help please!! state the possible rational zeros for each function. then find all zeros. y=x^3-13x^2-x+13 show all the step please .thanks

i know

look at the grouping:

x^2(x-13) - (x-13)
(x^2-1)(x-13)
(x-1)(x+1)(x-13)

To find the possible rational zeros of a polynomial function, you can use the Rational Zero Theorem. According to this theorem, the possible rational zeros of a polynomial function are the ratios of the factors of the constant term (in this case, 13) to the factors of the leading coefficient (in this case, 1).

The factors of 13 are 1 and 13, while the factors of 1 are 1 and -1. Therefore, the possible rational zeros of the given function (y=x^3-13x^2-x+13) are:

±1, ±13

To find the actual zeros of the function, you can use synthetic division or factoring. Let's try synthetic division:

1 | 1 -13 -1 13
| 1 -12 -13
_____________________
| 1 -12 -13 0

Since the remainder is zero, it means that (x-1) is a factor, and we're left with a quadratic equation. So the zeros of the function are x = 1 and the solutions of the quadratic equation:

1x^2 - 12x - 13 = 0

You can solve this quadratic equation using factoring or the quadratic formula:

(x-13)(x+1) = 0

Setting each factor to zero:

x-13 = 0 -> x = 13
x+1 = 0 -> x = -1

Therefore, the zeros of the function are x = 1, x = 13, and x = -1.

To find the possible rational zeros of the function y = x^3 - 13x^2 - x + 13, we can use the Rational Root Theorem. According to this theorem, any rational zeros of the function can be expressed as p/q, where p is a factor of the constant term (here, 13) and q is a factor of the leading coefficient (here, 1).

Step 1: Determine the factors of the constant term (13): ±1, ±13 \
Step 2: Determine the factors of the leading coefficient (1): ±1

The possible rational zeros can be calculated by taking all the possible combinations of p and q:

p/q = ±1/±1, ±1/±13, ±13/±1, ±13/±13

Simplifying gives the following possible rational zeros: ±1, ±13.

Now, let's find the actual zeros of the function:

Step 1: Start with the possible rational zeros and try them one by one.

When we substitute x = 1 into the function:
y = 1^3 - 13(1^2) - 1 + 13
= 1 - 13 - 1 + 13
= 0

Thus, x = 1 is a zero of the function.

Step 2: To find additional zeros, we need to factor the polynomial. Divide the given polynomial (x^3 - 13x^2 - x + 13) by the zero we found, x = 1, using long division or synthetic division.

Performing synthetic division:
1 | 1 -13 -1 13
| 1 -12 -13
|_________________
1 -12 -13 0

The result is a quadratic expression, 1x^2 - 12x - 13, and a constant term of 0. The quotient is 1x^2 - 12x - 13.

Step 3: Solve the quadratic equation 1x^2 - 12x - 13 = 0 using factoring, quadratic formula, or completing the square.

Since this quadratic does not factor easily, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a.

For our equation, a = 1, b = -12, and c = -13.

x = (12 ± √((-12)^2 - 4(1)(-13))) / 2(1)
= (12 ± √(144 + 52)) / 2
= (12 ± √196) / 2
= (12 ± 14) / 2

This gives two possibilities:
1) x = (12 + 14) / 2 = 26 / 2 = 13
2) x = (12 - 14) / 2 = -2 / 2 = -1

Thus, x = 13 and x = -1 are the other two zeros of the function.

So, the zeros of the function y = x^3 - 13x^2 - x + 13 are:
x = 1, x = 13, and x = -1.