If 1 and -3 are the zeroes of a polynomial p(x) = x^3 - ax^2 - 13x + b ,

find the values of a and b.

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Solution

p(x) = x^3 - ax^2 - 13x + b

p(1) =0
1^3 - a x 1^2 - 13 x 1 + b = 0
1-a-13+b =0 , b-a=12 ---(1)

p(-3) =0
(-3)^3 - a(-3)^2 - 13(-3) + b = 0
-27 -9a +39 +b =0 b-9a = -12----(2)

(2)-(1)---- b-9a-(b-a)=-12-12

-8a = -24 so a =3

substitute a in eqn (1)
b= 12+a =12+3=15

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