Question

P(x) = x2 + (3 - √2 ) x - 3√2, one zero is √2 find other zero​

Answer 1

Answer:

Given, √2 is one of the zero of the cubic polynomial.

Then, (x-√2) is one of the factor of the given polynomial p(x) = 6x³+√2x²-10x- 4√2.

Divide p(x) by x-√2

x-√2) 6x³+√2x²-10x-4√2 (6x² +7√2x

   6x³-6√2x²

(-)   (+)

  ----------------------------

7√2x² -10x-4√2

7√2x² -14x

 (-)       (+)

-------------------------

4x   - 4√2

4x   - 4√2

(-)    (+)

---------------------

0

6x³+√2x²-10x-4√2= (x-√2) (6x² +7√2x + 4)

= (x-√2) (6x² +4√2x + 3√2x + 4)

[By splitting middle term]

= (x-√2) [ 2x(3x+2√2) + √2(3x+2√2)

= (x-√2) (2x+√2)   (3x+2√2)

For zeroes of p(x), put p(x)= 0

(x-√2) (2x+√2)  (3x+2√2)= 0

x= √2 , x= -√2/2 ,x= -2√2/3

x= √2 , x= -1 /√2 ,x= -2√2/3

[ Rationalising second zero]

Hence, the other two zeroes of p(x) are -1/√2 and -2√2/3.

HOPE THIS WILL HELP YOU....