Question

A. use this function to answer questions a-d f(x)=3x^3-7x^2-14x 24 using the fundamental theorem of algebra. how many possible complex roots with this polynomial have? explain. b. using descartes rule determine how many possible positive roots and how many possible negative roots this function will have c. use the same function f(x)=3x^3-7x^2-14x 24 using the rational roots theorem list all the possible rational roots for the function d. using possible roots you found on the previous question find all the possible zeros for this function (hint: you will use remainder theorem and synthetic division)

Answer 1

Below in bold.

Explanation:

I have assumed that it's + 24.

a. This is a cubic expression (highest power is 3).

By the theorem there are 3 roots with 2 possible complex roots.

b. f(x) = 3x^3-7x^2-14x+ 24

Using Descartes Rule of signs, the signs from left to right are:

+ - - +

Here there are 2 sign changes (+ - and - +)

So, the maximum number of positive real roots is 2.

f(-x)= 3(-x)^3 - 7(-x)^2 - 14(-x) + 24

Sign changes are:

- - + +

There is one change of sign so maximum number of possible negative real roots is 1.

c. Using the Rational roots theorem, the list of all possible rational roots is:

+/- (factors of 24 / factors of 3)

= +/- [1, 2, 3, 4, 6, 8, 12, 24) / (1, 3) ]

= +/- 1, +/-2, +/-3, +/- 4, +/-6, +/-12, +/- 8, +/-24, +/- 1/3, +/-2/3, +/-4/3, +/-8/3

d. Possible roots, using the Remainder theorem:

x = 1:

f(1) = 3 - 7 - 14 + 24 = 8 so 1 is not a root

f(-1) = 28 - so x = -1 not a root

f(2) = -8 so 2 not a root

f(-2) = 0 So, x = -2 is a root

So, x + 2 is a factor of f(x) so we divide

x + 2)3x^3 - 7x^2 - 14x + 24(3x^2 - 13x + 12 <--- Quotient

3x^3 + 6x^2

- 13x^2 - 14x

- 13x^2 - 26x

12x + 24

12x + 24

..............

So, we solve the following equation to find the other 2 roots:

3x^2 - 13x + 12 = 0

3x^2 - 9x - 4x + 12 = 0

3x(x - 3) - 4(x - 3) = 0

3x - 4 = 0 or x - 3 = 0

x = 4/3, 3.

The 3 roots are -2, 4/3, 3.