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The roots of [tex]y = x^{5}-27\cdot x^{2}[/tex] are 0 (multiplicity 2), -3 (multiplicity 1), [tex]-\frac{3}{2}+i\,\frac{3\sqrt{3}}{2}[/tex] (multiplicity 1) and [tex]-\frac{3}{2}-i\,\frac{3\sqrt{3}}{2}[/tex] (multiplicity 1).
The characteristics of the polynomial can be derived from algebraic techniques. Roots and multiplicity can be found by factoring the polynomial. The multiplicity is represented by a power binomial of the form:
[tex](x-r_{i})^{m},\,m \le n[/tex] (1)
Where [tex]n[/tex] is the grade of the polynomial.
Now we proceed to factor the formula:
[tex]y = x^{5}-27\cdot x^{2}[/tex]
[tex]y = x^{2}\cdot (x^{3}-27)[/tex]
[tex]y = x^{2}\cdot (x^{2}+3\cdot x +9)\cdot (x-3)[/tex]
Please notice that [tex]x^{2}+3\cdot x + 9[/tex] have two complex roots.
[tex]y = x^{2}\cdot \left(x+\frac{3}{2}-i\,\frac{3\sqrt{3}}{2} \right)\cdot \left(x+\frac{3}{2}+i\,\frac{3\sqrt{3}}{3} \right)\cdot (x-3)[/tex]
The roots of [tex]y = x^{5}-27\cdot x^{2}[/tex] are 0 (multiplicity 2), -3 (multiplicity 1), [tex]-\frac{3}{2}+i\,\frac{3\sqrt{3}}{2}[/tex] (multiplicity 1) and [tex]-\frac{3}{2}-i\,\frac{3\sqrt{3}}{2}[/tex] (multiplicity 1).
We kindly invite to see this question on polynomials: https://brainly.com/question/1218505