How do you use the rational roots theorem to find all possible zeros of #f(x)=3x^5-2x^4-15x^3+10x^2+12x-8 #?

1 Answer
Mar 3, 2016

Answer:

Zeros of #f(x)# are #{-2, -1, 1, 2/3. 2}#

Explanation:

Rational root theorem states that if a polynomial #a_0x^n+a_1x^(n-1)+a_2x^(n-2)+...+a_n#, has rational roots #p/q#, where #p# and #q# are integers, then #q# is a factor of #a_0# and #p# is a factor of #a_n#.

Hence to find all possible zeros of #f(x)=3x^5−2x^4−15x^3+10x^2+12x−8#,

we have to find roots of the equation #3x^5−2x^4−15x^3+10x^2+12x−8=0#

For this we start by factors of constant term #-8# such as #{1,-1,2,-2,4,-4,8,-8}#.

It is observed that #x=1# makes #f(x)=0# and hence #(x-1)# is a factor of #f(x)# as #f(1)=3-2-15+10+12-8=0#.

Similarly for #x=-1# makes #f(x)=0# and hence #(x+1)# is a factor of #f(x)# as #f(1)=-3-2+15+10-12-8=0#.

#x=-2# makes #f(x)=0# as #f(2)=3*32-2*16-15*8+10*4+12*2-8=96-32-120+40+24-8=0# and hence #(x-2)# is a factor of #f(x)#.

#x=--2# makes #f(x)=0# as #f(2)=3*(-32)-2*16-15*(-8)+10*4+12*(-2)-8=-96-32+120+40-24-8=0# and hence #(x+2)# is a factor of #f(x)#.

As four factors of #f(x)# have been found, we can have fifth factor by dividing #f(x)# by these and result would be #(3x-2)# and hence #(3x-2)=0# or #x=2/3# is another zero of #f(x)#

Hence, zeros of #f(x)# are #{-2, -1, 1, 2/3. 2}#