x² + 9x + 4

Answer 1

In order to find rational zeros of a cubic, we lean on the Rational Root Theorem. It says any rational zero has to be a factor of the constant term divided by a factor of the leading coefficient. In both of these the leading coefficient is 1, which makes life easy, because it means we only have to look at the factors of the constant.

For the first one, x^3 - 10x^2 + 23x - 14, the constant is -14, so our candidates are plus or minus 1, 2, 7 and 14. Now we just test them, and I always start with 1 because it is the fastest to check mentally:

1 - 10 + 23 - 14 = 0

That works, so x = 1 is a zero and (x - 1) is a factor. Dividing the cubic by (x - 1), either with synthetic division or long division, leaves us with a quadratic:

x^2 - 9x + 14

That factors the ordinary way. We need two numbers that multiply to 14 and add to -9, which are -2 and -7:

(x - 1)(x - 2)(x - 7)

So the zeros are x = 1, 2 and 7.

Remember, the whole strategy is the same for the second one. For x^3 + 6x^2 + 9x + 4 the constant is 4, so the candidates are plus or minus 1, 2 and 4. Since every sign in the polynomial is positive, no positive number can ever make it zero, so we only need to try the negatives. Starting with -1:

-1 + 6 - 9 + 4 = 0

So x = -1 is a zero and (x + 1) is a factor. Dividing gives:

x^2 + 5x + 4

which factors into (x + 1)(x + 4). Notice that (x + 1) showed up a second time:

(x + 1)^2 (x + 4)

So the zeros are x = -1, which is a double root, and x = -4.

So, our final answers are 1, 2 and 7 for the first function, and -1 (multiplicity 2) and -4 for the second.

Answer 2

We will use the Rational Root Theorem, or p/q test, for both problems.

Given: q*x^n + ... + p

Possible Rational Roots will be ±[factors of p / factors of q]

We will then use synthetic division to see if they are roots.

============================================================

Problem #1:

x³ - 10x² + 23x - 14

p = 14

q = 1

p/q = 14/1

Possible Rational Roots will be ±[(14 * 1, 7 * 2) / 1] = ±{1,2,7,14} = 1, -1, 2, -2, 7, -7, 14, -14

Synthetic Division #1:

1 ⌋ 1 -10 23 -14

1 -9 14

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

1 -9 14 ⌉ 0

Therefore, x = 1 is a root

Synthetic Division #2:

7 ⌋ 1 -9 14

7 -14

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

1 -2 ⌉ 0

Therefore, x = 7 is a root

Synthetic Division #3:

2 ⌋ 1 -2

2

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

1 ⌉ 0

Therefore, x = 2 is a root

The Rational Roots for problem #1 are x = 1, 2, 7

============================================================

Problem #2:

x³ + 6x² + 9x + 4

p = 4

q = 1

p/q = 4/1

Possible Rational Roots will be ±[(4 * 1, 2 * 2) / 1] = ±{1,2,4} = 1, -1, 2, -2, 4, -4

Synthetic Division #1:

-1 ⌋ 1 6 9 4

-1 -5 -4

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

1 5 4 ⌉ 0

Therefore, x = -1 is a root

Synthetic Division #2:

-4 ⌋ 1 5 4

-4 -4

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

1 1 ⌉ 0

Therefore, x = -4 is a root

Synthetic Division #3:

-1 ⌋ 1 1

-1

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

1 ⌉ 0

Therefore, x = -1 is a root

The Rational Roots for problem #2 are x = -1, -1, -4

NOTE: -1 is a double root