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3 years ago

Simplify each expression

1 answer:

3 0

1. -4+5x^3-(-3y)^0

-1+-3=-4
3x^2+ 2x^1= 5x^3
9y^1-6y^1=14y^0

2. 2y-(-3x^5)+10

5y-3y=2y
-3x^1+-2x^2=-5x^3
-5x^3+2x^2=-3x^5

3. 3x+3y+8

4x^2+2x=6x^3
6x^3-3x^2=3x
7y-2y=4y
4y-y=3y
-5+(-3)=8

4. 6x^3+4y+2

3x^2+3x^1=6x^3
7y-2y=5y
5y-y=4y

I hope this helped

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3 years ago

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3 years ago

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Rainbow [258]

Answer:

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Step-by-step explanation:

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3 years ago

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

Nuetrik [128]

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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