Question

Verify that 3, –1, – (1/3) are the zeros of the cubic polynomial p(x) = 3x3 – 5x2 – 11x – 3, and then verify the relationship between the zeros and the coefficients.

Answer 1

Cubic:

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

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

Thanks

With Regards

Answer 2

Let given cubic polynomial be

p(x) = 3x³ - 5x² - 11x - 3

i ) If x = 3 , then

p(3) = 3(3)³ - 5(3)² - 11(3) - 3

= 81 - 45 - 33 - 3

= 81 - 81

= 0

ii ) If x = -1 , then

p(-1) = 3(-1)³ - 5(-1)² - 11(-1) - 3

= -3 - 5 + 11 - 3

= -11 + 11

= 0

iii ) If x = -1/3 , then

p(-1/3) = 3(-1/3)³ - 5(-1/3)² - 11(-1/3) - 3

= -3/27 - 5/9 + 11/3 - 3

= -1/9 - 5/9 + 11/3 - 3

= ( -1 - 5 + 33 - 27 )/9

= ( -33 + 33 )/9

= 0

Therefore ,

p(3) = p(-1) = p(-1/3) = 0 .

So, 3 , -1 , -1/3 are zeroes of given

cubic polynomial 3x³ - 5x² - 11x - 3 .

Compare the coefficients of given

cubic polynomial 3x³ - 5x² - 11x - 3

with ax³ + bx² + cx + d , we get

a = 3 , b = -5 , c = -11 , d = -3

Let the zeroes of the given cubic

polynomial p(x) are p =3 , q=-1, r=-1/3

Now ,

p+q+r = 3 - 1 - 1/3

= 2 - 1/3

= ( 6 - 1 )/3

= 5/3

= -b/a

pq+qr+rp

= 3(-1)+(-1)(-1/3)+(-1/3)(3)

= -3 + 1/3 - 1

= -4 + 1/3

= ( -12 + 1 )/3

= -11/3

= c/a

pqr = 3 × ( -1 ) × ( -1/3 )

= 1

=- d/a

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