Question

find all the zeros of the polynomial (2x⁴-9x³+5x²+3x-1) of two its zeros are (2+√3) and (2-√3)

Answer 1

I think this will hlp u

Answer 2

Answer:

Heya !!

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p(x) = (2x⁴-9x³+5x²+3x-1)

Since, (2+√3) and (2–√3) are zeroes of given polynimial.

x = 2+√3 or x = 2–√3

=> (x–2–√3) (x–2+√3) = 0

=> [(x–2)–√3] [(x–2)+√3] = 0

We know that, a²–b² = (a+b)(a–b)

(x–2)² – (√3)²

=> x²+4–4x–3

=> x²–4x+1

So, x²–4x+1 is a factor of given polynomial.

Refer to the attachment for Division.

Now, (2x⁴-9x³+5x²+3x-1) = (2+√3)(2–√3)(2x²–x–1)

=> (2x²–2x+x–1) = 0

=> 2x(x–1) +(x–1) =0

=> (2x+1) (x–1) = 0

=> x = –1/2 or x = 1

Thus, the zeroes of polynomial (2x⁴-9x³+5x²+3x-1) are (2+√3), (2–√3),– 1/2 and 1

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Hope my ans.'s satisfactory.☺