How do you find the rational roots of #x^4+5x^3+7x^2-3x-10=0#?
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Use the rational roots theorem to find rational roots #x=1# and #x=-2#.
Let #f(x) = x^4+5x^3+7x^2-3x-10#
The rational roots theorem tells us that all rational roots of #f(x) = 0# must be of the form #p/q# where #p# and #q# are integers, #q != 0#, #p# is a divisor of the constant term #10# and #q# is a divisor of the coefficient #1# of the highest order term #x^4#.
So the possible rational roots are:
#+-1#, #+-2#, #+-5# and #+-10#
Note that the sum of the coefficients of #f(x)# is #0#:
#1+5+7-3-10 = 0#
So #f(1) = 0#.
That is #x=1# is a root of #f(x) = 0# and #(x-1)# is a factor of #f(x)#
We also find
#f(-2) = 16-40+28+6-10 = 0#
So #x=-2# is a root of #f(x) = 0# and #(x+2)# is a factor of #f(x)#
#(x-1)(x+2) = x^2+x-2#
We can long divide #f(x)# by #x^2+x-2# to find:
#f(x)/(x^2+x-2) = x^2+4x+5#
A quick check of the discriminant of this quotient polynomial finds:
#Delta = 4^2-(4xx1xx5) = 16-20 = -4#
So there are no more real roots and no more linear factors with real coefficients.
