Let P(x)=x^3+a x^2+b x be a polynomial whose roots are non-negative and are in arithmetic progression. If the sum of coefficients of P(x) is 10 , then:

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Correct answer: D. the value of (b-a) is equal to 27 .

P(x)=x^3+a x^2+b x Note that x=0 is one of the root. Therefore 3 roots in A.P. can be taken as 0, d, 2 d (where d>0 ) Now, sum of the root =3 d=-a ...(1) sum taken 2 at a time =2 d^2=b ...(2) Also given 1+a+b=10 ...(3) From eqns. (1), (2) and (3), we get a+b=9 array llll Hence, & 2 d^2-3 d=9 & & 2 d^2-3 d-9=0 & 2 d^2-6 d+3 d-9=0 & & (d-3)(2 d+3)=0 array d=3 Hence, roots are 0,3,6 . Hence, a=-9 and b=18b-a=27 Sum of the roots of P(x)=-a=9

Let P(x)=x^3+a x^2+b x be a polynomial whose roots are non-negative and are in arithmetic progression. If the sum of coefficients of P(x) is 10 , then:

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