First, we use the Rational Root Theorem to determine all possible roots. To do this, we'll divide the constant term by the co-efficient of the highest term and factor this value.
#(-24)/1=24#
#24=2^3*3#
Possible roots are #+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-24#.
Now, it's just a matter of plugging each one in to see if the root works. If it does, then we can use synthetic division to divide out that root and keep going.
Let's start with #1#.
#(1)^4+4*(1)^3-7*(1)^2-34*(1)+24 #
#= 1+4-7-34-24 #
#= -60#
Since that isn't #0#, #1# is not a root.
Now, let's move on to #-1#.
#(-1)^4+4*(-1)^3-7*(-1)^2-34*(-1)-24 #
#= 1-4-7+34-24 #
#= 0#
This means that #-1# is a root. In other words, the binomial #x+1# divides into the original expression. Now that we know this root, we can divide it out of the expression using synthetic division.
The first step in using Synthetic Division is to find out what makes the binomial #0#. In this case, it is #-1#. Now, you put that number in a sort of secluded area like so:
#-1" "|#
Then, you determine the co-efficients for every term in your polynomial (you MUST also include co-efficients of #0#. However, there are none in this problem). Then, you make a list like this:
#-1" "|" "1" "" "4" "-7" "-34" "-24#
Then, you do this little trick where you bring down the first number (#1#), multiply by the secluded number (#-1#), put the answer down under the next number (#4#), and then add. After its all done, it'll look like this:
#-1" "|" "1" "" "4" "-7" "-34" "-24#
#color(white)(xxxxxxx)darrcolor(white)(x)-1" " -3" "" "10" "" "24#
#color(white)(xxxxx)stackrel("--------------------------------------------------------------------")#
#color(white)(xxxxxxx)1" "color(white)(xx)3color(white)(xij)-10" "-24|color(white)(xx)0#
Now, the remainder has the co-efficients on the third line (#1#, #3#, #-10#, and #-24#), making the polynomial #x^3+3x^2-10x-24#.
We now use this polynomial to check for more roots. If you repeat this process, you'll see that the roots are:
#x^4+4x^3-7x^2-34x-24=(x+1)(x+2)(x-3)(x+4)#
