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# Verify that the numbers given alongside of the cubic polynomial are their zeros. Also verify the relationship between the zeroes and the coefficients in each case :

x^{3} + 5x^{2} + 7x + 3 ; – 3, 2 – 1, – 1

Answer :

Let p(x) = x^{3} + 5x^{2} + 7x + 3.

Then, p( – 1) = ( – 1)^{3} + 5( – 1)^{2} + 7( – 1) + 3

= – 1 + 5 – 7 + 3

= 0

p( – 3) = ( – 3)^{3} + 5( – 3)^{2} + 7( – 3) + 3

= – 27 + 45 – 21 + 3

= 0

Hence, – 1, – 1 and – 3 are the zeroes of the given polynomial x^{3} + 5x^{2} + 7x + 3.

Now, Let α = – 1 , β = – 1 and γ = – 3

Then, α + β + γ = – 1 + ( – 1) + ( – 3) = – 5

αβ + βγ + γα = ( – 1)( – 1) + ( – 1)( – 3) + ( – 3)( – 1)

= 1 + 3 + 3

= 7

and αβγ = ( – 1) × ( – 1) × ( – 3)

= – 3

Thus, the relationship between the zeroes and the coefficients is verified.

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