How do you find all the zeros of #f(x)= x^4 - 4x^3 - 20x^2 + 48x#?

1 Answer
Feb 27, 2016

Answer:

Factors of #f(x)=x^4−4x^3−20x^2+48x#) are #x(x-2)(x+4)(x-6)# and zeros of #f(x)# are #{0,2, -4, 6}#

Explanation:

As there is no independent in #f(x)=x^4−4x^3−20x^2+48x# and hence#x# is factor of #f(x)# and hence #)# is first zero among the zeros of #f(x)#. Factorizing #f(x)#

#f(x)=x(x^3-4x^2-20x+48)#

Now identify factors of #48# i.e. #{1,-1,2,-2,3,-3,4,-4,6,-6,...}#, who make #(x^3-4x^2-20x+48)=0#. As #x=2#, makes #f(x)=0#, and hence #(x-2)# is a factor of #f(x)# and #2# is another zero of #f(x)#.

Factorizing #(x^3-4x^2-20x+48)=(x-2)(x^2-2x-24)#

Now #(x^2-2x-24)#, can be easily factorized as #-24# can be exactly factorized into #-6# and #4#, and hence factorization can proceed as under.

#x^2-2x-24=x^2-6x+4x-24=x(x-6)+4(x-6)=(x+4)(x-6)#

Hence complete factors of #f(x)=x^4−4x^3−20x^2+48x#) are #x(x-2)(x+4)(x-6)# and zeros of #f(x)# are #{0,2, -4, 6}#