Question 458892: Given the polynomial function f(x), find the rational zeros, then the other zeros(that is, solve the equation f(x)=0), and factor f(x) into linear factors.)
f(x)=3x^4-10x^3+20x^2-40x+32
I know that the first zero is 2 and that their should be 4 zeros in all.
thanks for the help.
Found 2 solutions by richwmiller, Edwin McCravy:
Answer by richwmiller(17219)   (Show Source):
Answer by Edwin McCravy(20059)  (Show Source):
You can put this solution on YOUR website!
The other tutor did not factor it completely into
linear factors.
f(x)=3x^4-10x^3+20x^2-40x+32
You say you know that the first zero is 2,
so we do this synthetic division:
2|3 -10 20 -40 32
| 6 -8 24 -32
3 -4 12 -16 0
So we have now partially factored f(x) as
f(x) = (x-2)(3x³-4x²+12x-16)
Now we will factor this
3x³-4x²+12x-16
by grouping. Out of the first two terms
factor out x²
x²(3x-4)+12x-16
Out of the last two terms factor out +4
x²(3x-4)+4(3x-4)
Factor out (3x-4) and get
(3x-4)(x²+4)
So we have now further partially factored f(x) as
f(x) = (x-2)(3x-4)(x²+4)
Now since i² = -1, -i² = 1, and since 4 = 4*1,
4*1 also equals 4(-i²) or -4i²,
so replace 4 by -4i²
f(x) = (x-2)(3x-4)(x²-4i²)
Now we can completely factor f(x) by factoring that
last factor as the difference of two squares:
f(x) = (x-2)(3x-4)(x-2i)(x+2i)
To solve it we set each of the factors = 0
x-2=0, 3x-4=0, x-2i=0, x+2i=0
x=2, 3x=4, x=2i x=-2i
x=4/3
So the solutions are 2, 4/3, 2i, and -2i.
Edwin
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