can someone please help me find all the zeros of the following polynomial. write the polynomial in factored form.
f(x)=x^3-3x^2+16x-48
can you please show all work
Group two by two
X^2(x-3)+16(x-3)
(X-3)(x^2+16)
For zeros
X-3=0
X=3
X^2+16=0
X^2=-16
X=+or- 4i
To find the zeros of the given polynomial and write it in factored form, we can use the method of polynomial factorization.
Step 1: Checking for Rational Roots
First, let's check for any rational roots using the Rational Root Theorem. According to the theorem, the possible rational roots are the factors of the constant term (in this case, -48) divided by the factors of the leading coefficient (in this case, 1).
The factors of -48 are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, ±48.
The factors of 1 are ±1.
Therefore, the possible rational roots are:
±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, ±48.
Step 2: Testing the Possible Rational Roots
Let's further examine each possible rational root using synthetic division or polynomial long division to check if any of them are actual roots.
We will start with ±1 and substitute them into the polynomial f(x) = x^3 - 3x^2 + 16x - 48.
For f(1):
1 | 1 -3 16 -48
|______1_____-2____14____-34
0 -2 14 -34
Since the remainder is zero, 1 is a root of f(x). Therefore, (x - 1) is a factor of f(x). We can now divide the polynomial by (x - 1) to simplify it.
Dividing f(x) by (x - 1):
x^2 - 2x + 14
Now we need to find the remaining zeros of the simplified quadratic polynomial, x^2 - 2x + 14.
To determine if there are any real solutions for this quadratic equation, we can calculate the discriminant:
Discriminant (D) = b^2 - 4ac
where a = 1, b = -2, and c = 14.
D = (-2)^2 - 4(1)(14)
D = 4 - 56
D = -52
Since the discriminant is negative, the quadratic equation does not have any real roots. Instead, it has two complex roots.
By applying the quadratic formula, x = (-b ± √D) / 2a, we can determine the complex roots of the quadratic equation.
x = (2 ± √(-52)) / (2)
x = (2 ± 2i√13) / 2
x = 1 ± i√13
Hence, the zeros of the given polynomial are:
x = 1 (with multiplicity 1)
x = 1 + i√13 (complex conjugate with multiplicity 1)
x = 1 - i√13 (complex conjugate with multiplicity 1)
Therefore, the factored form of the polynomial f(x) = x^3 - 3x^2 + 16x - 48 is:
f(x) = (x - 1)(x - (1 + i√13))(x - (1 - i√13))