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

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The discussion focuses on simplifying the polynomial 8a^3 + 27b^3 + 2a + 3b using the factor by grouping method. The initial grouping separates the terms into two groups, leading to a factorization of (2a + 3b)(4a^2 - 6ab + 9b^2 + 1). While the grouping approach is valid, it is pointed out that the final expression must maintain equivalence with the original polynomial. A clarification is made regarding the introduction of the +1 in the final factorization. Overall, the factor by grouping method effectively simplifies the polynomial while ensuring all terms are accounted for.
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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?
 
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I would factor as follows:

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

You grouping is valid, however they have no common factors, and so your factored expression isn't equivalent to the original expression.
 
In your final answer, where did + 1 come from?
 
RTCNTC said:
In your final answer, where did + 1 come from?

Think of it like this:

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

It's the same as writing:

$$xy+x=x(y+1)$$
 
Great work!
 
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