Obtain all zeros of polynomial 2x^4+x^3-14x^2-19x-6, if two of its zeroes are -2 & -1.
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Answered by
81
P(x) = 2 x⁴ + x³ - 14 x² - 19 x - 6
The number of sign changes in the coefficients : of P(x) = 1
the number of sign changes in the coefficients of P(-x) = 3
There are atmost 3 negative real roots and 1 positive real root.
two roots or zeros are : -2 and -1.
So (x+2) and (x+1) are the factors of P(x).
P(x) = (x² + 3 x + 2) (a x² + b x + c) = 2 x⁴ + x³ - 14 x² - 19 x - 6
comparing both sides we can say that:
a = 2 and c = - 3.
the coefficient of x³ on both sides are : 3 a + b = 1
b = -5
The roots of : 2 x² - 5 x - 3 = 0
(2 x + 1 ) ( x - 3 ) = 0
x = - 1/2 or 3
The number of sign changes in the coefficients : of P(x) = 1
the number of sign changes in the coefficients of P(-x) = 3
There are atmost 3 negative real roots and 1 positive real root.
two roots or zeros are : -2 and -1.
So (x+2) and (x+1) are the factors of P(x).
P(x) = (x² + 3 x + 2) (a x² + b x + c) = 2 x⁴ + x³ - 14 x² - 19 x - 6
comparing both sides we can say that:
a = 2 and c = - 3.
the coefficient of x³ on both sides are : 3 a + b = 1
b = -5
The roots of : 2 x² - 5 x - 3 = 0
(2 x + 1 ) ( x - 3 ) = 0
x = - 1/2 or 3
sanjays2402:
Tanx!
Answered by
9
Answer:x=-1/2 or 3
Step-by-step explanation:
P(x) = 2 x⁴ + x³ - 14 x² - 19 x - 6
The number of sign changes in the coefficients : of P(x) = 1
the number of sign changes in the coefficients of P(-x) = 3
There are atmost 3 negative real roots and 1 positive real root.
two roots or zeros are : -2 and -1.
So (x+2) and (x+1) are the factors of P(x).
P(x) = (x² + 3 x + 2) (a x² + b x + c) = 2 x⁴ + x³ - 14 x² - 19 x - 6
comparing both sides we can say that:
a = 2 and c = - 3.
the coefficient of x³ on both sides are : 3 a + b = 1
b = -5
The roots of : 2 x² - 5 x - 3 = 0
(2 x + 1 ) ( x - 3 ) = 0
x = - 1/2 or 3
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