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 "1980 AHSME Problems/Problem 2"

m (→‎Solution 2)
(→‎Solution 1)
 
(2 intermediate revisions by the same user not shown)
Line 7: Line 7:
 
== Solution 1==
 
== Solution 1==
  
−
It becomes <math> (x^{8}+...)(x^{9}+...) </math> with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or <math>\boxed{(D)}</math>
+
Expanding each factor yields <math>(x^{8}+...)(x^{9}+...)</math>. The degree of the first factor is <math>8</math>, while the degree of the second factor is <math>9</math>. Therefore, the degree of the polynomial is <math>8 + 9 = \boxed{(\textbf{D})\ 17}</math>
  
 
==Solution 2==
 
==Solution 2==
  
−
First note that given a polynomial <math>P(x)</math> and a polynomial <math>Q(x)</math>:
+
Let <math>\deg{P(x)}</math> be the degree of a polynomial in <math>x</math>.  Recall that for any polynomials <math>P(x)</math> and <math>Q(x)</math>, and nonnegative integer <math>n</math>, <math>\deg{(P(x)^n)} = n \deg{P(x)}</math> and <math>\deg{(P(x)Q(x))} = \deg {P(x)} + \deg {Q(x)}</math>.
  
 +
So <math>\deg{((x^2+1)^4(x^3+1)^3)}</math>  <math>= \deg{((x^2+1)^4)} + \deg{((x^3+1)^3)}</math>  <math>= 4 \deg{(x^2+1)} + 3 \deg{(x^3+1)} </math> <math>= 4 \cdot 2 + 3 \cdot 3 </math> <math>= \boxed{(\textbf{D})\ 17}</math>.
  
−
<math>deg(P(x))^n = ndeg(P(x))</math> and <math>deg(P(x)Q(x)) = deg(P(x))+deg(Q(x))</math>.
+
-mihirb, edited by j314andrews
−
 
 
−
 
 
−
 
 
−
We let <math>x^2+1=P(x)</math> and <math>x^3+1 = Q(x)</math>.
 
−
 
 
−
Hence <math>deg(P(x)) = 2</math> and <math>deg(Q(x)) = 3</math>
 
−
 
 
−
So <math>deg(P(x))^4) = 4\cdot2 = 8</math> and <math>deg(Q(x)^3) = 3\cdot3 = 9</math>
 
−
 
 
−
 
 
−
Now we let <math>P(x)^4 = R(x)</math> and <math>Q(x)^3 = S(x)</math>
 
−
 
 
−
We want to find <math>deg(R(x)S(x)) = deg(R(x))+deg(S(x)) = 9+8 = 17</math>.
 
−
 
 
−
So the answer is '''(D)''' 17.
 
−
 
 
−
*Solution 2 by mihirb
 
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1980|num-b=1|num-a=3}}
 
{{AHSME box|year=1980|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:14, 14 July 2025

Problem

The degree of $(x^2+1)^4 (x^3+1)^3$ as a polynomial in $x$ is

$\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72$

Solution 1

Expanding each factor yields $(x^{8}+...)(x^{9}+...)$. The degree of the first factor is $8$, while the degree of the second factor is $9$. Therefore, the degree of the polynomial is $8 + 9 = \boxed{(\textbf{D})\ 17}$

Solution 2

Let $\deg{P(x)}$ be the degree of a polynomial in $x$. Recall that for any polynomials $P(x)$ and $Q(x)$, and nonnegative integer $n$, $\deg{(P(x)^n)} = n \deg{P(x)}$ and $\deg{(P(x)Q(x))} = \deg {P(x)} + \deg {Q(x)}$.

So $\deg{((x^2+1)^4(x^3+1)^3)}$ $= \deg{((x^2+1)^4)} + \deg{((x^3+1)^3)}$ $= 4 \deg{(x^2+1)} + 3 \deg{(x^3+1)}$ $= 4 \cdot 2 + 3 \cdot 3$ $= \boxed{(\textbf{D})\ 17}$.

-mihirb, edited by j314andrews

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_2&oldid=254197"