Question

factor completely.

\\2y - 2y^2 + 6y^4 - 6y^3\\

Explanation:

⚡ Using what you learned: Factoring by Grouping

Step 1: Factor out the greatest common factor (GCF)

First, look at all four terms to find the greatest common factor:

$$ 2y - 2y^2 + 6y^4 - 6y^3 $$

Every term is divisible by \( 2y \). Factor \( 2y \) out of the entire expression:

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

Step 2: Group the terms inside the parentheses

Group the four terms inside the parentheses into two pairs:

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

Step 3: Factor each group

Factor out the GCF from each group:

• The first group \( (1 - y) \) has no common factor other than \( 1 \):

$$ 1(1 - y) $$

• The second group \( (3y^3 - 3y^2) \) has a GCF of \( 3y^2 \):

$$ 3y^2(y - 1) $$

This gives:

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

Step 4: Align the binomial factors

Notice that \( (1 - y) \) and \( (y - 1) \) are opposites. Rewrite \( 3y^2(y - 1) \) by factoring out a negative sign to make the binomials match:

$$ 3y^2(y - 1) = -3y^2(1 - y) $$

Now substitute this back into the expression:

$$ 1(1 - y) - 3y^2(1 - y) $$

Step 5: Factor out the common binomial

Factor out the common binomial factor \( (1 - y) \):

$$ (1 - y)(1 - 3y^2) $$

Step 6: Combine with the overall GCF

Multiply by the overall GCF \( 2y \) factored out in Step 1:

$$ 2y(1 - y)(1 - 3y^2) $$

Answer:

$$ 2y(1 - y)(1 - 3y^2) $$