Question

Olivia was asked to factor the following expression completely:

x^3-x+3x^2y=3y
x(x^2-1)+3y(x^2-1)
(x+3y)(x^2-1)
How can you check Olivia’s work to show the answer is or is not correct. If Olivia is not correct, explain to Olivia where her mistake is and how to fix it.

Answer 1

Explanation:

To check Olivia's work, we can multiply the factors she obtained to see if they result in the original expression. Let's perform the multiplication:

(x + 3y)(x^2 - 1) = x(x^2 - 1) + 3y(x^2 - 1)

Distributing the terms:

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

= x^3 - x + 3yx^2 - 3y

As we compare this with the original expression:

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

We can see that Olivia's factored expression, (x + 3y)(x^2 - 1), does not match the original expression. Olivia made a mistake in the step where she distributed the terms.

To correct the mistake, we need to distribute the terms correctly. Let's go through the factoring process again:

Starting with the original expression: x^3 - x + 3x^2y = 3y

Rearranging the terms: x^3 + 3x^2y - x - 3y = 0

Now, we can factor by grouping:

x^2(x + 3y) - 1(x + 3y) = 0

Notice that we have a common factor of (x + 3y). Factoring it out:

(x + 3y)(x^2 - 1) = 0

Now we have the correct factored expression.

Answer 2

Olivia's work is not correct.

The correct factorization of the expression is: (x-1)(x+1)(x+3y)

Explanation:

In order to check Olivia's work, we can expand the two factors she gave:

x(x^2-1)+3y(x^2-1)

x^3-x+3x^2*y-3xy

This is not equal to the original expression, so Olivia's factorization is incorrect.

To help Olivia find the correct factorization, we can first factor out a common factor of x from the first two terms:

x(x^2-1)+3y(x^2-1)

x(x^2-1)+3y(x^2-1)

Now, we can factor the quadratic expression x^2-1:

x(x-1)(x+1)+3y(x-1)(x+1)

Finally, we can factor out a common factor of (x-1)(x+1) from the two terms:

(x-1)(x+1)(x+3y)

This is the correct complete factorization of the expression.