Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 1, 2011"
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Let <math>P(x) = (x+1)(x+2)(x+3)(x+4) - 120</math>. Then <math>P(1) = 0</math> and <math>P(-6)=0</math> (since <math>120 = -5\cdot -4\cdot -3\cdot -2</math>). Therefore I have: | Let <math>P(x) = (x+1)(x+2)(x+3)(x+4) - 120</math>. Then <math>P(1) = 0</math> and <math>P(-6)=0</math> (since <math>120 = -5\cdot -4\cdot -3\cdot -2</math>). Therefore I have: | ||
<cmath>P(x) = (x-1)(x+6)Q(x)</cmath> | <cmath>P(x) = (x-1)(x+6)Q(x)</cmath> | ||
− | where <math>Q</math> is a quadratic. Simplifying yields <math>Q(x)=x^2+5x+16</math> as before, which is irreducible as <math>16 > 5^2 | + | where <math>Q</math> is a quadratic. Simplifying yields <math>Q(x)=x^2+5x+16</math> as before, which is irreducible as <math>16 > \frac{5^2}{4}</math>. |
Latest revision as of 19:20, 1 July 2011
Problem
AoPSWiki:Problem of the Day/July 1, 2011
Solution 1
To factor , we should try to find a way to create a quadratic in disguise. There, in fact, is a way!
Expand and separately:
We notice that there is a in both of these terms! Treat as a single quantity, .
Our original quantity was .
We can factor the original polynomial as .
can be factored as .
Our final factorization is .
Solution 2
Let . Then and (since ). Therefore I have: where is a quadratic. Simplifying yields as before, which is irreducible as .