Question

Please help.

The graph shows a cubic function:

1. How many real roots does the function have?
A. 0
B. 1
C. 2
D. 4
E. 3

2. Complete the equation of the graphed function. y= (x - ?)(x + ?)^2
The choices are 0,3,1,2,4

Answer 1

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

Explanation:

Answer 2

The polynomial has three real roots (of which one counts twice). It's formula is
`y= (x - 1)(x + 2)^2`

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,
`x=-2` and
`x = 1`.

But now let's face the problem in an analytic way: if
`x_0` is a solution for the polynomial
`p(x)`, then
`p(x)` can be factored as
`(x-x_0)r(x)` where
`r(x)` is some remainder.

So, since we are sure that
`x=-2` and
`x = 1` are solutions, we know that we can write our polynomial as

`p(x) = (x-1)(x+2)r(x)`

But since the polynomial is a cubic (i.e. degree 3),
`r(x)` must be a polynomial of degree 1, i.e.
`r(x)=x-k` for some
`k \in \mathbb{R}`

But then we would have

`p(x) = (x-1)(x+2)(x-k)`

and so we would have a third solution
`x=k`, but we can see from the graph that the only solutions are
`x=-2` and
`x = 1`.

This means that
`r(x)` is actually a copy of one of the two other factors, so that it doesn't add new solutions. So, we have two cases:

`r(x) = x-1 \implies p(x) = (x-1)^2(x+2)`

or

`r(x) = x+2 \implies p(x) = (x-1)(x+2)^2`

The answer lies in the shape of the polynomial: you can see that the graph is tangent to the x-axis in
`x=-2`, whereas it crosses the x-axis at
`x=1`. This tells us that the double solution is
`x=-2`, and thus the correct solution is
`p(x) = (x-1)(x+2)^2`