Question

-√2and √ 2 are two of the zeroes of p( x) 2x4 +7 x3 -8x2 -14 x +8 . find the remaining zeroes of p( x)​

Answer 1

Step-by-step explanation:

√2and √ 2 are two of the zeroes of p( x) 2x4 +7 x3 -8x2 -14 x +8

+√2 is zero so (x-√2)are factors of p(x)

-√2 is zero so ((x+√2) are factors of p(x)

their product is also (x-√2)(x+√2) are factor of p(x)

so( x^2-2) is actor of p(x)

x^2 - 2 ) 2x^4+7x^3 -8x^2-14x+8 ( 2x^2 + 7x - 4

2x^4 -4x^2

_______________

7x^3-4x^2-14x

7x3 -14x

_____________

-4x^2+8

-4x^2+8

_________

0

_________

the remaining two zeroes are zeroes of quotient 2x^2 +7x-4

2x^2 +7x-4=(2x-1) (X+4) = 0

2x- 1= 0 X+4=0

2x= 1 X = -4

X =1/2

remaining two zeroes are 1/2, -4