Question

Find all the zeroes of the polynomial x⁴-7x²+12 whose two zeroes are √3 and -√3​

Answer 1

Step-by-step explanation:

Given :-

Two zeroes √3 and -√3 are of the polynomial is x^4-7x^2+12.

To find :-

Find all the zeroes of x^4-7x^2+12 ?

Solution :-

Given that

The bi-quadratic polynomial P(x) = x^4-7x^2+12.

Given two zeroes = √3 and -√3

If √3 is the zero of the P(x) then by Factor theorem , (x-√3) is a factor of P(x).

If -√3 is the zero of the P(x) then by Factor theorem , (x+√3) is a factor of P(x).

=> (x-√3)(x+√3) is also a factor of P (x).

=> x^2-(√3)^2 is also a factor of P (x)

Since (a+b)(a-b)=a^2-b^2

=>x^2-3 is a factor of P(x).

To get Zeroes we write P(x) = 0

=> x^4-7x^2+12 = 0

=> x^4-3x^2-4x^2+12 = 0

=> x^2(x^2-3)-4(x^2-3) = 0

=> (x^2-3)(x^2-4) = 0

=> x^2-3 = 0 or x^2-4 = 0

=> x^2=3 or x^2 = 4

=> x=±√3 or x=±√4

=>x=±√3 or x=±2

x = √3,-√3 ,2 and -2

or

To get another zeroes then we have to divide P(x) by x^2-3

(x^4-7x^2+12)÷(x^2-3) = 0

=> (x^4-3x^2-4x^2+12 )÷(x^2-3)= 0

=> [x^2(x^2-3)-4(x^2-3)]÷(x^2-3) = 0

=>[ (x^2-3)(x^2-4) ]÷(x^2-3)= 0

=> x^2-4 = 0

=> x^2 = 4

=> x=±√4

=>x=±2

=> x = 2 and -2

Answer:-

All the zeroes of the given bi-quadratic polynomial are √3 , -√3 , 2 and -2

Used formulae:-

Factor Theorem:-

• Let P(x) be a polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial if x-a is a factor of P (x) then P(a) = 0 vice-versa.

• (a+b)(a-b)=a^2-b^2