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find all zeroes of the polynomials 2x⁴+7x³-19x²-14x+30 if two of its zeroes are √2,-√2

 

Two of its zeroes are √2,-√2, so two factors are, (x – √2) and (x + √2).

The product of these factors is also a factor of the original expression.

i.e. (x – √2)( x + √2) = x² - 2 divides into 2x⁴+7x³-19x²-14x+30.

Now we do a long division.

x² - 2) 2x⁴ + 7x³ - 19x² - 14x + 30 (2x² + 7x - 15

           2 x⁴          - 4x²

                     7x³ - 15x²

                     7x³            - 14x

                           - 15x²            + 30

                           - 15x²            + 30

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

So, we can now write our expression as: 2x⁴ + 7x³ - 19x² - 14x + 30 = (x² - 2)( 2x² + 7x - 15)

And the 2nd quadratic factorises as: 2x² + 7x – 15 = (2x – 3)(x + 5)

All the zeroes are: -5, 3/2, -√2, +√2

by Level 11 User (80.9k points)

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