Asked by swetha
Obtain all the zeroes of the polynomial of p(x) if sum of all the polynomials are given : (1). P (x) = x^4-8x^3+23x^2-28x+12 if 2 zeroes are 1 & 2.(2). p (x)=96x^4-128x^3+62x^2-13x+1 & 2 zeroes are 1/2 & 1/4
Answers
Answered by
Reiny
#1
checking for the given roots
P(1) = 1 - 8 + 23 - 28 + 12 = 0
yup, x-1 is a factor
P(2) = 16 - 64 + 92 - 56 + 12 = 0
yup, x-2 is a factor
now divided x^4-8x^3+23x^2-28x+12 by (x-1) to get a cubic, then divide that cubic by (x-2),
leaving you with a quadratic expression
set that equal to zero and solve to get two more solutions.
Hint: you will get two more integers, one of them will be the same as one of those given.
Do #2 the same way
hint: the factors you need for the divisions are
(2x-1) and (4x-1)
hint#2: all zeroes will be simple fractions
checking for the given roots
P(1) = 1 - 8 + 23 - 28 + 12 = 0
yup, x-1 is a factor
P(2) = 16 - 64 + 92 - 56 + 12 = 0
yup, x-2 is a factor
now divided x^4-8x^3+23x^2-28x+12 by (x-1) to get a cubic, then divide that cubic by (x-2),
leaving you with a quadratic expression
set that equal to zero and solve to get two more solutions.
Hint: you will get two more integers, one of them will be the same as one of those given.
Do #2 the same way
hint: the factors you need for the divisions are
(2x-1) and (4x-1)
hint#2: all zeroes will be simple fractions
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