Question

factor the polynomial 3x^3+4x^2-35x-12

Answer 1

Give me a formula to use for any equation of this kind

Answer 2

To factor the polynomial \( 3x^3 + 4x^2 - 35x - 12 \), we can use the Rational Root Theorem to test for possible rational roots. The factors of the constant term (-12) and the leading coefficient (3) will give us candidates to test.

The possible rational roots can be found by taking the factors of \(-12\) and dividing by the factors of \(3\):

Possible rational roots:
\[
\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}, \pm \frac{12}{3} \quad \text{(which simplifies to \( \pm 4 \))}
\]

Now we will test some of these values:

1. Testing \(x = 1\):
\[
3(1)^3 + 4(1)^2 - 35(1) - 12 = 3 + 4 - 35 - 12 = -40 \quad \text{(not a root)}
\]

2. Testing \(x = -1\):
\[
3(-1)^3 + 4(-1)^2 - 35(-1) - 12 = -3 + 4 + 35 - 12 = 24 \quad \text{(not a root)}
\]

3. Testing \(x = 2\):
\[
3(2)^3 + 4(2)^2 - 35(2) - 12 = 24 + 16 - 70 - 12 = -42 \quad \text{(not a root)}
\]

4. Testing \(x = -2\):
\[
3(-2)^3 + 4(-2)^2 - 35(-2) - 12 = -24 + 16 + 70 - 12 = 50 \quad \text{(not a root)}
\]

5. Testing \(x = -3\):
\[
3(-3)^3 + 4(-3)^2 - 35(-3) - 12 = -81 + 36 + 105 - 12 = 48 \quad \text{(not a root)}
\]

6. Testing \(x = -4\):
\[
3(-4)^3 + 4(-4)^2 - 35(-4) - 12 = -192 + 64 + 140 - 12 = 0 \quad \text{(is a root)}
\]

Since \(x = -4\) is a root, we can use synthetic division to divide the polynomial by \(x + 4\).

Performing synthetic division of \(3x^3 + 4x^2 - 35x - 12\) by \(x + 4\):

\[
\begin{array}{r|rrrr}
-4 & 3 & 4 & -35 & -12 \\
& & -12 & 32 & 12 \\
\hline
& 3 & -8 & -3 & 0 \\
\end{array}
\]

The result is \(3x^2 - 8x - 3\).

Now we need to factor \(3x^2 - 8x - 3\). To factor this quadratic, we look for two numbers that multiply to \(3 \cdot (-3) = -9\) and add to \(-8\). The numbers are \(-9\) and \(1\).

We can rewrite the quadratic:
\[
3x^2 - 9x + x - 3 = 3x(x - 3) + 1(x - 3) = (3x + 1)(x - 3)
\]

Therefore, our complete factorization of the original polynomial is:
\[
3x^3 + 4x^2 - 35x - 12 = (x + 4)(3x + 1)(x - 3)
\]

So, the fully factored form is:
\[
\boxed{(x + 4)(3x + 1)(x - 3)}
\]

Answer 3

For factoring cubic polynomials of the form \( ax^3 + bx^2 + cx + d \), there isn't a one-size-fits-all formula, but there is a general approach that you can use, combining several techniques, including synthetic division, the Rational Root Theorem, and polynomial long division. Here’s a structured procedure you can follow for any cubic polynomial:

### Steps to Factor a Cubic Polynomial

1. **Identify the Coefficients**: Write down the polynomial in standard form \( ax^3 + bx^2 + cx + d \), identifying \( a, b, c, \) and \( d \).

2. **Use the Rational Root Theorem**:
- List all possible rational roots, which can be calculated as the factors of the constant term \( d \) divided by the factors of the leading coefficient \( a \).
- The possible rational roots are of the form \( \frac{p}{q} \), where \( p \) is any factor of \( d \) and \( q \) is any factor of \( a \).

3. **Test Possible Roots**: Substitute the possible rational roots into the polynomial to determine if any of them are actual roots (i.e., make the polynomial equal to zero).

4. **Synthetic Division**: If you find a root \( r \):
- Use synthetic division to divide the cubic polynomial by \( (x - r) \).
- This will leave you with a quadratic polynomial.

5. **Factor the Quadratic Polynomial**:
- Try to factor the resulting quadratic polynomial using methods such as factoring by grouping, completing the square, or using the quadratic formula if necessary.

6. **Combine Factors**: The final factorization will be of the form:
\[
ax^3 + bx^2 + cx + d = (x - r)(\text{quadratic factor}),
\]
where the quadratic can be expressed in a factored form if possible.

### Notes

- **Quadratic Formula**: If you cannot factor the quadratic polynomial directly, remember you can always use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), to find the roots. The quadratic can be factored as:
\[
ax^2 + bx + c = a\left(x - r_1\right)\left(x - r_2\right),
\]
where \( r_1 \) and \( r_2 \) are the roots found using the quadratic formula.

- **Multiple Roots and Complex Numbers**: If the cubic polynomial has complex or repeated roots, you may need to consider additional techniques/advanced algebraic methods.

### Example

For a polynomial \( 2x^3 - 3x^2 - 8x + 4 \):
1. Coefficients are \( a=2, b=-3, c=-8, d=4 \).
2. Possible rational roots: \( \pm1, \pm2, \pm4, \pm\frac{1}{2}, \pm\frac{2}{2}, \pm\frac{4}{2} \).
3. Test these roots using substitution.
4. Upon finding a root (e.g., \( x = 2 \)), use synthetic division to reduce to a quadratic.
5. Factor the quadratic as needed.

This approach can systematically guide you through the process of factoring any cubic polynomial.