Respuesta :

Problem 1

Answer: The roots are x = 0, x = 2, x = -3

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Explanation:

The term "root" is another way of saying "zeros of a function" or "x intercept".

To find the roots, set each factor equal to zero and solve for x.

x-2 = 0 solves to x = 2, while x+3 = 0 solves to x = -3

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Problem 2

Answer: Roots are x = 1, [tex]x = i\sqrt{5}[/tex] and [tex]x = -i\sqrt{5}[/tex]

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Explanation:

Factor by grouping

x^3 - x^2 + 5x - 5 = 0

(x^3 - x^2) + (5x - 5) = 0

x^2(x - 1) + 5(x - 1) = 0

(x^2 + 5)(x - 1) = 0

x^2 + 5 = 0 solves to the roots [tex]x = i\sqrt{5}[/tex] and [tex]x = -i\sqrt{5}[/tex]

x-1 = 0 solves to x = 1

Note: [tex]i = \sqrt{-1}[/tex]

lvvies

1) Wat are the real zeros of the function y=x(x-2)(x+3):

Answer: [tex]x_{1} = -3 ; x_{2} = 0; x_{3} =2[/tex]

To find the x-intercept, set y = 0:

x( x -2 ) ( x + 3) = 0

Apply the Zero Product Property:

x - 2 = 0 or x + 3 = 0

x = 2 or x = -3

Find the union of solutions:

x = 0 or x = 2 or x = -3

2) Find all of the solutions of the equation

x^3 -x^2 +5x - 5 = 0

Answer: x = 1 or no real solution

Apply grouping:

[tex](x^{3} -x^{2} ) + *(5x-5)=0[/tex]

Factor out common factor:

[tex]x^{2} (x-1) +5(x-1)=0[/tex]

Factor out --> [tex](x-1)(x^{2} +5) =0[/tex]

Apply zero product property:

[tex]x - 1 =0[/tex] or [tex]x^{2} +5 =0[/tex]

x - 1 = 0

Rearrange variables to the left side of the equation--> x=1 or no real solutions.

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