Question

If two zeroes of the polynomial 2x4 - 12x3 - 52x2 + 276x - 70 are 2 + √3 and 2+ √3.Find the zeroes.

Answer 1

Answer:

The remaining two zeros are -5 and 7

Step-by-step explanation:

Given zeros are

$2+\sqrt3\:and\:2-\sqrt3$

First we find a quadratic polynomial

having these zeros

sum of the zeros

$=2+\sqrt3+\:2-\sqrt3$

$=4$

product of the zeros

$=(2+\sqrt3)(2-\sqrt3)$

$=2^2-{\sqrt3}^2$

$=4-3$

$=1$

Corresponding factor is

$x^2-4x+1$

Now,

$2x^4-12x^3-52x^2+ 276x-70=(x^2-4x+1)(2x^2+px-70)$

Equating coefficient of x on both sides we get

276=280+p

p=-4

other factor is

$2x^2-4x-70$

$2(x^2-2x-35)$

$2(x-7)(x+5)$

corresponding zeros are -5, 7