The polynomial has three real roots (of which one counts twice). It's formula is [tex] y= (x - 1)(x + 2)^2 [/tex]
The roots of a polynomial are the points in which the polynomial inersects the x-axis. So, you might be tempted to say that there are two roots, [tex] x=-2 [/tex] and [tex] x = 1 [/tex].
But now let's face the problem in an analytic way: if [tex] x_0 [/tex] is a solution for the polynomial [tex] p(x) [/tex], then [tex] p(x) [/tex] can be factored as [tex] (x-x_0)r(x) [/tex] where [tex] r(x) [/tex] is some remainder.
So, since we are sure that [tex] x=-2 [/tex] and [tex] x = 1 [/tex] are solutions, we know that we can write our polynomial as
[tex] p(x) = (x-1)(x+2)r(x) [/tex]
But since the polynomial is a cubic (i.e. degree 3), [tex] r(x) [/tex] must be a polynomial of degree 1, i.e. [tex] r(x)=x-k [/tex] for some [tex] k \in \mathbb{R} [/tex]
But then we would have
[tex] p(x) = (x-1)(x+2)(x-k) [/tex]
and so we would have a third solution [tex] x=k [/tex], but we can see from the graph that the only solutions are [tex] x=-2 [/tex] and [tex] x = 1 [/tex].
This means that [tex] r(x) [/tex] is actually a copy of one of the two other factors, so that it doesn't add new solutions. So, we have two cases:
[tex] r(x) = x-1 \implies p(x) = (x-1)^2(x+2) [/tex]
or
[tex] r(x) = x+2 \implies p(x) = (x-1)(x+2)^2 [/tex]
The answer lies in the shape of the polynomial: you can see that the graph is tangent to the x-axis in [tex] x=-2 [/tex], whereas it crosses the x-axis at [tex] x=1 [/tex]. This tells us that the double solution is [tex] x=-2 [/tex], and thus the correct solution is [tex] p(x) = (x-1)(x+2)^2 [/tex]
Answer:
We are to solve the cube root equation, ∛x – 1 + 2 = 0
Solve like terms; subtract 1 from 2 to get -1 (-1 + 2 = +1)
Therefore, ∛x + 1 = 0
Move +1 to the right side of the equation, since it does not contain the variable to solve for.
∛x + 1 = 0
Subtract 1 from both sides of the equation
∛x + 1 – 1 = 0 – 1
∛x = -1
Cube both sides of the equation in order to remove the radical on the left side
(∛x)^3 = (-1)^3
x = -1We are to solve the cube root equation, ∛x – 1 + 2 = 0
Solve like terms; subtract 1 from 2 to get -1 (-1 + 2 = +1)
Therefore, ∛x + 1 = 0
Move +1 to the right side of the equation, since it does not contain the variable to solve for.
∛x + 1 = 0
Subtract 1 from both sides of the equation
∛x + 1 – 1 = 0 – 1
∛x = -1
Cube both sides of the equation in order to remove the radical on the left side
(∛x)^3 = (-1)^3
x = -1We are to solve the cube root equation, ∛x – 1 + 2 = 0
Solve like terms; subtract 1 from 2 to get -1 (-1 + 2 = +1)
Therefore, ∛x + 1 = 0
Move +1 to the right side of the equation, since it does not contain the variable to solve for.
∛x + 1 = 0
Subtract 1 from both sides of the equation
∛x + 1 – 1 = 0 – 1
∛x = -1
Cube both sides of the equation in order to remove the radical on the left side
(∛x)^3 = (-1)^3
x = -1We are to solve the cube root equation, ∛x – 1 + 2 = 0
Solve like terms; subtract 1 from 2 to get -1 (-1 + 2 = +1)
Therefore, ∛x + 1 = 0
Move +1 to the right side of the equation, since it does not contain the variable to solve for.
∛x + 1 = 0
Subtract 1 from both sides of the equation
∛x + 1 – 1 = 0 – 1
∛x = -1
Cube both sides of the equation in order to remove the radical on the left side
(∛x)^3 = (-1)^3
x = -1We are to solve the cube root equation, ∛x – 1 + 2 = 0
Solve like terms; subtract 1 from 2 to get -1 (-1 + 2 = +1)
Therefore, ∛x + 1 = 0
Move +1 to the right side of the equation, since it does not contain the variable to solve for.
∛x + 1 = 0
Subtract 1 from both sides of the equation
∛x + 1 – 1 = 0 – 1
∛x = -1
Cube both sides of the equation in order to remove the radical on the left side
(∛x)^3 = (-1)^3
x = -1
Step-by-step explanation:
