## anonymous 5 years ago find a polynomial equation with integer coefficients, for which 5 is a double root and 1+ square root of 2 and 1- square root of 3i are solutions. I dont know how to solve this or write it out.. help?

1. anonymous

if A, B, C , D are roots of a polynomial, the the polynomial is the product of (x -A)(x-B)(x-C)(x-D) You have the four roots as 5, 5, 1 + root2 and 1-root 3i

2. anonymous

k.then what do i do.

3. anonymous

In the above equation, replace A, B, C and D by the given roots and take their product to get the polynomial

4. anonymous

so do (x-5)(x-5)(x-1+square root of 2)(x-1-square root of 3i)?

5. anonymous

It will be (x-5)(x-5)[x-(1+root2)][x- (1- root of 3i)} Which becomes (x-5)(x-5)[x-1 - root2][x-1+ root of 3i}

6. anonymous

ok

7. anonymous

see if the co-efficients are integer and real then complex conjugate roots must exist. and if $1+\sqrt{2}$ exists then $1-\sqrt{2}$also exits. let $\sqrt{3i}=ki$ whre k is a no. $(x-5)^{2}(x-1-\sqrt{2})(x-1+\sqrt{2})(x-1-ki)(x-1+ki)$this gives $(x-5)^{2}((x-1)^{2}-2)((x-1)^{2}+3)$ if you don't consider the conjugate part then you can't get integral coefficient....