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)
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.
4 answers
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.
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).
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).
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.
final factored form is (x-3)(x^2-4x+13)=0.
The solution set was not required.