The roots of $y = x^{5}-27\cdot x^{2}$ are 0 (multiplicity 2), -3 (multiplicity 1), $-\frac{3}{2}+i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1) and $-\frac{3}{2}-i\,\frac{3\sqrt{3}}{2}$ (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:

$(x-r_{i})^{m},\,m \le n$ (1)

Where $n$ is the grade of the polynomial.

Now we proceed to factor the formula:

$y = x^{5}-27\cdot x^{2}$

$y = x^{2}\cdot (x^{3}-27)$

$y = x^{2}\cdot (x^{2}+3\cdot x +9)\cdot (x-3)$

Please notice that $x^{2}+3\cdot x + 9$ have two complex roots.

$y = x^{2}\cdot \left(x+\frac{3}{2}-i\,\frac{3\sqrt{3}}{2} \right)\cdot \left(x+\frac{3}{2}+i\,\frac{3\sqrt{3}}{3} \right)\cdot (x-3)$

The roots of $y = x^{5}-27\cdot x^{2}$ are 0 (multiplicity 2), -3 (multiplicity 1), $-\frac{3}{2}+i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1) and $-\frac{3}{2}-i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1).

We kindly invite to see this question on polynomials: brainly.com/question/1218505

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2 years ago