honestly hard questions. onnly top contributors can figure out probably. please help, nobody else can help me?

Question

please help! i hate listing zeroes and stuff and i have actually tried but i dont get it

For the following polynomials list the possible rational zeros. (Do not find the zeros!)

P(x) = x^5 + 3x^3 + 2x - 6

P(x) = 3x^4 + x^3 + 2x^2 - 2

P(x) = 2x^5 + 3x^4 + 2x^2 - 2

Answer 1

Hi,

Possible rational roots are

...(factors of the constant)
+-__________________________
....(factors of the leading coefficient)

P(x) = x^5 + 3x^3 + 2x - 6
.....factors of 6
+/-.---------------- = +/-1,+/-2,+/-3,+/-6
......factors of 1


P(x) = 3x^4 + x^3 + 2x^2 - 2
.....factors of 2.......1.or.2
+/-.----------------.=.----------.=.+/-1,+/-2,+/-1/3,+/-2/3
......factors of 3.......1.or.3


P(x) = 2x^5 + 3x^4 + 2x^2 - 2
.....factors of 2.......1.or.2
+/-.----------------.=.----------.=.+/-1,+/-2,+/-1/2
......factors of 2.......1.or.2

I hope that helps!! :-)

Answer 2

Here's the rule for this:
If a/b is a rational zero of P(x) = a_nx^n + ... + a_1x + a_0
a must divide a_0 and b must divide a_n.
Moreover we may take b positive without
loss of generality.
So let's see what we have:
1. P(x) = x^5 + 3x^3 + 2x - 6.
3. Here we may take b = 1. So the possible rational zeros are all the divisors of -6,
namely -6,-3,-2,-1,1,2,3 and 6.

2. P(x) = 3x^4 + x^3 + 2x^2 - 2
Here b is 1 or 3, so the possible rational zeros are
-2, -1, 1, 2, -2/3, -1/3, 1/3, and 2/3.

3.
P(x) = 2x^5 + 3x^4 + 2x^2 - 2
Now b = 1 or 2, so the possible rational zeros are
-1, -2,1, 2, -1/2 and 1/2.

Answer 3

Comparing the highest and lowest co-efficients will give you all the possible rational factors.

1) The only possible rational factors are ±1, ±2, ±3, ±6.

2) The only possible rational factors are ±1, ±2, ±1/3, ±2/3.

3) The only possible rational factors are ±1, ±2, ±1/2.

Answer 4

you use p's and q's, where p represents factors of the last term and q represents factors of the coefficient of the first term. all the possible roots are p's over q's(all the answers are plus or minus)

1. p's: 1, 2, 3, 6
q's: 1
possible roots: +/- 6, 3, 2, 1
2. p's: 1, 2
q's: 1, 3
possible roots: +/- (2/3), (1/3), 2, 1
3. p's: 1, 2
q's: 1, 2
possible roots: +/- 1, 2, (1/2)

Answer 5

ok so if u are finding possible values...
(if i remember correctly)
isn't it (P/Q)
The P being all factors of the constant, in the first problem the negative six
and the Q being all factors of the coefficient of the largest power, again in the first problem, the invisible number 1 in front of x^5
so find all possible values (ex. -6,-3,-2,-1,1,2,3,6)

Answer 6

1) zero is at 1
2) (-.8004308, 0) and (.69479505, 0)
3) (-1.723337, 0) and (-.8274684, 0) and (.69605952, 0)

I just used a graphing calculator, I don't know if you need to find this manually, which I am probably guessing you do... get it equal to zero and solve.

Answer 7

So are you supposed to estimate where the zeros occur or something? Graph them and see where they cross the x-axis.