Answer:
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):
Answer: x = -5 or x = i sqrt(3) or x = -i sqrt(3)
Answer:
C. x = −5, 1.7i, −1.7i
Step-by-step explanation:
Quick answer:
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 answer:
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 [tex]\pm[/tex]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 [tex]\pm[/tex]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
