Please somebody help me solve these problems. I have triedunsuccessfully to arri
ID: 3093881 • Letter: P
Question
Please somebody help me solve these problems. I have triedunsuccessfully to arrive at the right answers! I will appreciateyour efforts and your precious time spent. 1a) How do I solve this;2x3-23x2+58x+35=0 given that 5 is a zero off(x)=2x3-23x2+58x+35? b) I want to find the pressure of a gas- the pressure ofthe gas varies jointly as the amount of the gas(measured in moles)and the temperature and inveserly as the volume of the gas. If thepressure is 1014kiloPascals when the number of moles is 7, thetemperature is 2600kelvin, and the volume is 560cc. Howdo I find the pressure when the number of moles is 6, thetemperature is 3300K, and the volume is 720cc? 2a) I tried this also and my answer was wrong. So, how do Iuse the Rational Zero Theorem to list all posible rational zerosfor this function- f(x) = -2x3+3x2-4x +8? b) In this case, I want to use Descarte's Rule of Signs todetermine the possible number of positive and negative real zerosfor this function- f(x) = -7x9+x5-x2 + 6. Please help. Thanking you all for your assistance. Please somebody help me solve these problems. I have triedunsuccessfully to arrive at the right answers! I will appreciateyour efforts and your precious time spent. 1a) How do I solve this;2x3-23x2+58x+35=0 given that 5 is a zero off(x)=2x3-23x2+58x+35? b) I want to find the pressure of a gas- the pressure ofthe gas varies jointly as the amount of the gas(measured in moles)and the temperature and inveserly as the volume of the gas. If thepressure is 1014kiloPascals when the number of moles is 7, thetemperature is 2600kelvin, and the volume is 560cc. Howdo I find the pressure when the number of moles is 6, thetemperature is 3300K, and the volume is 720cc? 2a) I tried this also and my answer was wrong. So, how do Iuse the Rational Zero Theorem to list all posible rational zerosfor this function- f(x) = -2x3+3x2-4x +8? b) In this case, I want to use Descarte's Rule of Signs todetermine the possible number of positive and negative real zerosfor this function- f(x) = -7x9+x5-x2 + 6. Please help. Thanking you all for your assistance.Explanation / Answer
1a) How do I solve this;2x3-23x2+58x+35=0 given that 5 is a zero off(x)=2x3-23x2+58x+35?I think we can assume that (x-5) is a factor of this3rd-degree polynomial. So let's factor it out!
(2x3-23x2+58x+35) / (x - 5)
We should be able to take out a 2x2 to get ridof the cube in the numerator.
Factor out 2x2:
Remainder: (2x3 - 23x2 + 58x + 35) -2x2 * (x - 5)
=2x3 - 23x2 + 58x + 35 - 2x3 +10x2
= -13x2 + 58x + 35
Factor out -13x:
Remainder: (-13x2 + 58x + 35) - (-13x) * (x - 5)
= -13x2 + 58x + 35 +13x2 - 65x
= -7x + 35
Factor out -7:
Remainder: (-7x + 35) - (-7) * (x - 5)
= -7x + 35 + 7x - 35
= 0
We factored out 2x2 - 13x - 7 with no remainder,so now we just have to solve for the roots of 2x2 - 13x- 7!
roots = ( -b +/- sqrt(b2 - 4ac) ) / 2a
= ( 13 +/- sqrt(169 - 4(2)(-7)) ) / 2(2)
= ( 13 +/- sqrt(169 + 56) ) / 4
= ( 13 +/- sqrt(225) ) / 4
= ( 13 - 15 ) / 4 -and- ( 13 + 15 ) / 4
= -2/4 -and- 28/4
= -1/2 -and- 7
b) I want to find the pressure of a gas- thepressure of the gas varies jointly as the amount of thegas(measured in moles) and the temperature and inveserly as thevolume of the gas. If the pressure is 1014kiloPascals when thenumber of moles is 7, the temperature is 2600kelvin, andthe volume is 560cc. How do I find the pressure when the number ofmoles is 6, the temperature is 3300K, and the volume is720cc?
PV = nRT, where R is the universal gasconstant. Let's solve for it...
(1014 kPa )(560 cc) = (7 mol)(R)(260 K)
R = (1014 kPa )(560 cc) / (7)(260 K)
R = 312 (kPa-cc/mol-K are the units of this number)
Now that we have R, we can do the second half, with only 1variable to solve for... pressure!
P(720 cc) = (6 mol)(R)(330 K)
P = (6 mol)(R)(330 K) / (720 cc)
P = (6 * 330 / 720) * R
P = (11 / 4) * (312)
P = 858 kPa
Sorry I can't help you with the other problem. I have never heardof it before but it's also time to do my own homework =p
I tried it quick but it didn't make sense. The only zero I foundwas 1.70393... are you sure you have the equation right?