How many and of which kind of roots does this equation have?
2 answers:
Answer:
B
Step-by-step explanation:
The polynomial is of degree 5, thus will have 5 roots either real or complex or both.
Using Descartes' rule of signs.
Count the number of sign changes in f(x)
There are no sign changes thus there are no real positive roots.
Find f(- x) then count the number of sign changes.
f(- x) = - + 3
- 8x³ + 14x² - 16x + 8
- / + / - / + / - / + ← 5 sign changes
Thus there are 5 negative real roots
Thus f(x) has 5 real roots → B
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