Question

If two zeros of the polynomial x4-6x3-26x2+138x-35 are (2+root3) (2-root 3) find other zeros

Answer 1

since two zeroes of the given polynomial p(x) is 2-√3 and 2+√3 so we divided p(x) by {x-(2+√3)}{x-(2-√3)}. Then we find factor of quotient found.

Answer 2

Answer:

The remaining zeros are -5 and 7.

Step-by-step explanation:

The given polynomial is

$P(x)=x^4-6x^3-26x^2+138x-35$

It is given that $2+\sqrt{3}$ and $2-\sqrt{3}$ are two zeros. It means $(x-2-\sqrt{3})$ and $(x-2+\sqrt{3})$ are factors of the given polynomial.

$(x-2-\sqrt{3})(x-2+\sqrt{3})=(x-2)^2-3=x^2-4x+4-3=x^2-4x+1$

Divide the given polynomial by (x^2-4x+1).

$\frac{x^4-6x^3-26x^2+138x-35}{x^2-4x+1}=x^2-2x-35$

It means the remaining factor of P(x) is $x^2-2x-35$.

$x^2-2x-35=0$

$x^2-7x+5x-35=0$

$x(x-7)+5(x-7)=0$

$(x+5)(x-7)=0$

Equate each factor equal to 0.

$x=-5,x=7$

Therefore the remaining zeros are -5 and 7.