NMTC Practice Part 7
If 3 zeros of the polynomial f ( x ) = x 4 + a x 2 + b x + c are 2 , − 3 , and 5 , then find the value of a + b + c .
This prob is a part of the set NMTC Practice Problems
The answer is 79.
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4 solutions
Beautiful approach. Thanks..
Excuse me, I didn't understand how we can realize that a + b +c = f(1) - 1 . Could you explain, if you may? Thanks in advance!
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f(1) allways gives the sum of all coefficiens. In his problem, f(1)=1+0+a+b+c, So f(1)-1=a+b+c. f(0) allways gives the constant term. Say here f(0)=c..
polynomial has 4 roots 2,-3,5 and let 4th root be d.
then poly=
f(x)= (x-2)(x+3)(x-5)(x-d)
multiply this and equate coefficient of x^2 to 0. you will get d= -4. put this in f(x) and get a= -27, b= -14 and c=120
a+b+c =79
1 5 1 6 f ( 2 ) + 1 0 1 f ( − 3 ) − 6 1 f ( 5 ) = − 7 9 + a + b + c = 0 ⟹ a + b + c = 7 9
From vieta we can find the fourth root which is − 4 . Now f ( x ) = ( x − 2 ) ( x + 3 ) ( x − 5 ) ( x + 4 ) We observe that f ( 1 ) = ( − 1 ) ( 4 ) ( − 4 ) ( 5 ) = 8 0 = 1 + a + b + c ⇒ a + b + c = 7 9
Sir, How did you find out that the 4th root is -4?
The polynomial is ( x − 2 ) ( x + 3 ) ( x − 5 ) ( x + 4 ) by Vieta. Then a + b + c = f ( 1 ) − 1 = ( 1 − 2 ) ( 1 + 3 ) ( 1 − 5 ) ( 1 + 4 ) − 1 = 7 9 . No need to actually find a , b , c .