Question

Hi guys......If√3 and -√3 are the zeros of p(x)=x^4+4x^3-8x^2-12x+15,then find the remaining zeros of p(x)​

Answer 1

Answer:

1,-5 are the remaining zeros

As √3, -√3ate zeros

x=√3 and x=-√3

therefore (x-√3) and (x+√3) are factors of equation that means (x-√3)(x+√3) is also a factor of the equation

now, we do ( x^4+4x^3-8x^2-12x+15 )÷{(x-√3)(x+√3)} to get other factor of equation which is x^2+4x-5 On solving this we get x=1 and x=-5

Answer 2

Answer:√3,-√3,1,-5

Hint- divide p(x) by x^2-3

And Factorize the quotient ,answer will come