What are the possible rational zeros of f ( x ) = 2 x 4 + 5 x 3 − 2 x 2 − 11 x − 6 ?

Answer 1

The Rational Zeros Theorem (aka Rational Roots Theorem) says that IF there are any rational zeros (aka: roots), they will be (plus or minus) the factors of the constant term divided by the factors of the leading coefficient.

If we call the constant term "p" and the leading coefficient "q", then the possible rational zeros are ± (factors of p)/(factors of q):

The factors of the constant term (p) are 6, 3, 2, and 1.

The factors of the leading coefficient (q) are 2 and 1.

So the possible rational zeros are ±6/2. ±6/1, ±3/2, ±3/1, ±2/2, ±2/1, ±1/2, ±1/1

If you simplify these and throw out the duplicates you get the possible rational zeros are:

±6, ±3, ±2, ±3/2, ±1, ±1/2

Answer 2

2x^4 +5x^3 -2x^2 -11x -6

has 4 zeros, either 4, 2 or 0 real

4, 2 or 0 imaginary

although some may repeat, to get 4 zeros

use Descartes rule of signs

there's only 1 sign change, so maximum 1 positive zero

replace x with -x and count the sign changes

3 sign changes, so maximum 3 negative zeros

that leaves

a few possibilities:

1 positive zero, 3 negative zeros

1 positive, 1 negative, 2 imaginary

0 positive 2 negative, 2 imaginary

0 positive, 0 negative, 4 imaginary

try some simple integers, 0, 1 and 2

f(0) =-6

f(1) =-12

f(2) = 36

the sign change from -12 to +36 means there's a zero between 1 and 2, maybe in the middle

that means either 1 real positive zero and 3 negative zeros

or 1 real positive, 1 negative and 2 imaginary zeros

try x=-1

f(-1) = 0

-1 is one negative real zero,

that leaves two possibilities

1 +, 1 -, 2 imaginary

or 1+, 3 -

try f(-2) = 0

bingo another zero

that means 1+ and 3- zeros

divide (x+2)(x+1) into the 4th degree polynomial to get a quadratic polynomial

then factor, complete the square or use the quadratic formula to find the remaining

two zeros, one positive, one negative

2x^2 -x -3 is the quotient

it factors

(2x-3)(x+1) = 0

x = -1 and x= 3/2

the zeros are -2, -1, -1 and 1.5

all real, 3 negative, 1 positive

or

if you don't count -1 twice

then 3 zeros, -2, -1 and +1.5

the repetition of -1 means if you graph it that point is tangent to the x axis

((x-1.5)(x+2)(x+1)^2 = 2x^4+5x^3-2x^2 -11x -6

if you have a graphing caluculator just graph the 4th degree polynomial and see where it intersects or touches the x axis. those points are the zeros, the roots, x intercepts or solutions to the 4th degree polynomial set =0

sometimes though, when asked for the "possible zeros," it really means list the +/-q/p fractions

with p = factors of the coefficient of the highest powered term and q =factors of the the constant term

p=2, q=-6. factors of 2 are 1 and 2, factors of -6 are 1,2,3, and 6, all either - or +

q/p= -1, -2 and 3/2, the actual zeros

but other "possible" zeros might be listed as

6/1, 6/2, 3/1, 2/1 and 1/2 with both signs

= +/- 1/2, 2, 3 and 6. in addition to the real zeros, -2,-1 and 1.5

also -3/2

they aren't really real zeros, except for the actual ones, but in some trivial perspective you could call them "possible zeros" depending on what the problem is trying to get at, or what your instructor was expecting