Find a polynomial of degree n that has the given zeros and degree. Answers may vary.

zeros: x = 0,-4,5,9

Degree n = 7

Apply the Rational Zero Test to find a zero of the function. Use the result to find the remaining zeros and sketch a graph of the polynomial (no idea what the rational zero test is)

f(x)= x^3 - 4x^2 + x + 6

Guest Apr 18, 2020

#1**+1 **

If the zeros are 0, -4, 5 and 9,

there must be factors of: (x - 0), (x + 4), (x - 5), and (x - 9).

If the degree is 7, you must have 7 factors (some can be repeated; in this case some must be repeated).

You can also have a constant as a multiplier.

Thus, one possible polynomial function is: y = -7(x)^{2}(x + 4)(x - 5)^{3}(x - 9)

The constant in front (the -7) can be any number except 0.

The exponents on the factors are your choice, they just must have a total of 7.

Rational Zero Test.

If there is a rational root, like 7 or -2/3, the numerator must be a divisor of the last term (in this case, the 6) and the denominator must be a divisor of the coefficient of the first term (in this case, the 1, the coefficient of the x^{3} term).

There will be several possibilities, including all positive and negatives.

For this problem: y = x^{3} - 4x^{2} + x + 6

To get the numerator, we have to factor the 6, and get 1, 2, 3, and 6.

To get the denominator, we have to factor the 1, and get just 1.

So, the possibilities are 1/1 = 1, 2/1 = 2, 3/1 = 3, and 4/1 = 4.

We also need the negatives: -1, -2, -3, and -4.

One by one, we try these in the function:

1: 1^{3} - 4·1^{2} + 1 + 6 = 4 (this doesn't work; the answer must be 0 for it to work)

2: 2^{3} - 4·2^{2} + 2 + 6 = 0 (this works, therefore, x - 2 is a root).

After you find a root divide the function by this factor to get a smaller function:

(x^{3} - 4x^{2} + x + 6)) divided by (x - 2) gives a quotient of x^{2} - 2x - 3

Factoring this quotient into (x - 3)(x + 1) gives you the other roots of 3 and -1.

You now have the three roots: 2, 3, and -1.

geno3141 Apr 18, 2020