Question

find all the zeroes of x^4 - 3x^3 + 6x _ 4 , if two of its zeroes are root 2 and -root 2

Answer 1

Step-by-step explanation:

${x}^{4} - 4 - 3 {x}^{3} + 6x = >$

$( {x}^{2} - 2)( {x}^{2} + 2) - 3x( {x}^{2} - 2) = >$

$( {x}^{2} - 3x + 2)( {x}^{2} - 2)$

$= > ( {x}^{2} - 2x - x + 2)( {x}^{2} - 2)$

$= > (x(x - 2) - 1(x - 2))( {x}^{2} - 2)$

$= > (x - 1)(x - 2)(x - \sqrt{2})(x + \sqrt{2})$

So zeroes of this polynomial are 1,2, +root 2 and - root 2. Hope it helps you.