Q. 6.79 - Intermediate Algebra

Question

Factor completely: $4a^{5}b-64ab$

Step-by-Step Solution

Answer

The required factors are $4ab(a^2+4)(a-2)(a+2)$

Step 1. Given Information

The given expression is $4a^{5}b-64ab$

Step 2. Calculation

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

$$\begin{gathered} 4a^{5}b-64ab=4ab(a^4-16) \\ =4ab({\left(a^2\right)}^2-4^2) \\ =4ab(a^2-4)(a^2+4) \\ =4ab(a^2+4)(a-2)(a+2) \end{gathered}$$

Step 3. Check

We will verify by multiplying the factors.

$$\begin{gathered} 4ab(a^2+4)(a-2)(a+2)=4ab(a^2-4)(a^2+4) \\ =4ab({\left(a^2\right)}^2-4^2) \\ =4ab(a^4-16) \\ =4a^{5}b-64ab \end{gathered}$$