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Answer 1

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Answer 2

2x^4+x^3-14x^2-19x-6, if two of its zeroes are -2 & -1.

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