# Solve 6x^4 + 5x^3 -24x^2 - 15x+18=0 if sum of two of its roots are zero.

`f(x) =6x^4 + 5x^3 -24x^2 - 15x+18`

Let A and B are two roots of f(x) = 0 where A+B = 0

so we can say B = -A

Since A and B are roots of f(x) then;

`6A^4 + 5A^3 -24A^2 - 15A+18 = 0` ----(1)

`6B^4 + 5B^3 -24B^2 - 15B+18 = 0`

but B = -A

`6(-A)^4 + 5(-A)^3 -24(-A)^2 - 15(-A)+18 = 0`

`6A^4 - 5A^3 -24A^2 +15A+18 = 0` ----(2)

(1)-(2)

`10A^3-30A = 0`

`A(A^2-3) = 0 `

A = 0 or `A^2-3 = 0 `

f(0) = 18 therefore A = 0 is not a root.

`A^2-3 = 0 `

`A = +-sqrt3`

`f(sqrt3) = 0`

`f(-sqrt3)= 0 `

so `sqrt 3` and `-sqrt3` are roots of f(x) = 0

So we can write

`f(x) = (x-sqrt3)(x+sqrt3)(Px^2+Qx+R)`

`(x-sqrt3)(x+sqrt3)(Px^2+Qx+R)`

`= (x^2-3)(Px^2+Qx+R)`

`= Px^4+Qx^3+(R-3P)x^2-3Qx-3R`

`Px^4+Qx^3+(R-3P)x^2-3Qx-3R =6x^4 + 5x^3 -24x^2 - 15x+18`

By comparing components;

P = 6

Q = 5

R-3P = -24 so R = -24+3*6 = -6

So `Px^2+Qx+R` contains other two roots.'

`Px^2+Qx+R = 0`

`6x^2+5x-6 = 0`

By solving this you will get x =-3/2 and x = 2/3

So the other roots are x = -1.5 and x = 2/3

`f(x) =6x^4 + 5x^3 -24x^2 - 15x+18`

**Roots of f(x) = 0 are** `x = sqrt3,-sqrt3,-1.5,2/3`