Question

if two zeroes of a polynomial x^4- 6x^3 - 26x^2 + 138x - 35 are 2,+/- root 3, find other zeroes

Answer 1

It's zeroes are -5, 7 , 2+√3 & 2-√3..
and sorry for handwriting

Answer 2

Answer:

The other zeroes 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 $2\pm \sqrt{3}$ are two zeroes. It means $(x-2+\sqrt{3})\text{  and  }(x-2-\sqrt{3})$ are two 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 $x^2-4x+1$, to find the remaining factors.

Using long division method we get

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

Equate the remaining factor equal to 0, to find the other zeroes.

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

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

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

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

Equate each factor equal to 0.

$x=7$

$x=-5$

Therefore the other zeroes are -5 and 7.