Question

how to find the x intercepts for f(x) = 3x^3+7x^2+3x+1
please show your steps

Answer 1

I guess you didn't accept my previous reply to your question.

Well, there is one intercept at x = -1.899
rounded off to the third decimal.

As I told you before, it does not factor with rational numbers,
if you solve 3x^3+7x^2+3x+1 = 0 you get the real root that I stated, plus two imaginary roots.

Answer 2

The x intercepts are where y = f(x) = 0
I don't see any easy solutions here. Whatever roots (x-values) there are will be negative. I suggest a graphical or an iteration solution. One root is approximately x = -1.90

Answer 3

The exact solution is not difficult to obtain, though.

The equation is

x^3+7/3 x^2+x+1/3 = 0

Get rid of the quadratic term by substituting x = y - 7/9:


y^3 - 22/27 y + 362/729 = 0 (1)


This can be solved by comparing with the identity:

(a+b)^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3

You can rewrite this as:

(a+b)^3 = 3ab(a+b) + a^3 + b^3

This means that a solution of the equation:

y^3 - 3ab y - (a^3 + b^3) = 0

is y = a + b

So, to solve (1) we can extract a and b:

3 a b = 22/27 (2)

a^3 + b^3 = -362/729 (3)

Take the third power of (2) and define

A = a^3, B = b^3:

A B = 22^3/3^12 (4)

And (3) can be written as:

A + B = -362/3^6 (5)

Solving (4) and (5) amounts to solving a quadratic equation, A and B are then the two solutions:

A = -181/3^6 + 1/81 sqrt(273)

B = -181/3^6 - 1/81 sqrt(273)

The real solution is then obtained as:

y = a + b

with

a = -[181/3^6 - 1/81 sqrt(273)]^1/3

b = -[181/3^6 + 1/81 sqrt(273)]^1/3


The complex solutions are obtained from the other cube roots of A and B. If we multiply a by exp(2 pi i n/3) to obtain another cube root of A, then we must multiply b by exp(-2 pi i n/3), because eq. (2) needs to be satisfied.