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

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)