How can the factor by grouping method simplify polynomials with multiple terms?

[mathdad]

Factor 8a^3 + 27b^3 + 2a + 3b.

8a^3 + 2a + 27b^3 + 3b

2a(4a^2 + 1) = Group A

3b(9b^2 + 1) = Group B

(4a^2 + 1)(9b^2 + 1)(2a + 3b)

Correct?

 

[MarkFL]

I would factor as follows:

\(\displaystyle 8a^3+27b^3+2a+3b=(2a)^2+(3b)^3+2a+3b=(2a+3b)\left(4a^2-6ab+9b^2\right)+(2a+3b)=(2a+3b)\left(4a^2-6ab+9b^2+1\right)\)

You grouping is valid, however they have no common factors, and so your factored expression isn't equivalent to the original expression.

 

[mathdad]

In your final answer, where did + 1 come from?

 

[MarkFL]

In your final answer, where did + 1 come from?

Think of it like this:

\(\displaystyle (2a+3b)\left(4a^2-6ab+9b^2\right)+(2a+3b)=(2a+3b)\left(4a^2-6ab+9b^2\right)+(2a+3b)\cdot1=(2a+3b)\left(4a^2-6ab+9b^2+1\right)\)

It's the same as writing:

\(\displaystyle xy+x=x(y+1)\)

 

[mathdad]

Great work!