Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 1, 2011"

m
(→‎Solution 2)
 
Line 20: Line 20:
 
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 > 5^2/4</math>.
+
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 $(n+1)(n+2)(n+3)(n+4) - 120$, we should try to find a way to create a quadratic in disguise. There, in fact, is a way!

Expand $(n+1)(n+4)$ and $(n+2)(n+3)$ separately:

$(n+1)(n+4)(n+2)(n+3) = (n^2+5n+4)(n^2+5n+6)$

We notice that there is a $n^2 + 5n$ in both of these terms! Treat $n^2 + 5n$ as a single quantity, $x$.

Our original quantity was $(x + 4)(x+6)-120 = x^2+10x+24-120=x^2+10x-96=(x-6)(x+16)$.

We can factor the original polynomial as $[(n^2+5n)-6][(n^2+5n)+16] = (n^2+5n-6)(n^2+5n+16)$.

$n^2+5n-6$ can be factored as $(n-1)(n+6)$.

Our final factorization is $\boxed{(n-1)(n+6)(n^2+5n+16)}$.

Solution 2

Let $P(x) = (x+1)(x+2)(x+3)(x+4) - 120$. Then $P(1) = 0$ and $P(-6)=0$ (since $120 = -5\cdot -4\cdot -3\cdot -2$). Therefore I have: \[P(x) = (x-1)(x+6)Q(x)\] where $Q$ is a quadratic. Simplifying yields $Q(x)=x^2+5x+16$ as before, which is irreducible as $16 > \frac{5^2}{4}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/July_1,_2011&oldid=40155"