# 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

#### Explanation:

It iseasy to find

By the rational root theórem we get

Dividing

See explanation...

#### Explanation:

Given:

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

By the rational zeros theorem, any rational zeros of

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

Further, note that the pattern of signs of the coefficients of

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

#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

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