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MC IIB Algebra Advanced Factoring Prob 8 part b)
factor 1 + 2a + 3a^2 + 4a^3 + 5a^4 + 6a^5 + 5a^6 + 4a^7 + 3a^8 + 2a^9 + a^10:
The solution just says we need to use grouping to get (1 + a + a^2 + a^3 + a^4 + a^5)^2, how do we use grouping to get that?
Do we just seperate out 1 + a^2 + a^4 + .. + a^10 and see that the rest of the terms are of the form 2ab where a and b are from 1 + a^2 + a^4 + .. + a^10?
thanks
It's probably easiest to recognize a factoring pattern here: $$\begin{aligned} (a+1)^2 &= a^2 + 2a + 1, \\ (a^2 + a + 1)^2 &= a^4 + 2a^3 + 3a^2 + 2a + 1, \\ (a^3+a^2+a+1)^2 &= a^6 + 2a^5 + 3a^4 + 4a^3 + 3a^2 + 2a + 1, \\ &\ldots \end{aligned}$$
To prove this by grouping it's easier to do a smaller example. Consider $a^6 + 2a^5 + 3a^4 + \cdots + 2a + 1$. This can be written as $$\begin{array}[llllll] \\ a^6 & +a^5 & +a^4 & + a^3 & & & \\ & +a^5 & +a^4 & +a^3 & + a^2 & & \\ & & +a^4 & +a^3 & + a^2 & +a & \\ & & & +a^3 & + a^2 & +a & +1\\ \end{array}$$ Thus grouping the terms gives $(a^3+a^2+a+1)^2$ as needed.
An identical idea works for $(1+a+a^2+a^3+a^4+a^5)^2$, where we group in sets of $6$ terms.
Hope this helps!
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