How do you find all the zeros of # 2x^3+9x^2+6x-8#?

1 Answer
George C.
Aug 13, 2016

This cubic has zeros #-2# and #1/4(-5+-sqrt(57))#

Explanation:

#f(x) = 2x^3+9x^2+6x-8#

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-8# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2, +-1, +-2, +-4, +-8#

We find:

#f(-2) = 2(-8)+9(4)+6(-2)-8 = -16+36-12-8 = 0#

So #x=-2# is a zero and #(x+2)# a factor:

#2x^3+9x^2+6x-8=(x+2)(2x^2+5x-4)#

The remaining quadratic is in the form #ax^2+bx+c# with #a=2#, #b=5#, #c=-4#.

We can solve this using the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#=(-5+-sqrt(5^2-4(2)(-4)))/(2*2)#

#=1/4(-5+-sqrt(25+32))#

#=1/4(-5+-sqrt(57))#

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