# Verify the relationship between zeroes and coefficients of the cubic polynomial p(x) = 3x^{3} - 5x^{2} - 11x - 3p(x)=3x3−5x2−11x−3.

### Asked by bibhudeepanp | 1st Jul, 2021, 10:03: PM

Expert Answer:

### P(x) = 3 x^{3} - 5 x^{2} -11 x -3 ............................. (1)
If we put x = -1 , we get P(-1) = 0
Hence (x+1) is a factor of P(x)
After dividing P(x) by (x+1) , we get , P(x) = (x+1) ( 3x^{2 }-8x -3 )
After factorising ( 3x^{2 }-8x -3 ) , we get, P(x) = (x+1) (x-3)(3x+1)
Cubic polynomial with zeros of polynomials as -a, -b and -c is given as
(x+a) (x+b) ( x+c) = x^{3} + ( a +b +c )x^{2} + ( ab + bc + ca ) x + (abc) ......................(2)
Given polynomial P(x) can be written in terms of factors as
P(x) = (1/3) ( x+1) ( x-3 ) [ x + (1/3) ] = (1/3) [ x^{3} - (5/3)x^{2} -(11/3)x -1 ] ..................(3)
we get from the factors of polynomial , a = 1 , b = -3 and c = (1/3)
By comparing eqn.(2) and (3) , we should get , a+b+c = -5/3 . This can be verified by substituting values of a, b and c
Similarly ( ab+bc+ca ) = -5/3 . This can be verified by substituting values of a, b and c
Similarly ( a b c ) = 1 × (-3) × (1/3) = -1

^{3}- 5 x

^{2}-11 x -3 ............................. (1)

^{2 }-8x -3 )

^{2 }-8x -3 ) , we get, P(x) = (x+1) (x-3)(3x+1)

^{3}+ ( a +b +c )x

^{2}+ ( ab + bc + ca ) x + (abc) ......................(2)

^{3}- (5/3)x

^{2}-(11/3)x -1 ] ..................(3)

### Answered by Thiyagarajan K | 18th Sep, 2021, 03:49: PM

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