Difference between revisions of "1965 AHSME Problems/Problem 19"
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By ~Ahmed_Ashhab | By ~Ahmed_Ashhab | ||
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| + | == Solution 2 == | ||
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| + | Notice that to obtain the <math>x^4</math> term one must multiply <math>x^4+4x^3+6px^2+4qx+r</math> by some linear function of the form <math>x-a</math>. Looking at the <math>x^3</math> term, it is clear that <math>a</math> must equal <math>1</math>. Therefore by multiplying <math>x^4+4x^3+6px^2+4qx+r</math> by <math>x+1</math>, the product will be <math>x^4+4x^3+12x^2+12x+3</math>. Therefore <math>p=2</math>, <math>q=3</math>, <math>r=3</math>. Thus <math>(2+3)3=\fbox{15}</math> | ||
== See Also == | == See Also == | ||
Revision as of 08:40, 26 November 2022
Contents
Problem 19
If 
is exactly divisible by 
, the value of 
is:
Solution 1
Let 
and 
.
Let 3 roots of 
be 
and 
. As 
, 3 roots of 4 roots of 
will be same as roots of . Let the 4th root of be 
. By vieta's formula
In
In
so 
By ~Ahmed_Ashhab
Solution 2
Notice that to obtain the 
term one must multiply 
by some linear function of the form 
. Looking at the 
term, it is clear that 
must equal 
. Therefore by multiplying by 
, the product will be 
. Therefore 
, 
, 
. Thus 
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by 20 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
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