Question

find the other zeros of the polynomial x^4-2x^3-26x^2+54x-27, if two of its zeros are 3√3 and -3√3.

Answer 1

Hiii friend,

(3✓3) and (-3✓3) are the zeros of the polynomial X⁴-2X³-26X²+54X-27.


(X-3✓3)(X+3✓3) = (X)²-(3✓3)² = X²-27


G(X) = X²-27

P(X) = X⁴-2X³-26X²+54X-27


On dividing P(X) by G(X) we get,


REMAINDER = 0


QUOTIENT = X²-2X+1

Factories the Quotient then we will get the two other zeros of the polynomial P(X).


=> X²-2X+1


=> X²-X-X+1


=> X(X-1) -1(X-1)


=> (X-1) (X-1)


=> (X-1) = 0 OR (X-1) = 0

=> X = 1 OR X = 1


Hence,

1,1,3✓3 and -3✓3 are the four zeros of the polynomial X⁴-2X³-26X+54X-27




HOPE IT WILL HELP YOU...... :-)


On dividing

Answer 2

Hey dear!!!

____________________________

==> In the example, we have given that,

p(x) = x⁴ - 2x³ -26x² + 54x - 27

Also we have given that 3√3 and -3√3 are the two zeroes of the given polynomial .

We have to find another zeroes of the given polynomial .

$(x + 3 \sqrt{3} )(x - 3 \sqrt{3} ) = x {}^{2} - 27$

x² - 27 is the factor of the given polynomial .

Let divide the given polynomial by x² - 27 we get,

Refer from the attachment which I have provided .

In this attachment you can see we obtained x² - 2x + 1 as Quotient .

Now, we can easily find the other zeros by splitting the middle term of quotient .

=> Given :

=> x² - 2x + 1

=> x² - x - x + 1

=> x(x - 1) - 1(x - 1)

=> ( x - 1) (x - 1)

∴ (x - 1) = 0

∴ x = 1

Also,

∴ (x -1) = 0

∴ x = 1

Therefore,1, 1 , 3√3 and -3√3 are the zeroes of the given polynomial .

Thanks !!!!

[ Be Brainly ]