Question

given that √2 is a zero of the cubic polynomial 6x+√2xpower2-10x-4√2,find its other two zeroes

Answer 1

√2 is zero of 6x³ +√2x² -10x -4√2 ,
So, (x - √2) is a factor of 6x³ + √2x² -10x -4√2 .

6x³ + √2x² -10x -4√2
= 6x³ -6√2x² + 7√2x² -14x + 4x - 4√2
= 6x²(x - √2) + 7√2x(x -2) + 4(x -√2)
= {6x² + 7√2x + 4}(x - √2)
= {6x² + 3√2x + 4√2x + 4}(x -√2)
= {3√2x(√2x + 1) + 4(√2x +1)}(x -√2)
= (3√2x +4)(√2x +1)(x - √2)

Hence, two other zeros are -4/3√2, and -1/√2

Answer 2

Step-by-step explanation:

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

= (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

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