How do you use the rational root theorem to find the roots of #10x^4 + x^3 + 7x^2 + x -3 = 0#?

1 Answer
George C.
Nov 4, 2015

See explanation...

Explanation:

#f(x) = 10x^4+x^3+7x^2+x-3#

By the rational root theorem, any rational roots of #f(x) = 0# are expressible in the form #p/q#, where #p, q in ZZ#, #q != 0#, #p# and #q# have no common factor apart from #+-1#, #p# is a divisor of the constant term #-3# and #q# a divisor of the coefficient, #10#, of the leading term.

So the possible rational roots are:

#+-1/10#, #+-1/5#, #+-3/10#, #+-1/2#, #+-3/5#, #+-1#, #+-3/2#, #+-3#

Let's try to narrow down the search a little.

#f(0) = -3#
#f(1) = 10+1+7+1-3 = 16#

So there is a Real root in the interval #(0, 1)#. Since #f(x)# has only one change of sign, this will be the only positive root.

Let us calculate #f(3/10)# to split our remaining positive rational possibilities:

#f(3/10) = 10(3/10)^4+(3/10)^3+7(3/10)^2+(3/10)-3#

#=81/1000+27/1000+63/100+3/10-3#

#=(81+27+630+300-3000)/1000#

#=-1962/1000 < 0#

So the remaining possible positive rational roots are #1/2# and #3/5#.

#f(1/2) = 10(1/2)^4+(1/2)^3+7(1/2)^2+1/2-3#

#=5/8+1/8+7/4+1/2-3#

#=(5+1+14+4-24)/8 = 0#

So #x=1/2# is a root and #(2x-1)# is a factor of #f(x)#

If we separate out this factor, it's easy to spot the the remaining cubic factor will factor easily by grouping:

#10x^4+x^3+7x^2+x-3 = (2x-1)(5x^3+3x^2+5x+3)#

#=(2x-1)((5x^3+3x^2)+(5x+3))#

#=(2x-1)(x^2(5x+3)+(5x+3))#

#=(2x-1)(x^2+1)(5x+3)#

So the remaining rational root is #-3/5# and the two other roots are #+-sqrt(-1) = +-i#.

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