Question

Find the zeros of Cubic Polynomial and verify that the relationship between the zeros and the coefficients.

$\bf \large \rightarrow \: 3 {x}^{3} \: - \: 5 {x}^{2} \: - \: 11x \: - \: 3$


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Answer 1

= 3x³-5x²-11x-3

3x³-5x²-11x-3= 2x²-11x-3

3x³-5x²-11x-3= 2x²-11x-3=9x-3

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Answer 2

Step-by-step explanation:

Given :-

3x³-5x²-11x-3

To find :-

Find the zeros of Cubic Polynomial and verify that the relationship between the zeros and the coefficients?

Solution :-

Finding the zeroes :-

Given Cubic Polynomial is 3x³-5x²-11x-3

Let P(x) = 3x³-5x²-11x-3

It can be written as

=> P(x) = 3x³-8x²+3x²-8x-3x-3

=> P(x) = 3x³+3x²-8x²-8x-3x-3

=> P(x) = 3x²(x+1)-8x(x+1)-3(x+1)

=> P(x) = (x+1)(3x²-8x-3)

=> P(x) = (x+1)(3x²-9x+x-3)

=> P(x) = (x+1)[(3x(x-3)+1(x-3)]

=> P(x) = (x+1)(x-3)(3x+1)

To get zeroes of P(x) , we write it as

P(x) = 0

=> P(x) = (x+1)(x-3)(3x+1) = 0

=>x+1 = 0 or x-3 = 0 or 3x+1 = 0

=> x = -1 or x = 3 or 3x = -1

=> x = -1 or x = 3 or x = -1/3

The zeroes are -1 , -1/3 , 3

Verifying the relationship between the zeroes and the coefficients of P(x):-

P(x) = 3x³-5x²-11x-3

On comparing this with the standard Cubic Polynomial ax³+bx²+cx+d then

a = 3

b = -5

c = -11

d = -3

The zeroes of P(x) = -1 , -1/3 , 3

Let α = -1 , β = -1/3 and γ = 3

Relation-1:-

Sum of the zeroes = α+β+γ

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

=> (-3-1+9)/3

=> (9-4)/3

=> 5/3

We know that

Sum of the zeroes = α+β+γ

=> -(Coefficient of x²)/Coefficient of x³

=> -b/a

=> -(-5)/3

=> 5/3

Therefore, Sum of the zeroes = -b/a

Relation -2:-

Sum of the product of the two zeroes taken at a time = αβ+βγ+αγ

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

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

=> (1-3-9)/3

=> (1-12)/3

=> -11/3

We know that

αβ+βγ+αγ = Coefficient of x/ Coefficient of x³

=> c/a

=> -11/3

Therefore, Sum of the product of the two zeroes taken at a time = c/a

Relation-3:-

Product of the zeroes = αβγ

=> (-1)(-1/3)(3)

=> (3/3)

=> 1

We know that

Product of the zeroes =αβγ

=> - Coefficient of x/ Coefficient of x³

=> -d/a

=> -(-3)/3

=> 3/3

=> 1

Therefore, Product of the zeroes= -d/a

Verified the relationship between the zeroes and the coefficients of the given cubic polynomial.

Answer:-

The zeroes of the given polynomial are -1 , -1/3 and 3

Used formulae:-

• The standard Cubic Polynomial is ax³+bx²+cx+d
• Sum of the zeroes =α+β+γ =

-(Coefficient of x²)/Coefficient of x³

= -b/a

• Sum of the product of the two zeroes taken at a time = αβ+βγ+αγ = Coefficient of x/ Coefficient of x³

= c/a

• Product of the zeroes = αβγ =

- Coefficient of x/ Coefficient of x³

= -d/a