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Why is $$(1+x+x^2)^2 = 1+x^2+(x^2)^2+2(x+x^2+x^3)?$$

My professor wrote that in the class but he didn't explain why.

Thank you

Rick
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zeeks
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2 Answers2

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He/she applied the multinomial formula: $$\Bigl(\sum_{i=1}^na_i\Bigr)^2=\sum_{i=1}^na_i^2+2\!\!\!\sum_{1\le i<j\le n}\!\!\!a_i a_j. $$

Bernard
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Here is a really long but elementary solution to the problem (does require basic knowledge of factoring by grouping):

$$(a+b)^2 = (a+b)(a+b) = a(a+b) + b(a+b) = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2$$ Now follow the same logic to get $(a+b+c)^2$: $$(a+b+c)^2 $$ $$= (a+b+c)(a+b+c)$$$$ = a(a+b+c) + b(a+b+c) + c(a+b+c)$$ $$ = (a^2 + ab + ac) + (ab+b^2+bc) + (ac+bc+c^2)$$ $$ = a^2+ 2ab + 2ac + b^2 + 2bc + c^2$$ Make the substitutions $a=1$, $b=x$, and $c=x^3$ to get $$(1)^2 + 2(1)(x) + 2(1)(x^2) + (x)^2 + 2(x)(x^2) + (x^2)^2$$ $$= 1 + 2x + 2x^2 + x^2 + 2x^3 + x^4$$ $$= 1 + x^2 + (x^2)^2 + 2(x+x^2+x^3)$$ As desired.

  • thank you for the answer :) – zeeks Sep 15 '15 at 22:52
  • @zeeks Sure thing! Since you were unable to understand the problem, I thought it would be better to give an elementary solution than present you with an even harder equation that is much harder to prove. The multinomial formula is great, but can be difficult to grasp for beginners in this subject, and is usually part of a general study of polynomials – Brevan Ellefsen Sep 15 '15 at 23:09