In a cubic polynomial# f(x)#, the coefficient of #x^3# is 1. The roots of #f(x)# = 0 are #-1, 2k and k^2#, where #k# is a positive integer. When #f(x)# is divided by #x-5#, the remainder is 24.. (a) Show that #2k^3-5k^2 -10k+21=0#, (b) Find of #k# ?

1 Answer
George C.
Mar 3, 2018

#k=3#

Explanation:

A zero #a# of a polynomial corresponds to a linear factor like #(x-a)#.

Since #f(x)# has zeros #-1#, #2k# and #k^2#, it can be written:

#f(x) = (x+1)(x-2k)(x-k^2)#

The remainder when #f(x)# is divided by #a# is #f(a)#. So we have:

#24 = f(5)#

#color(white)(24) = (color(blue)(5)+1)(color(blue)(5)-2k)(color(blue)(5)-k^2)#

#color(white)(24) = 6(2k^3-5k^2-10k+25)#

Divide both ends by #6# to get:

#4 = 2k^3-5k^2-10k+25#

Subtract #4# from both sides and transpose to find:

#2k^3-5k^2-10k+21 = 0#

By the rational roots theorem, any rational roots of this cubic are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #21# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2, +-1, +-3/2, +-3, +-7/2, +-7, +-21/2, +-21#

Trying these, we find:

#2(color(blue)(3))^3-5(color(blue)(3))^2-10(color(blue)(3))+21 = 54-45-30+21 = 0#

So #k=3# is a zero and #(k-3)# a factor:

#2k^3-5k^2-10k+21 = (k-3)(2k^2+k-7)#

Using the quadratic formula, the zeros of the remaining quadratic factor are:

#k = (-1+-sqrt(1^2-4(2)(-7)))/(2*2) = -1/4+-sqrt(57)/4#

However, note that we were told that #k# is a positive integer, so we can ignore these roots.

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