Find all the zeros of the equation.

Question

-x^3-3x^2=6x+4

Answer 1

To find the zeros of the equation, we need to rearrange it to have one side equal to zero. Let's do that:

-x^3 - 3x^2 = 6x + 4

Adding 6x and subtracting 4 from both sides:

-x^3 - 3x^2 - 6x - 4 = 0

Now, let's combine like terms:

-x^3 - 3x^2 - 6x - 4 = 0

The equation is a cubic equation. To solve it, we can use various methods, such as factoring, synthetic division, or the rational root theorem.

Let's try using the rational root theorem to find any rational zeros. According to the theorem, any rational zero of the equation must be of the form p/q, where p is a factor of the constant term -4, and q is a factor of the leading coefficient -1.

The factors of -4 are ±1, ±2, and ±4.
The factors of -1 are ±1.

Therefore, the possible rational zeros are ±1/1, ±2/1, and ±4/1, which simplify to ±1, ±2, and ±4.

We can substitute these values into the equation to check if any of them are zeros.

1:
(-1)^3 - 3(-1)^2 - 6(1) - 4 = -1 - 3 - 6 - 4 = -14 (not a zero)

-1:
(-(-1))^3 - 3(-(-1))^2 - 6(-1) - 4 = -1 - 3 + 6 - 4 = -2 (not a zero)

2:
(-2)^3 - 3(-2)^2 - 6(2) - 4 = -8 - 12 - 12 - 4 = -36 (not a zero)

-2:
(-(-2))^3 - 3(-(-2))^2 - 6(-2) - 4 = -8 - 12 + 12 - 4 = -12 (not a zero)

3:
(-3)^3 - 3(-3)^2 - 6(3) - 4 = -27 - 27 - 18 - 4 = -76 (not a zero)

-3:
(-(-3))^3 - 3(-(-3))^2 - 6(-3) - 4 = -27 - 27 + 18 - 4 = -40 (not a zero)

4:
(-4)^3 - 3(-4)^2 - 6(4) - 4 = -64 - 48 - 24 - 4 = -140 (not a zero)

-4:
(-(-4))^3 - 3(-(-4))^2 - 6(-4) - 4 = -64 - 48 + 24 - 4 = -92 (not a zero)

None of the possible rational zeros satisfy the equation, so there are no rational zeros.

Therefore, the equation has no real zeros.