Question

What are the possible number of positive, negative, and complex zeros of f(x) = 3x4 - 5x3 - x2 - 8x + 4?

Answer 1

I am using the Descartes Rule of Signs.

f(x) = 3x^4 - 5x³ - x² - 8x + 4

positive zero: 2 or 0.
x = + - - +
There are 2 sign changes : + to - ; - to +.

negative zero: 1.

f(-x) = 3(-x^4) - 5(-x³) - (-x²) - 8(-x) + 4
f(-x) = 3x^4 + 5x³ - x² + 8x + 4

-x = + + - + +

there is only 1 change of sign. - to +.

There are 2 or 0 positive solutions and 1 negative solution. 

Answer 2

Answer:

The possible number of positive, negative and complex zeros of the function are:

1)two positive real zeros and two negative real zeros.

2)two pairs of conjugate complex roots.

3)two positive real zeros and a pair of conjugate complex roots.

4)two negative real zeros and a pair of conjugate complex roots.

Step-by-step explanation:

We have a quartic function that means we could have up to four different roots.

We can use Descartes' rule of signs that states that in a single variable polynomial with real coefficients ordered by descending variable exponent, the number of positive real zeros is either equal to the number of changes in sign of the coefficients or less than that number by an even integer.

$f(x)=3x^{4} -5x^{3}-x^{2}-8x+4$

The sign sequence is $(+ - - - +)$

There are two sign changes: one between the first and second term and the other between the fourth and fifth term. That means there could be two or zero positive real roots.

To find the number of negative real zeros, we change the signs of the coefficients of the terms with odd exponents. We apply  Descartes' rule of signs to the polynomial $g(x)=f(-x)$.

Then $g(x)=f(-x)=3x^{4} +5x^{3}-x^{2}+8x+4$

the sign sequence is $(+ + - + +)$

There are two sign changes: one between the second and third term and the other between the third and fourth term. That means there could be two or zero negative real roots.

Keeping in mind that complex roots come in pairs, that means that if $x=a+ib$ is a zero of the function then the conjugate $x=a-ib$ is also a zero of the function,  we get the possible number of roots stated above.