How do you find all zeros with multiplicities of #f(x)=-2x^3+19x^2-49x+20#?

1 Answer
Mar 19, 2017

Answer:

The zeros of #f(x)# are #1/2#, #4# and #5#, all with multiplicity #1#.

Explanation:

#f(x) = -2x^3+19x^2-49x+20#

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 #20# 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, +-5/2, +-4, +-5, +-10, +-20#

in addition, notice that the signs of the coefficients are in the pattern #- + - +#. So by Descartes' Rule of Signs, since there are #3# changes of sign, this cubic has #1# or #3# positive real zeros. Also the signs of the coefficients of #f(-x)# are in the pattern #+ + + +#. So with no changes of sign, this cubic has no negative real zeros.

So the only possible rational zeros are:

#1/2, 1, 2, 4, 5, 10, 20#

Trying each in turn we first find:

#f(1/2) = -2(1/8)+19(1/4)-49(1/2)+20#

#color(white)(f(1/2)) = (-1+19-98+80)/4#

#color(white)(f(1/2)) = 0#

So #x=1/2# is a zero and #(2x-1)# a factor:

#-2x^3+19x^2-49x+20 = (2x-1)(-x^2+9x-20)#

To factor the remaining quadratic note that #4+5 = 9# and #4*5 = 20#.

Hence:

#-x^2+9x-20 = -(x-4)(x-5)#

So the remaining two zeros are #x=4# and #x=5#.

All of the zeros have multiplicity #1#.