Q. 6.85 - Intermediate Algebra

Question

Factor completely: $4x^2-12xy+9y^2-25$

Step-by-Step Solution

Answer

The required factors are $(2x-3y-5)(2x-3y+5)$

Step 1. Given Information

The given expression is $4x^2-12xy+9y^2-25$

Step 2. Calculation

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

$$\begin{gathered} 4x^2-12xy+9y^2-25={\left(2x\right)}^2-2\times 2x\times 3y+{\left(3y\right)}^2-25 \\ ={[2x-3y]}^2-5^2 \\ =(2x-3y-5)(2x-3y+5) \end{gathered}$$

Step 3. Check

We will verify by multiplying the factors.

$$\begin{gathered} (2x-3y-5)(2x-3y+5)={[2x-3y]}^2-5^2 \\ ={\left(2x\right)}^2-2\times 2x\times 3y+{\left(3y\right)}^2-25 \\ =4x^2-12xy+9y^2-25 \end{gathered}$$