Question

Find all zeroes of the polynomials x⁴+x³-14x²-2x+24 if two zeros of its zeros are √2and -√2​

Answer 1

Step-by-step explanation:

Given:-

x^4+x^3-14x^2-2x+24 and it's two zeroes are √2 and -√2.

To find:-

Find the all zeroes of the given polynomial?

Solution:-

Given bi-quadratic polynomial is

P(x) = x^4+x^3-14x^2-2x+24

Given two zeroes = 2 and -√2

If √2 is a zero then by Factor Theorem (x-√2) is a factor of P(x)

and

If -√2 is a zero then by Factor Theorem (x+√2) is a factor of P(x)

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

It is in the form of (a+b)(a-b)

Where a = x and b=√2

we know that

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

=> (x+√2)(x-√2) =x^2-(√2)^2=x^2-2

Therefore,x^2-2 is a factor of P (x).

Now to get other two zeores then we have to divide P(x) by x^2-2

=> P(x)÷(x^2-2)

=>(x^4+x^3-14x^2-2x+24)÷(x^2-2)

=> (x^4+x^3-12x^2-2x^2-2x+24)÷(x^2-2)

=>[ (x^4+x^3-12x^2)+(-2x^2-2x+24)]÷(x^2-2)

=>[x^2(x^2+x-12)-2(x^2+x-12)]÷(x^2-2)]

=> (x^2+x-12)(x^2-2) ÷(x^2-2)

=> x^2+x-12

We get quotient = x^2+x-12

So to get other zeroes we write x^2+x-12=0

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

=>x(x+4)-3(x+4)=0

=>(x+4)(x-3)=0

=>x+4=0 or x-3=0

=>x=-4 or x=3

The zeroes are -4 and 3

Answer:-

All zeroes of the given Polynomial are √2 ,-√2, -4 and 3

Used formulae:-

Factor Theorem:-

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.