According to Factor theorem, for a polynomial #p(x)# of degree greater than or equal to one, #x-alpha# is a factor of #p(x)#, if #p(alpha)=0#, then #(x-alpha)# is a factor of #p(x)#, where #alpha# is a real number. In fact #alpha# is also a factor of the constant term in the polynomial #p(x)#.
Hence, if we have a polynomial say #p(x)=x^3+lx^2+mx+n# and #p(alpha)=0#, then #alpha# is a factor of #n#. For example if the constant term is say #6#, then our #alpha#’s could be #+-1,+-2,+-3,+-6#. Hence, we seek a number among these for which #p(alpha)=0# and then we have #(x-alpha)# a factor of #p(x)#.
Observe that coefficient of highest power of #x# in #p(x)# is #1#. In case, the coefficient of highest power of #x# in #p(x)# is more than #1#, say #a# like in polynomial #ax^3+bx^2+cx+d#, then our #alpha# is a factor of #d/a#.
In practice, Factor Theorem is used for factorising polynomials completely. So our steps in factorising a polynomial #p(x)# would be
1 Check the constant term in #p(x)#
2. Find its all possible factors.
3. Take one of the factors, say #alpha# and replace #x# by this factor in the given polynomial #p(x)#.
4. Try more factors whose number should be equal to the degree of polynomial.
And you have got all the factors.
Coming to your question, we have #2ln(4x-5)+ln(x+1)=3ln3#, which can be written as #ln{(4x-5)^2(x+1)}=ln27#
or #(16x^2-40x+25)(x+1)}=ln27#
or #16x^3-24x^2-15x-2=0#
#16x^3-24x^2-15x-2=0#
This is the same as in #(b)# so we use factor theorem as follows.
We should have zeros among #+-2,+-1,+-1/2,+-1/4,+-1/8#.Observe that for #x=2,-1/4#, it is #0# and hence #x-(-1/4)# or #x+1/4# i.e. #(4x+1)# and #(x-2)# are two factors. We do not get any other number for which it is #0#, but dividing #16x^3-24x^2-15x-2# by #(4x+1)# and #(x-2)#, we get #(4x+1)#, hence factors are
#(4x+1)^2(x-2)=0#
Note that it is easier to check whether #+-1# are zeros or not, as for #+1#, we should have sum of all coefficients as #zero# and for #-1#, we change sign of alternate coefficient and find if sum is zero or not. As it is not so in given polynomial one can avoid using them.
