To factor the polynomial x^3 - 3x^2 + 4x - 12, we will first check if any of the possible rational zeros of the function are actual zeros.
The possible rational zeros are given by the equation: ±(factors of constant term)/(factors of leading coefficient)
The constant term is -12 and its factors are ±1, ±2, ±3, ±4, ±6, ±12.
The leading coefficient is 1 and its factors are ±1.
Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±12.
We can test each of these values by substituting them into the polynomial to see if any of them give a result of 0.
f(1) = (1)^3 - 3(1)^2 + 4(1) - 12 = 1 - 3 + 4 - 12 = -10
f(-1) = (-1)^3 - 3(-1)^2 + 4(-1) - 12 = -1 - 3 - 4 - 12 = -20
f(2) = (2)^3 - 3(2)^2 + 4(2) - 12 = 8 - 12 + 8 - 12 = -8
f(-2) = (-2)^3 - 3(-2)^2 + 4(-2) - 12 = -8 - 12 - 8 - 12 = -40
f(3) = (3)^3 - 3(3)^2 + 4(3) - 12 = 27 - 27 + 12 - 12 = 0
Therefore, one zero of the polynomial is x = 3.
To factor the polynomial, we can use synthetic division with x = 3 as a zero:
3 | 1 - 3 4 -12
| 3 0 12
| ---------
| 1 0 4 0
The resulting quotient is x^2 + 4. This means that the factored form of the polynomial is:
(x - 3)(x^2 + 4)
The zeros of the polynomial are x = 3, x = 2i, x = -2i (where i is the imaginary unit).
factor completely and then find all zeros of the polynomial function:
x^3-3x^2+4x-12
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