Question

2 zeroes are √3 and -√3 .find two other zeroes of x4-3x3-x2+9x-6​

Answer 1

Answer:

Step-by-step explanation:

Let p (x) = x⁴ - 3x³ - x² + 9x - 6

Given that,

2 zeroes of p (x) are √3 and - √3.

So,the zeroes are ( x - √3) and ( x + √3 )

∴p ( x ) = x⁴ - 3x³ - x² + 9x - 6

= ( x - √3 ) ( x + √3 )

= x² - 3

Now we dividing p ( x ) by ( x² - 3 ), we get

x⁴ - 3x³ - x² + 9x - 6 ÷ (x² - 3 )

= x² - 3x + 2

= x² - 2x - x + 2

= x (x - 2) - 1 ( x - 2)

= (x - 1) (x - 2)

∴The other two zeroes are 1 and 2.