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Chapter 6: Q. 6.81 (page 610)

Factor completely:6x2-12xc+6bx-12bc

Short Answer

Expert verified

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

Step by step solution

01

Step 1. Given Information  

The given expression is6x2-12xc+6bx-12bc

02

Step 2. Calculation 

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

6x2-12xc+6bx-12bc=6(x2-2xc+bx-2bc)=6[x(x-2c)+b(x-2c)]=6(x-2c)(x+b)

03

Step 3. Check

We will verify by multiplying the factors.

6(x-2c)(x+b)=6[x(x-2c)+b(x-2c)]=6(x2-2xc+bx-2bc)=6x2-12xc+6bx-12bc

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Most popular questions from this chapter

Maribel factored y2-30y+81as (y-9)2. Was she right or wrong? How do you know?

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