Q. 6.80 - Intermediate Algebra

Question

Factor completely: $7x{y}^5-7xy$

Step-by-Step Solution

Answer

The required factors are $7xy(y^2+1)(y-1)(y+1)$

Step 1. Given Information

The given expression is $7x{y}^5-7xy$

Step 2. Calculation

We will factor the expression by taking the common factor outside the paranthesis and then apply the property.

$$\begin{gathered} 7x{y}^5-7xy=7xy(y^4-1) \\ =7xy({\left(y^2\right)}^2-1^2) \\ =7xy(y^2-1)(y^2+1) \\ =7xy(y^2+1)(y-1)(y+1) \end{gathered}$$

Step 3. Check

We will verify by multiplying the factors.

$$\begin{gathered} 7xy(y^2+1)(y-1)(y+1)=7xy(y^2-1)(y^2+1) \\ =7xy({\left(y^2\right)}^2-1^2) \\ =7xy(y^4-1) \\ = 7x{y}^5-7xy \end{gathered}$$