# college algebra

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find the complete zeros of the polynomial function. write f in factored form.
f(x)=x^3-7x^2+25x-39
please show all work. I do not understand this.

• college algebra - ,

we would let f(x)=x^3-7x^2+25x-39 = 0.

we would factor to get (x-3)(x^ 2-4x+13).

find the solutions to the second part (x^2-4x+13)using the quadratic formula. (the first part is x=3)

• college algebra - ,

i am not understanding why you put 0 for the first answer. it doesn't seem like you added or did any type of math to get the 0. can you explain more. this is all confusing to me.

• college algebra - ,

F(x) = Y = x^3 - 7x^2 + 25x - 39 = 0.
The real solutions to the Eq are the x-intercepts or the points where the graph crosses the x-axis. The value of Y
where the curve crosses the x-axis is 0.
That is why Y is set to 0.

All real values of X that satisfy the Eq
is a real solution. It was found by trial and error that 3 gives a zero output(3,0).

X = 3
x-3 = 0 and it is a factor.
Using long-hand division, we divide the
Eq by x-3 and get x^2-4x+13. Now we have(x-3)(x^2-4x+13) = 0

x = 4 +-sqrt(16-52)/2
X = 4 +-sqrt(-36)/2
x = (4 +-6i)/2 = 2 +- 3i

Solution set:X = 3,(2+3i), and(2-3i).

• college algebra - ,

NOTE: There is only 1 zero(x=3),the
final factored form is (x-3)(x^2-4x+13)=0.

The solution set was not required.

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