College algebra help!
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College algebra help!

[From: ] [author: ] [Date: 12-10-07] [Hit: ]
-Careful with the complex roots:(x+4)(x-1/3)[x-(2+3i)][x-(2-3i)]we want to multiply it all out, but also, you need to consider that there may be a constant term out front, so put an a there:a(x+4)(x-1/3)[x-(2+3i)][x-(2-3i)]foil the last two together:a(x+4)(x-1/3)(x^2 - 4x + 13)Now you might want to plug in the 1 for x(from the f(1)=100 )f(1) = a(1+4)(1-1/3)(1^2 - 4*1 + 13)100 = a(5)(2/3)(10)100 = a(100/3)multiply by the reciprocal of 100/3 to get3 = aso f(x) = 3(x+4)(x-1/3)(x^2 -4x + 13)you should multiply the 3 in to the second set of parenthesis to get rid of the fractionf(x) = (x+4)*3*(x-1/3)(x^2 - 4x + 13)f(x) = (x+4)(3x-1)(x^2 -4x + 13)the foil outf(x) = (3x^2 + 11x - 4)(x^2 - 4x + 13)f(x) = 3x^4 - x^3 - 9x^2 + 159x - 52to check, plug in the x = 1 to get 100 for f(1)it checks out with f(1) = 100 so I likely multiplied everything out correctly-f(x)=k(x+4)(x-1/3)(2+3i)(2-3i) where I have made k a constant because it could exist in the problem and if it is 1 no problem and lets hope its not 0. Youre telling me you dont know how to work with fractions?......
Find an nth degree polynomial function w/ real coefficients satisfying the given conditions: n = 4; -4, 1/3, and 2 +3i are zeros ; f(1)=100


I'm not sure what to do with fractions

I get to

(x+4)(x-1/3)(2+3i)(2-3i)

and thats it!

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Careful with the complex roots:

(x+4)(x-1/3)[x-(2+3i)][x-(2-3i)]

we want to multiply it all out, but also, you need to consider that there may be a constant term out front, so put an "a" there:

a(x+4)(x-1/3)[x-(2+3i)][x-(2-3i)]

foil the last two together:

a(x+4)(x-1/3)(x^2 - 4x + 13)

Now you might want to plug in the 1 for x (from the f(1)=100 )

f(1) = a(1+4)(1-1/3)(1^2 - 4*1 + 13)

100 = a(5)(2/3)(10)

100 = a(100/3)

multiply by the reciprocal of 100/3 to get

3 = a

so f(x) = 3(x+4)(x-1/3)(x^2 -4x + 13)

you should multiply the 3 in to the second set of parenthesis to get rid of the fraction

f(x) = (x+4)*3*(x-1/3)(x^2 - 4x + 13)

f(x) = (x+4)(3x-1)(x^2 -4x + 13)

the foil out

f(x) = (3x^2 + 11x - 4)(x^2 - 4x + 13)

f(x) = 3x^4 - x^3 - 9x^2 + 159x - 52

to check, plug in the x = 1 to get 100 for f(1)

it checks out with f(1) = 100 so I likely multiplied everything out correctly

-
f(x)=k(x+4)(x-1/3)(2+3i)(2-3i) where I have made k a constant because it could exist in the problem and if it is 1 no problem and let's hope it's not 0. You're telling me you don't know how to work with fractions? I doubt it.

So go to that site and learn how to simplify this expression, or multiply everything out.

Next, you have 100= f(1), so in your simplified equation, you will find the value of the constant k. Again, see handy website. With k found, you plug that back into the equation to find the desired function which you expect is an nth degree polynomial.

Edit: make sure you use the right factors as scott said.
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