state the possible positive real zeros, negative real zeros, and imaginary zeros of each function.?

Question

f(x)= 7x³+5x-1


It is not factor directly,
we can not find one factor that the (x - k) form . i also try the method "synthetic substitution" but i could not approproate my answer correctly.


I can draw graph of the cubic equation f(x)= 7x³+5x-1, it give the value x = 0.20 approx.

it is the cubic equation and it will have three roots, the one root is x = 0.20, not find the other two roots

Answer 1

The possible rational roots are:

1, -1, 1/7, -1/7

To actually determine which one it is, we plug them into our function, looking for a value of 0.

f(1) = 7 + 5 - 1 = 11
f(-1) = 7(-1)^3 + 5(-1) - 1 = -7 - 5 - 1 = -13
f(1/7) = 7(1/7)^3 + 5(1/7) - 1 = 1/49 + 5/7 - 1 = [nonzero]
f(-1/7) = 7(-1/7)^3 + 5(-1/7) - 1 = -1/49 + 35/49 - 49/49 = -15/49

There are no rational roots for this function.

To determine how many possible positive real zeros there are, use Descartes' rule of signs.
f(x) = 7x^3 + 5x - 1. The sign of the coefficients alternate only once, so there is one positive zero.

The negative real zeros are found using the same method, but for f(-x)
f(-x) = -7x^3 - 5x - 1. The signs of the coefficients don't alternative, so there are no negative real zeros.

This question's answer is irrational.

Answer 2

This equation can't be solved using a formula, but only with algorithms. Using an Excel spreadsheet, I found one of the roots is approximately 0.190397193. Now, what you have to do is divide 7x³+5x-1 by (x - 0.190397193). You'll get a polynomial of degree 2, and it's roots, real or not, can be determined using Bhaskara.

Answer 3

1 real positive 2 imaginary 1 real negative