How do you find all zeros with multiplicities of #f(x)=36x^4-12x^3-11x^2+2x+1#?

1 Answer
Mar 5, 2017

Answer:

#f(x)# has zeros #1/2# with multiplicity #2# and #-1/3# with multiplicity #2#

Explanation:

Given:

#f(x) = 36x^4-12x^3-11x^2+2x+1#

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 #1# and #q# a divisor of the coefficient #36# of the leading term.

That means that the only possible rational zeros are:

#+-1/36, +-1/18, +-1/12, +-1/9, +-1/6, +-1/4, +-1/3, +-1/2, +-1#

Note that #36 > 26 = 12+11+2+1#, so #+-1# are not zeros.

#f(1/2) = 36(color(blue)(1/2))^4-12(color(blue)(1/2))^3-11(color(blue)(1/2))^2+2(color(blue)(1/2))+1#

#color(white)(f(1/2)) = 36/16-12/8-11/4+2/2+1#

#color(white)(f(1/2)) = (9-6-11+4+4)/4#

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

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

#36x^4-12x^3-11x^2+2x+1 = (2x-1)(18x^3+3x^2-4x-1)#

Trying #x=1/2# with the remaining cubic factor, we find:

#18(color(blue)(1/2))^3+3(color(blue)(1/2))^2-4(color(blue)(1/2))-1 = 18/8+3/4-4/2-1#

#color(white)(18(1/2)^3+3(1/2)^2-4(1/2)-1) = (9+3-8-4)/4#

#color(white)(18(1/2)^3+3(1/2)^2-4(1/2)-1) = 0#

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

#18x^3+3x^2-4x-1 = (2x-1)(9x^2+6x+1)#

#color(white)(18x^3+3x^2-4x-1) = (2x-1)(3x+1)^2#

So the remaining zero is #x=-1/3# with multiplicity #2#.

How did I get from #9x^2+6x+1# to #(3x+1)^2# ?

Note that #961 = 31^2# and the multiplication #31*31 = 961# involves no carrying of digits. So it's just like #9x^2+6x+1 = (3x+1)^2# with #x=10#.

graph{36x^4-12x^3-11x^2+2x+1 [-1.2, 1.2, -0.74, 1.76]}