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December 20, 2014

December 20, 2014

Posted by **Heather** on Tuesday, November 20, 2012 at 5:16pm.

f(x)=x^3-7x^2+25x-39

please show all work. I do not understand this.

- college algebra -
**confused**, Tuesday, November 20, 2012 at 5:22pmwe 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 -
**confused**, Tuesday, November 20, 2012 at 5:39pmi 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 -
**Henry**, Thursday, November 22, 2012 at 4:35pmF(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 -
**Henry**, Thursday, November 22, 2012 at 4:48pmNOTE: 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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