Find all the zeros of the polynomial (2x^{4} ‒ 11x^{3} + 7x^{2} + 13x ‒ 7), it being given that two of its zeros are (3 + √3) and (3 ‒ √3)

Let us assume f (x) = 2x^{4} – 11x^{3} + 7x^{2} + 13x - 7

As (3 + √2) and (3- √2) are the zeros of the given polynomial therefore each one of (x + 3 + √2) and (x + 3 - √2) is a factor of f (x)

Consequently, [(x – (3 + √2)][(x – (3 -√2)

= [(x – 3) -√2] [(x – 3) + ]

= [(x – 3)^{2} – 2] = x^{2} – 6x + 7 is a factor of f (x)

Now, on dividing f (x) by (x^{2} – 6x + 7) we get:

f (x) = 0

2x^{4} - 11x^{3} + 7x^{2} + 13x – 7 = 0

(x^{2} - 6x + 7) (2x^{2} + x – 7) = 0

(x + 3 + √2) (x + 3 -√2) (2x – 1) (x + 1) = 0

∴ x = - 3 - √2 or x = - 3 + √2 or x = 1/2 or x = - 1

Hence, all the zeros of the given polynomial are (-3 -√2), (-3 + √2), 1/2 and – 1

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