 Hello and welcome to the session. Today I'm there to help you with polynomials. Let us discuss the following question. If two zeros of the polynomial x4-6x cube-26x square-138x-35r2 plus minus root 3 find the other zeros. Let us write the key idea that we will be using in the question. A real number alpha is a zero of a polynomial p of x if p of alpha is equal to zero. Let us write the solution given to us is p of x is equal to x4 minus 6x cube minus 26x square plus 138x minus 35 and given zeros are 2 plus root 3 and 2 minus root 3. Now therefore x-2 plus root 3 x-2 minus root 3 minus root 3 into x-2 plus root 3 is a factor of solving this root 3 x cuts off and minus 2 root 3 plus root 3 cuts off. That is minus 4 so we get 4 minus 3 so we have our divisor as x square minus 4x plus 1 and our dividend as x4 minus 6x cube minus 26x35. Now solving this by long division method we get. So divide by x square do not forget to change the sign by 2 which gives us minus. Now again solving it further we get by division I have got p of x equal to g of x into q of x plus x4 minus 26x square plus 138x35 equal to x square minus 4x plus 1 into x square minus 2x minus 35. The other 2 those of the polynomial p of x are the zeros of the polynomial x square minus 2x minus 35. So solving it we get x square minus 2x minus 35 is equal to 0 which implies now splitting the middle term so we get x square minus 7x plus 5x minus 35 is equal to 0 which implies x into x minus 7 plus 5 into x minus 7 is equal to 0 which implies x plus 5 into x minus 7 is equal to 0 which implies x is equal to minus 5 and x is equal to 7. Therefore required zeros of p of x are 7 comma minus 5. Hope you understood the problem. Bye and have a nice day.