Question

solve this on paper please

Answer 1

Heya !!

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p(x) = (2x⁴-9x³+5x²+3x-1)

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

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.☺