Find all zeros of the polynomial. (Enter your answers as a comma-separated list.)

P(x) = x3 − 11x2 + 33x − 35

type exponents this way:

P(x) = x^3 - 11x^2 + 33x - 35
try values of x = ±1, ±5, ±7

arggh, the last one P(7) worked
P(7) = 343 - 11(49) + 33(7) - 35 = 0

so x-7 is a factor
after sysnthetic division:
P(x) = (x-7)(x^2 - 4x + 5) = 0
x = 7 or by formula, x = (4 ± √-4)/2
= 2 ± 2i

x = 7, 2 ± 2i

don't know if you want only real roots

To find the zeros of the polynomial P(x) = x^3 - 11x^2 + 33x - 35, we need to solve the equation P(x) = 0.

Step 1: Start by factoring out any common factors, if possible. In this case, there are no common factors.

Step 2: Use synthetic division or long division to find one zero, which can be used to factor the polynomial. We can try substituting some small values (usually integers) into the polynomial to see if we get a zero. One possible zero could be x = 1. Let's test it using synthetic division:

1 | 1 -11 33 -35
|_________
| 1 -10 23
|_________
1 -10 23 -12

The remainder is not zero, so x = 1 is not a zero of the polynomial.

Step 3: Now, we can use the Rational Root Theorem to find possible rational zeros. According to the Rational Root Theorem, the possible rational roots of P(x) are of the form p/q, where p is a factor of the constant term (-35) and q is a factor of the leading coefficient (1). The factors of -35 are ±1, ±5, ±7, ±35, and the factors of 1 are ±1.

So, the possible rational zeros are: ±1, ±5, ±7, ±35.

Step 4: We can now test each of these possible rational zeros using synthetic division to see if we get a zero remainder. Let's start with x = 1 again:

1 | 1 -11 33 -35
|_________
| 1 -10 23
|_________
1 -10 23 -12

The remainder is not zero, so x = 1 is not a zero.

Proceeding with the other possible rational zeros:

-1 | 1 -11 33 -35
|_________
| -1 12 -45
|_________
1 -12 45 10

The remainder is not zero, so x = -1 is not a zero.

5 | 1 -11 33 -35
|_________
| 5 -30 15
|_________
1 -6 3 -20

The remainder is not zero, so x = 5 is not a zero.

-5 | 1 -11 33 -35
|_________
| -5 80 -565
|_________
1 -16 113 -600

The remainder is not zero, so x = -5 is not a zero.

7 | 1 -11 33 -35
|_________
| 7 -28 35
|_________
1 -4 5 0

The remainder is zero, so x = 7 is a zero.

Hence, the zeros of the polynomial P(x) = x^3 - 11x^2 + 33x - 35 are: 7.

Therefore, the answer is 7.

To find the zeros of a polynomial, you need to solve the equation P(x) = 0.

For the given polynomial P(x) = x^3 - 11x^2 + 33x - 35, we can find the zeros by factoring or by using numerical methods such as the Rational Root Theorem or synthetic division.

Let's start by factoring.

First, we test possible factors of the constant term (-35) divided by the coefficients of the leading term (1). The factors of -35 are ±1, ±5, ±7, ±35.

We can start by trying x = 1 as a possible zero. Substituting x = 1 into the polynomial, we get:

P(1) = 1^3 - 11(1)^2 + 33(1) - 35 = 1 - 11 + 33 - 35 = -12

Since P(1) ≠ 0, x = 1 is not a zero.

Next, we try x = -1 as a possible zero. Substituting x = -1 into the polynomial, we get:

P(-1) = (-1)^3 - 11(-1)^2 + 33(-1) - 35 = -1 - 11 - 33 - 35 = -80

Since P(-1) ≠ 0, x = -1 is not a zero.

We continue testing the other possible factors until we find a zero or exhaust all the options.

Next, we try x = 5 as a possible zero. Substituting x = 5 into the polynomial, we get:

P(5) = 5^3 - 11(5)^2 + 33(5) - 35 = 125 - 275 + 165 - 35 = -20

Since P(5) ≠ 0, x = 5 is not a zero.

Next, we try x = -5 as a possible zero. Substituting x = -5 into the polynomial, we get:

P(-5) = (-5)^3 - 11(-5)^2 + 33(-5) - 35 = -125 - 275 - 165 - 35 = -600

Since P(-5) ≠ 0, x = -5 is not a zero.

Next, we try x = 7 as a possible zero. Substituting x = 7 into the polynomial, we get:

P(7) = 7^3 - 11(7)^2 + 33(7) - 35 = 343 - 539 + 231 - 35 = 0

Since P(7) = 0, x = 7 is a zero.

Now we have found one zero, x = 7. To find the remaining zeros, we can perform synthetic division or long division to factorize the polynomial.

Using synthetic division with x = 7 as the zero, we get:

7 | 1 - 11 33 -35
| 7 -28 35
--------------
1 -4 5 0

The quotient from synthetic division is 1x^2 - 4x + 5. This is a quadratic polynomial, which we can solve using the quadratic formula or by factoring.

The quadratic polynomial 1x^2 - 4x + 5 does not have real solutions because its discriminant (b^2 - 4ac) is negative. Therefore, the remaining zeros are complex numbers which can be found using the quadratic formula.

The quadratic formula is x = (-b ± sqrt(b^2 - 4ac)) / (2a).

For our quadratic polynomial, a = 1, b = -4, and c = 5. Substituting these values into the quadratic formula, we get:

x = (-(-4) ± sqrt((-4)^2 - 4(1)(5))) / (2(1))
x = (4 ± sqrt(16 - 20)) / 2
x = (4 ± sqrt(-4)) / 2

Since the square root of a negative number is imaginary, we have:

x = (4 ± 2i) / 2
x = 2 ± 1i

Therefore, the zeros of the polynomial P(x) = x^3 - 11x^2 + 33x - 35 are:
x = 7, x = 2 + i, and x = 2 - i.

So the zeros are: 7, 2 + i, 2 - i.