32. Identify all real and non-real zeros of the function f(x) = x^3 + 5x^2 + 3x + 15. options:

A. x = 0, −5, 1.7i, −1.7i

B. x = 0,−5, 1.7i

C. x = −5, 1.7i, −1.7i

D. x = 0,−3, −5

x = -5 or x = i sqrt(3) or x = -i sqrt(3)

Step-by-step explanation:

Solve for x:

x^3 + 5 x^2 + 3 x + 15 = 0

The left hand side factors into a product with two terms:

(x + 5) (x^2 + 3) = 0

Split into two equations:

x + 5 = 0 or x^2 + 3 = 0

Subtract 5 from both sides:

x = -5 or x^2 + 3 = 0

x = (0 ± sqrt(0^2 - 4×3))/2 = ( ± sqrt(-12))/2:

x = -5 or x = sqrt(-12)/2 or x = (-sqrt(-12))/2

sqrt(-12) = sqrt(-1) sqrt(12) = i sqrt(12):

x = -5 or x = (i sqrt(12))/2 or x = (-i sqrt(12))/2

sqrt(12) = sqrt(4×3) = sqrt(2^2×3) = 2sqrt(3):

x = -5 or x = (i×2 sqrt(3))/2 or x = (-i×2 sqrt(3))/2

(2 i sqrt(3))/2 = i sqrt(3):

x = -5 or x = i sqrt(3) or x = (-2 i sqrt(3))/2

(2 (-i sqrt(3)))/2 = -i sqrt(3):

x = -5 or x = i sqrt(3) or x = -i sqrt(3)

C. x = −5, 1.7i, −1.7i

Step-by-step explanation:

Quick

C. x = −5, 1.7i, −1.7i

explanation:

C. is the only answer option that does NOT have 0 as a root, which is impossible, because there is a constant term, which means that all roots are non-zero. In other words, we cannot extract x as a factor.

Complete

All odd degree polynomials have at least one real root.

By the real roots theorem, we know that

if there is a real root, it must be of the form p/q where q is any of the factors of the leading coefficient (1 in this case) and p is any factor of the constant term d (15 in this case).

Values of p/q are

On trial and error, using the factor theorem, we see that

f(-5) = 0, therefore -5 is a real root.  By long division, we have a quotient of x^2+3 = 0, which gives readily the remaining (complex) roots of +/- sqrt(5) i

The answer is {-5, +/- sqrt(5) i}, or again,

C. x = −5, 1.7i, −1.7i

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