How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descarte's rule of signs to list the possible positive/negative zeros of #f(x)=2x^4+x^3-7x^2-3x+3#?

2 Answers
May 31, 2018

Answer:

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

Explanation:

It iseasy to find
#x=-1#
By the rational root theórem we get
#x=1/2#
Dividing

#2x^4+x^3-7x^2-3x+3# by
#2x^2+x-1# we get
#x^2-3#

Aug 10, 2018

Answer:

See explanation...

Explanation:

Given:

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

By the rational zeros theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #3# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

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

In addition, note that the pattern of signs of the coefficients of #f(x)# is #+ + - - +#. With two changes of sign, Descartes' Rule of Signs tells us that #f(x)# has #2# or #0# positive real zeros.

Further, note that the pattern of signs of the coefficients of #f(-x)# is #+ - - + +#. With two changes of sign, Descartes' Rule of signs tells us that #f(x)# has #2# or #0# negative real zeros.

Trying each of the possible rational zeros in turn we find:

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

#color(white)(f(1/2)) = 2/16+1/8-7/4-3/2+3#

#color(white)(f(1/2)) = 0#

So #x=1/2# is a zero and #(2x-1)# is a factor:

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

#color(white)(2x^4+x^3-7x^2-3x+3) = (2x-1)(x^2(x+1)-3(x+1))#

#color(white)(2x^4+x^3-7x^2-3x+3) = (2x-1)(x^2-3)(x+1)#

So the other zeros are #x=+-sqrt(3)# and #x=-1#

graph{2x^4+x^3-7x^2-3x+3 [-2.5, 2.5, -6, 5]}