Question

if two zeros of The polynomial x cube minus 3 X square + 2 are 1 + root 3 and 1 minus root 3 then find the third zero.

Answer 1

x³ - 3x² + 2 = 0

Sum of roots = 3 

Let the roots be α,β,p

Given α = 1+√3 and β = 1-√3

α+β+p = 3
2 + p  = 3
p = 1

So, the other root is 1

Answer 2

Answer:

1

Step-by-step explanation:

Dividend =$x ^3 -3x^2+2$

The roots of this polynomial are $1+\sqrt{3}$ and $1- \sqrt{3}$

So, $(x-(1+\sqrt{3})(x-(1-\sqrt{3})$

$(x-1-\sqrt{3})(x-1+\sqrt{3})$

Property : $(a-b)(a+b)=a^2-b^2$

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

$x^2+1-2x-3$

$x^2-2x-2$

So, Divisor =  $x^2-2x-2$

$Dividend = (Divisor \times Quotient)+Remainder$

$x ^3 -3x^2+2 = (x^2-2x-2 \times x-1)+0$

So, the quotient = x-1

So, $x-1=0$

So, x =1

So, the third zero of the given polynomial is 1