What are the possible numbers of positive, negative, and complex zeros of f(x) = −x6 − x5 − x4 − 4x3 − 12x2 + 12

Positive roots = 1 or 0

Negative roots = 0, 2, or 4

Complex roots = 0, 2, 4, or 6

Step-by-step explanation:

We are given the polynomial:

f(x) = -x⁶ - x⁵ - x⁴ - 4x³ − 12x² + 12

Now, by inspection, the highest degree is 6 and as such the polynomial is therefore a polynomial with more than 2 as a degree.

Applying Descartes Rule of Signs to the polynomial :

f(x) = -x⁶ - x⁵ - x⁴ - 4x³ − 12x² + 12

Signs are: - - - - - +

There is only 1 sign change and thus, it means there is 1 or 0 positive roots

To find number of negative roots, we will use f(-x);

f(−x) = -(−x)⁶ - (−x)⁵ - (−x)⁴ - 4(−x)³ − 12(−x)² + 12(−x)

f(-x) = -x⁶ + x⁵ - x⁴ + 4x³ − 12x² - 12

Signs are: - + - + - -

There are 4 sign changes and thus, it means there are 4, 2 or 0 negative roots

Thus; Complex roots = 0, 2, 4, or 6

y - y1 = m(x- x1)

y - (-5) = 2/3(x -0)

y + 5 = 2/3x

y = 2/3x - 5

step-by-step explanation:

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