1965 AHSME Problems/Problem 19
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) | ||
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