The function #f(x) = x^3-x^2-14x+24# intersects the #x#-axis at the point #(k, 0)#. What is one possible value of #k#?

2 Answers
Jim G.
Dec 18, 2017

#k=2#

Explanation:

#"substitute "(k,0)" into "f(x)#

#f(k)=k^3-k^2-14k+24=0#

#"note that "k=2tof(2)=0#

#"hence one value of "k=2#

#"other values may be found as follows"#

#k=2rArr(k-2)" is a factor"#

#rArrk^3-k^2-14k+24#

#=color(red)(k^2)(k-2)color(magenta)(+2k^2)-k^2-14k+24#

#=color(red)(k^2)(k-2)color(red)(+k)(k-2)color(magenta)(+2k)-14k+24#

#=color(red)(k^2)(k-2)color(red)(+k)(k-2)color(red)(-12)(k-2)cancel(color(magenta)(-24))cancel(+24)#

#rArr(k-2)(k^2+k-12)=0#

#rArr(k-2)(k+4)(k-3)=0#

#rArrk=-4,k=2,k=3" are all possible values"#

George C.
Dec 18, 2017

#k = 2#, #k = -4# or #k = 3#

Explanation:

Given:

#f(x) = x^3-x^2-14x+24#

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

That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-24#

We find:

#f(2) = 8-4-28+24 = 0#

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

#x^3-x^2-14x+24 = (x-2)(x^2+x-12) = (x-2)(x+4)(x-3)#

So #k = 2#, #k = -4# or #k = 3#

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