Question

1.) WHat are the real zeros of the function

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


2.) Find all of the solutions of the equation
x^3 -x^2 +5x - 5 = 0

Answer 1

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

Answer:
`x_(1) = -3 ; x_(2) = 0; x_(3) =2`

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:

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

Factor out common factor:

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

Factor out -->
`(x-1)(x^(2) +5) =0`

Apply zero product property:

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

x - 1 = 0

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

Answer 2

Problem 1

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

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Step-by-step 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,
`x = i√(5)` and
`x = -i√(5)`

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Step-by-step 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
`x = i√(5)` and
`x = -i√(5)`

x-1 = 0 solves to x = 1

Note:
`i = √(-1)`