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

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

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]}