Question

verify that 3 -1 -1/3 are the zeroes of the cubic polynomial p(x) =3x cube -5x square -11x-3 and then verify the relationship between zeroes and the coefficients.......

Answer 1

Answer:

Step-by-step explanation:

The function is given as :

$p(x) = 3x^{3}-5x^{2}  -11x-3;$;

the values of zeroes is = 3, -1, -1/3

Now putting the values of zeros in that equation;

$p(3) = 3*3^{3}- 5* 3^{2}  -11*3-3$ ;

p(3) = 0;

p(-1) = 0;

p(-1/3) = 0; All the zeros satisfied;

Answer 2

Answer:

3 -1 -1/3 are the zeroes of the cubic polynomial p(x).

Step-by-step explanation:

The given polynomial is

$P(x)=3x^3-5x^2-11x-3$

If c is a zero of the polynomial, then f(c)=0.

Substitute x=3 in the given polynomial.

$P(3)=3(3)^3-5(3)^2-11(3)-3=0$

Substitute x=-1 in the given polynomial.

$P(3)=3(-1)^3-5(-1)^2-11(-1)-3=0$

Substitute x=-1/3 in the given polynomial.

$P(3)=3(-\frac{1}{3})^3-5(-\frac{1}{3})^2-11(-\frac{1}{3})-3=0$

Therefore 3 -1 -1/3 are the zeroes of the cubic polynomial p(x).

If α,β,γ are three zeros of the polynomial, then

$\alpha +\beta +\gamma=\frac{-b}{a}$

$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a}$

$\alpha \beta \gamma=\frac{-d}{a}$

Using these formula, the relation amount zeros is defined as

$3-1-\frac{1}[{3}=\frac{5}{3}$

$(3)(-1)+(-1)(-\frac{1}{3})+(-\frac{1}{3})(3)=-\frac{11}{3}$

$3\times (-1)\times (-\frac{1}{3})=\frac{3}{3}=1$