Q. 6.81 - Intermediate Algebra

Question

Factor completely: $6x^2-12xc+6bx-12bc$

Step-by-Step Solution

Answer

The required factors are  $6(x-2c)(x+b)$

Step 1. Given Information

The given expression is$6x^2-12xc+6bx-12bc$

Step 2. Calculation

We will factor the expression by taking the common factor outside the paranthesis.

$$\begin{gathered} 6x^2-12xc+6bx-12bc=6(x^2-2xc+bx-2bc) \\ =6\left[x\right(x-2c)+b(x-2c\left)\right] \\ =6(x-2c)(x+b) \end{gathered}$$

Step 3. Check

We will verify by multiplying the factors.

$$\begin{gathered} 6(x-2c)(x+b)=6\left[x\right(x-2c)+b(x-2c\left)\right] \\ =6(x^2-2xc+bx-2bc) \\ =6x^2-12xc+6bx-12bc \end{gathered}$$