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.
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)
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).
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.
To find the complete zeros of the polynomial function f(x) = x^3 - 7x^2 + 25x - 39 and write it in factored form, you can use a method called factoring by grouping or the Rational Root Theorem.
1. Start by checking if any of the possible rational roots divide evenly into the constant term (-39 in this case) and the leading coefficient (1 in this case). The possible rational roots can be found by taking all the positive and negative factors of the constant term and dividing them by the factors of the leading coefficient. In this case, the factors of -39 are ±1, ±3, ±13, ±39, and the factors of 1 are ±1, so the possible rational roots are ±1, ±3, ±13, ±39.
2. To check if a possible root is a zero, you can use synthetic division or substitute the value into the function and see if it gives you zero. Let's start with the possible root x = 1:
f(1) = (1)^3 - 7(1)^2 + 25(1) - 39
= 1 - 7 + 25 - 39
= -20
Since f(1) is not zero, x = 1 is not a zero of the function f(x). Repeat the process for the other possible rational roots until you find a zero or try to factor the polynomial.
3. Let's try the next possible root x = -1:
f(-1) = (-1)^3 - 7(-1)^2 + 25(-1) - 39
= -1 - 7 - 25 - 39
= -72
Since f(-1) is not zero, x = -1 is also not a zero of the function f(x).
4. Continue the process for the remaining possible roots. After trying all the possible rational roots, we find that none of them are zeros of the function f(x).
5. If none of the rational roots work, you can use numerical methods or a graphing calculator to approximate the zeros. By using a graphing calculator, I found that the zeros of the function are approximately x = 3 and x = 13.
Therefore, the complete zeros of the polynomial function f(x) = x^3 - 7x^2 + 25x - 39 are x = 3, x = 13 (approximate values).
To write the function in factored form, we can use the zeros we found:
f(x) = (x - 3)(x - 13)(x - k)
Here, k represents another zero that we haven't found. Since the degree of the polynomial is 3, we know that there must be a third zero. To find the value of k, we can divide the original polynomial by the known zeros.
Using long division, we can divide f(x) by (x - 3)(x - 13):
(x^3 - 7x^2 + 25x - 39) / ((x - 3)(x - 13))
Performing the division, we get:
x^2 - 4x + 12
The remainder is zero, which means that (x - 3)(x - 13) is a factor of f(x). Now, we can rewrite the polynomial as:
f(x) = (x - 3)(x - 13)(x - k)
To find the value of k, we can either divide the result x^2 - 4x + 12 by (x - 3) or (x - 13) or use other methods like synthetic division. By dividing x^2 - 4x + 12 by (x - 3), we get:
x^2 - 4x + 12 = (x - 3)(x - 4)
Therefore, the factored form of the polynomial function f(x) = x^3 - 7x^2 + 25x - 39 is:
f(x) = (x - 3)(x - 13)(x - 4)