# what are the zeroes inclucing irrational of this polynomial? 6x^4 - 5x^3 - 9x^2

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To find the zeros of the polynomial, first factor the `x^2` , then use factoring for the remaining quadratic.

`6x^4-5x^3-9x^2`

`=x^2(6x^2-5x-9)` the second term has zeros from the quadratic formula

`x={5+-sqrt{25+4(6)(9)}}/12`

`={5+-sqrt241}/12`

**The zeros are 0, `{5+-sqrt241}/12` .**

The zeroes of 6x^4 - 5x^3 - 9x^2 are the same as the solution of the equation 6x^4 - 5x^3 - 9x^2 = 0.

Solving the equation 6x^4 - 5x^3 - 9x^2 = 0

Factor out x^2.

x^2*(6x^2 - 5x - 9) = 0

x^2 = 0 gives x = 0

6x^2 - 5x - 9 = 0 can be solved as follows:

The roots of a quadratic equation ax^2 + bx + c = 0 are `(-b+-sqrt(b^2-4ac))/(2a)`

For 6x^2 - 5x - 9 = 0, a = 6, b = -5 and c = -9. The roots are:

`(5+-sqrt(25 + 216))/(12)`

= `5/12 +- sqrt241/12`

The zeroes of 6x^4 - 5x^3 - 9x^2 are 0, `5/12 + sqrt241/12` , `5/12 - sqrt241/12`