Question

19. Find all the zeros of the polynomial (2x4 - 11x + 7x2 + 13x - 7), it being
given that two of its zeros are (3 + 2) and (3 - 2).​

Answer 1

Answer:

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The given polynomial is f(x) = 2x^4 - 11x^3 + 7x^2 + 13x - 7

since , (3+√3) and (3-√2) is the zeroes of f(x)

it is follows that x - 3+√3 and x - 3- √2 are the factor of f ( x).

\begin{gathered}consequatly \: = (x - 3 + \sqrt{3})(x + 3 - \sqrt{2} ) \\ = {x}^{2} - 6x + 7 \: \: is \: a \: \: factor \: \: of \: \: p(x)\end{gathered}consequatly=(x−3+3)(x+3−2)=x2−6x+7isafactorofp(x)

Dividing p(x) by x^2 - 6x +7

we , get remainder is 0

=> f (x) =0

=> (x^2 - 6x + 7)(2x^2 + x - 1)

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

=>x = -3 -√2. , x = -3+√2 or x =1/2 or x = -1

HENCE , the zeroes of the given polynomial are ( -3-√2) , (-3+√2) , 1/2 and -1

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