Question

verify that 5, -2, 1/3 are the zeroes of cubic polynomial 3x cube - 10x square - 27x +10 and verify the relationship between its zeroes and coefficients

Answer 1

GIVEN :

Verify that 5, -2,  are the zeroes of cubic polynomial  and verify the relationship between its zeroes and coefficients

TO FIND :

That 5, -2,  are the zeroes of cubic polynomial  and relationship between the given zeroes and coefficients

SOLUTION :

Given that cubic polynomial is  $3x^3- 10x^2- 27x +10$

Let p(x) be the given cubic polynomial.

$p(x)=3x^3- 10x^2- 27x +10$

Now we have to verify that 5, -2, $\frac{1}{3}$ are the zeroes of cubic polynomial p(x) :

Put x=5 in p(x)

$p(5)=3(5)^3- 10(5)^2- 27(5) +10$

$=3(125)-10(25)-135+10$

=375-250-135+10

=385-385

=0

⇒ p(5)=0

∴ 5 is a zero of p(x).

Put x=-2 in p(x)

$p(-2)=3(-2)^3- 10(-2)^2- 27(-2) +10$

$=3(-8)-10(4)+54+10$

=-24-40+54+10

=64-64

=0

⇒ p(-2)=0

∴ -2 is a zero of p(x).

Put $x=\frac{1}{3}$ in p(x)

$p(\frac{1}{3})=3(\frac{1}{3})^3- 10(\frac{1}{3})^2- 27(\frac{1}{3}) +10$

$=\frac{1}{9}-\frac{10}{9}-9+10$

$=-\frac{9}{9}+1$

=-1+1

=0

⇒ $p(\frac{1}{3})=0$

∴ $\frac{1}{3}$ is a zero of p(x).

Hence 5, -2, $\frac{1}{3}$ are the zeroes of cubic polynomial p(x) is verified.

Now verify the relationship between the given zeroes and coefficients  :

For a cubic polynomial $p(x)=ax^3+bx^2+cx+d$ with the zeroes $\alpha$ , $\beta$ and $\gamma$ we have,

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

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

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

For $p(x)=3x^3- 10x^2- 27x +10$

5, -2, $\frac{1}{3}$ are the zeroes of cubic polynomial p(x)

Let $\alpha=5$ , $\bet-2$ and $\gamma=\frac{1}{3}$

Here a=3 , b=-10 , c=-27 and d=10

Substitute the values in the formulae  we get,

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

$5+(-2)+\frac{1}{3}=\frac{-(-10)}{3}$

$3+\frac{1}{3}=\frac{10}{3}$

$\frac{9+1}{3}=\frac{10}{3}$

$\frac{10}{3}=\frac{10}{3}$

LHS=RHS

∴ $\alpha+\beta+\gamma=\frac{-b}{a}$

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

$5(-2)+(-2)(\frac{1}{3})+(\frac{1}{3})(5)=\frac{-27}{3}$

$-10-\frac{2}{3})+\frac{5}{3}=-9$

$-10+\frac{3}{3}=-9$

-10+1=-9

-9=-9

LHS=RHS

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

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

$5(-2)(\frac{1}{3})=\frac{-10}{3}$

$\frac{-10}{3}=\frac{-10}{3}$

LHS=RHS

∴ $\alpha \beta \gamma=\frac{-d}{a}$

Hence the relationship between the given zeroes and coefficients is verified.

Answer 2

Answer:

here is your answer

Step-by-step explanation:

p(x) = (3x3 – 10x2 – 27x +verify-that-5-2-and-1-3-are-the-zeroes-of-the-cubic-polynomial-p-x-3x-3-10x-2-27x-10