.
#12x^4+11x^3-26x^2+x+2=0#
We can use The Rational Zeros Theorem which states:
If #P(x)# is a polynomial with integer coefficients and if #p/q# is a zero of #P(x)#, i.e. #P(p/q) = 0#, then #p# is a factor of the constant term of #P(x)# and #q# is a factor of the leading coefficient of #P(x)#.
The constant term of our equation is #2# and the leading coefficient is #12#.
Therefore:
#p=+-1, +-2#
#q=+-1, +-2, +-3, +-6, +-12#
Then, possible roots of the polynomial would be:
#p/q=+-(1,2)/(1,2,3,6,12)#
We now begin to try each combination of the numerator and the denominator for #p/q# to see which value of #p/q# is the root of the polynomial. The first combination is #(+1)/(+1)=1#.
Plugging this into the equation gives us:
#12(1)^4+11(1)^3-26(1)^2+1+2=12+11-26+1+2=0#
This means #1# is a root. Now , we can either use long division, synthetic division, or factoring to factor #(x-1)# out. Let's see if we can do it by factoring:
We will break up the terms as follows:
#12x^4-12x^3+23x^3-23x^2-3x^2+3x-2x+2=0#
#12x^3(x-1)+23x^2(x-1)-3x(x-1)-2(x-1)=0#
#(x-1)(12x^3+23x^2-3x-2)=0#
Repeating the same process for the remaining cubic function, we will find that #-2# is a root. Using any of the methods above, we can factor #(x+2)# out. Again, let's use factoring:
We will break up terms as follows:
#12x^3+24x^2-x^2-2x-x-2=0#
#12x^2(x+2)-x(x+2)-(x+2)=0#
Therefore, our polynomial becomes:
#(x-1)(x+2)(12x^2-x-1)=0#
Since the remaining quadratic can easily be factored, we end up with:
#(x-1)(x+2)(4x+1)(3x-1)#
