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!
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
(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
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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.
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.