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` .
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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`
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