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+3f(x)=2x4+x37x23x+3?

2 Answers
May 31, 2018

2x^4+x^3-7x^2-3x+3=(2x-1)(x+1)(x^2-3)2x4+x37x23x+3=(2x1)(x+1)(x23)

Explanation:

It iseasy to find
x=-1x=1
By the rational root theórem we get
x=1/2x=12
Dividing

2x^4+x^3-7x^2-3x+32x4+x37x23x+3 by
2x^2+x-12x2+x1 we get
x^2-3x23

Aug 10, 2018

See explanation...

Explanation:

Given:

f(x) = 2x^4+x^3-7x^2-3x+3f(x)=2x4+x37x23x+3

By the rational zeros theorem, any rational zeros of f(x)f(x) are expressible in the form p/qpq for integers p, qp,q with pp a divisor of the constant term 33 and qq a divisor of the coefficient 22 of the leading term.

That means that the only possible rational zeros are:

+-1/2, +-1, +-3/2, +-3±12,±1,±32,±3

In addition, note that the pattern of signs of the coefficients of f(x)f(x) is + + - - ++++. With two changes of sign, Descartes' Rule of Signs tells us that f(x)f(x) has 22 or 00 positive real zeros.

Further, note that the pattern of signs of the coefficients of f(-x)f(x) is + - - + ++++. With two changes of sign, Descartes' Rule of signs tells us that f(x)f(x) has 22 or 00 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))+3f(12)=2(12)4+(12)37(12)23(12)+3

color(white)(f(1/2)) = 2/16+1/8-7/4-3/2+3f(12)=216+187432+3

color(white)(f(1/2)) = 0f(12)=0

So x=1/2x=12 is a zero and (2x-1)(2x1) is a factor:

2x^4+x^3-7x^2-3x+3 = (2x-1)(x^3+x^2-3x-3)2x4+x37x23x+3=(2x1)(x3+x23x3)

color(white)(2x^4+x^3-7x^2-3x+3) = (2x-1)(x^2(x+1)-3(x+1))2x4+x37x23x+3=(2x1)(x2(x+1)3(x+1))

color(white)(2x^4+x^3-7x^2-3x+3) = (2x-1)(x^2-3)(x+1)2x4+x37x23x+3=(2x1)(x23)(x+1)

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

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