How do you solve this? f(x)= x^3- 6x^2 + 4x + 5?

Question

ahh.. our teacher never explained this well.. HELP!
so..far.. i came up with the solution, f'(x)= 3x^2 - 12x +4?
is this the right answer already??!

This Lesson is about derivatives!!! CALCULUS!! not.. factoring.. -__-"!!!

Answer 1

What is meant by "solve" ?
Do you wish to find f `(x)?

f `(x) = 3x² - 12x + 4 as you correctly state.

Answer 2

The rational root theorem tells you that the roots of a polynomial can be any one of the factors of the product of the first and last terms in the polynomial. This means that 1*5 = 5, so a root may be 1, 5, -1, or -5. By experimentation, it turns out that one of the roots is 5. Then, you can use long division or synthetic division to find the other two roots, and those are your answers.

The roots are x = 5, -(sqrt(5) -1)/2, (sqrt(5)+1)/2

Answer 3

Either by Remainder Theorem, or by grouping, we find that x - 5 is a factor of this polynomial.

x^3 - 6x^2 + 4x + 5 = x^3 - 5x^2 - x^2 + 5x - x + 5

= x^2(x - 5) - x (x - 5) - (x - 5)

= (x - 5)(x^2 - x - 1)................................................................. (1)

So that if you want solution to the "EQUATION"

x^3 - 6x^2 + 4x + 5 = 0 would be, from (1):

Either x - 5 = 0, giving x = 5, OR

x^2 - x - 1 = 0. Soving this quadratic x = ½*[1 + sq. root(5)]

or x = ½*[1 - sq. root(5)].

Answer 4

take a derivative of thw f(x) and place it equal to zero to get the values of X ..ull get 2 values ..use them to get the third..the 2 values of X are 3.36 and .37

Answer 5

x^3-6x^2+4x+5=(x-5)(x^2-x-1)

Answer 6

f(x)=4x(^3)-6^2+5

Answer 7

use factor theorem!!! hope it helps.

Answer 8

rght

but thts

f'(x)

YAY IM CORRECT