What are all the possible rational zeros for #f(x)=10x^3-15x^2-16x+12# and how do you find all zeros?

1 Answer
Aug 7, 2017

Answer:

The "possible" rational zeros are:

#+-1/10, +-1/5, +-3/10, +-2/5, +-1/2, +-3/5, +-4/5, +-1, +-6/5, +-3/2, +-2, +-12/5, +-3, +-4, +-6, +-12#

The actual zeros are:

#2" "# and #" "-1/4+-sqrt(265)/20#

Explanation:

Given:

#f(x) = 10x^3-15x^2-16x+12#

By the rational root 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 #12# and #q# a divisor of the coefficient #10# of the leading term.

That means that the only possible rational zeros are:

#+-1/10, +-1/5, +-3/10, +-2/5, +-1/2, +-3/5, +-4/5, +-1, +-6/5, +-3/2, +-2, +-12/5, +-3, +-4, +-6, +-12#

Trying a few of the simpler values, we eventually find:

#f(2) = 10(color(blue)(2))^3-15(color(blue)(2))^2-16(color(blue)(2))+12#

#color(white)(f(2)) = 80-60-32+12#

#color(white)(f(2)) = 0#

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

#10x^3-15x^2-16x+12 = (x-2)(10x^2+5x-6)#

The remaining quadratic factor is in the standard form #ax^2+bx+c# with #a=10#, #b=5# and #c=-6#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 5^2-4(10)(-6) = 25+240 = 265#

This is positive but not a perfect square. Hence we can deduce that the remaining zeros are real but irrational. We can find the zeros using the quadratic formula:

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

#color(white)(x) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x) = (-5+-sqrt(265))/(2*10)#

#color(white)(x) = (-5+-sqrt(265))/20#

#color(white)(x) = -1/4+-sqrt(265)/20#