The results for the polynomials are summarized below:
Case 7: Number of real roots: 5, Roots: 0 (Multiplicity 3), - 1, - 2.
Case 8: Number of real roots: 2, Roots: √3, - √3
Case 9: Number of real roots: 2, Roots: 0 (Multiplicity 2).
Case 10: Number of real roots: 6, Roots: 0 (Multiplicity 4), 2, - 2.
Case 11: Number of real roots: 6, Roots: 0 (Multiplicity 4), 2, 6.
Case 12: Number of real roots: 5, Roots: 0 (Multiplicity 3), 1.333, - 5.
How many real zeros does exists in a polynomial?
Six polynomials are found in the image set aside, the determination of real zeros can be done by factoring, which are applications of algebra properties. Now we show the results for each case:
Case 7:
f(x) = x⁵ + 3 · x⁴ + 2 · x³
f(x) = x³ · (x² + 3 · x + 2)
f(x) = x³ · (x + 2) · (x + 1)
Number of real roots: 5, Roots: 0 (Multiplicity 3), - 1, - 2.
Case 8:
f(x) = x⁴ + 4 · x² - 21
f(x) = (x² + 7) · (x² - 3)
f(x) = (x² + 7) · (x + √3) · (x - √3)
Number of real roots: 2, Roots: √3, - √3
Case 9:
h(x) = x⁴ - 4 · x³ + 32 · x²
h(x) = x² · (x² - 4 · x + 32)
Number of real roots: 2, Roots: 0 (Multiplicity 2).
Case 10:
h(x) = 9 · x⁶ - 36 · x⁴
h(x) = 9 · x⁴ · (x² - 4)
h(x) = 9 · x⁴ · (x + 2) · (x - 2)
Number of real roots: 6, Roots: 0 (Multiplicity 4), 2, - 2.
Case 11:
h(x) = x⁶ - 8 · x⁵ + 12 · x⁴
h(x) = x⁴ · (x² - 8 · x + 12)
h(x) = x⁴ · (x - 6) · (x - 2)
Number of real roots: 6, Roots: 0 (Multiplicity 4), 2, 6.
Case 12:
g(x) = 3 · x⁵ + 11 · x⁴ - 20 · x³
g(x) = x³ · (3 · x² + 11 · x - 20)
g(x) = x³ · (x - 1.333) · (x + 5)
Number of real roots: 5, Roots: 0 (Multiplicity 3), 1.333, - 5.