Math Problem

[JSchnett]

1. Find the values of p and q so that -2 and 3 are zeros of
f(x)= 2x^3 -x^2 +px +q

2. Find the remaining zero(s)


Any help would be greatly appreciated I can't figure this out.

Thanks
Ogre

[Borodog]

Something is wrong with the problem. Do you see why?

Perhaps you mean f(x)= 2x^3 -x^2 + px +q ?

[JSchnett]

[ QUOTE ]
Something is wrong with the problem. Do you see why?

Perhaps you mean f(x)= 2x^3 -x^2 + px +q ?

[/ QUOTE ]

yes you are right, I fixed it but i still dont get it.

[Borodog]

Just plug in your two values of x, set the resulting equations equal to zero. That gives you 2 equations and 2 unknowns. Solve.

[arahant]

[ QUOTE ]
[ QUOTE ]
Something is wrong with the problem. Do you see why?

Perhaps you mean f(x)= 2x^3 -x^2 + px +q ?

[/ QUOTE ]

yes you are right, I fixed it but i still dont get it.

[/ QUOTE ]
0 = 2(-2)^3 - -2^2 + p(-2) + q = -16-4-2p + q = q-2p -20
0 = 2(3)^3 - 2^2 + p(3) + q = 54 - 4 +3p +q = q + 3p + 50

-5p -70 = 0 => p = -14...=> q + 28 = 20 => q = -8

f(x) = 2x^3 - x^2 - 14x -8

need help finding the zeros? or is that enough...

[JSchnett]

I see where I was making my mistake. thanks alot arahant

[NLSoldier]

lol precalcaments

[bigpooch]

You made a substitution error:

The second equation should be

0 = 2(3^3) - 3^2 + 3p + q

Then, p = -13 and q = -6.

f(x) = 2(x^3) - x^2 - 13x - 6

Then dividing by (x+2)(x-3) gives (2x+1) so x = -1/2 is the
other zero.

[jason1990]

The linear algebra is a waste of time, since you have to do the polynomial division anyway.

[BruceZ]

[ QUOTE ]
1. Find the values of p and q so that -2 and 3 are zeros of
f(x)= 2x^3 -x^2 +px +q

2. Find the remaining zero(s)

[/ QUOTE ]

Here's the fast way to get the answer to both parts at once. You know f(x) is divisible by (x+2)*(x-3) = x^2 - x - 6. Divide this into f(x) by long division to give 2x + 1 with a remainder of (p+13)x + q + 6. To make this 0, we have p = -13, and q = -6, and x = -1/2 is the other root.

[arahant]

[ QUOTE ]
You made a substitution error:

The second equation should be

0 = 2(3^3) - 3^2 + 3p + q

Then, p = -13 and q = -6.

f(x) = 2(x^3) - x^2 - 13x - 6

Then dividing by (x+2)(x-3) gives (2x+1) so x = -1/2 is the
other zero.

[/ QUOTE ]

Oops. Hope I didn't cause him to miss the problem, fail the course, drop out of school, and become a reclusive serial killer.

That would suck.