Question

-√2 and √2 are two of the zeros of p(x)=2x⁴+7x³–8x²–14x+8. Find the remaining zeros of p(x).

Answer 1

Given p(x) = 2x^4 + 7x^3 - 8x^2 - 14x + 8.

Given Zeroes are $-\sqrt{2},\sqrt{2}$

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

= > x^2 - 2 are the zeroes of the polynomial.!

Now,

Here, we split the middle terms of p(x) in order to get x^2 - 2 as common factor.

= > 2x^4 + 7x^3 - 8x^2 - 14x + 8

= > 2x^4 - x^3 + 8x^3 - 4x^2 - 4x^2 + 2x - 16x + 8

= > 2x^4 - x^3 - 4x^2 + 2x + 8x^3 - 4x^2 - 16x + 8

= > x(2x^3 - x^2 - 4x + 2) + 4(2x^3 - x^2 - 4x + 2)

= > (x + 4)(2x^3 - x^2 - 4x + 2)

= > (x + 4)[x^2(2x - 1) - 2(2x - 1)]

= > (x + 4)(2x - 1)(x^2 - 2).

Therefore, the remaining zeroes of the given polynomial are,

$= > -\sqrt{2}, \sqrt{2}, \frac{1}{2}, -4$

Hope this helps!

Answer 2

p( x ) = ( x + √2 )( x - √2 )( 2x² + 7x - 4 )

= ( x + √2 )( x - √2 )[ 2x² + 8x - x - 4 ]

= ( x + √2 )( x - √2 )[2x(x + 4 )-1(x+4)]

= ( x + √2 )( x - √2 )( x + 4 )( 2x - 1 )

To find other zeroes of p( x ) ,

we have take x + 4 = 0 or 2x - 1 = 0

x = - 4 or x = 1/2

Therefore ,

Required two zeroes are -4 , 1/2

I hope this helps you.

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