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Irina18 [472]
2 years ago
5

Explain how to find each root, with its multiplicity, for the polynomial y=x^5-27x^2.

Mathematics
1 answer:
Nuetrik [128]2 years ago
7 0

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 <em>factoring</em> 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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