Question

verify that 3,-1,-1/3are the zeroes of the cubuc polynomial p(x)=3x3-5x2-11x-3​

Answer 1

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ax^{3} +bx^{2} +cx+d

Where, a,b,c,d are coefficients

Let p,q and r are the zeros of the polynomial.

Then, p+q+r= -b/a

And, pqr= -d/a

pq+qr+rp = c/a

i)3x^{3} -5x^{2} -11x-3

3*3^{3} -5*3^{2} -11*3-3

81-45-33-3=81-81=0

3*(-1)^{3} -5*(-1)^{2} -11*(-1)-3

-3-5+11-3=11-11=0

3*(-1/3)^{3} -5*(-1/3)^{2} -11*(-1/3)-3

\frac{-1}{9 } -\frac{5}{9} +\frac{11}{3} -3=-\frac{2}{3} +\frac{11}{3}-3=3-3=0

Verified.

Now,p+q+r= 3-1-(1/3)=5/3=-b/a

    pqr= 3*-1*(-1/3)=1=-d/a

pq+qr+rp= 3*(-1)+(-1)*(-1/3)+(-1/3)*(3)=-11/3=c/a

Hence,verified

Hope it helps

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Step-by-step explanation: