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:
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Testing \(x = 1\): \[ 3(1)^3 + 4(1)^2 - 35(1) - 12 = 3 + 4 - 35 - 12 = -40 \quad \text{(not a root)} \]
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Testing \(x = -1\): \[ 3(-1)^3 + 4(-1)^2 - 35(-1) - 12 = -3 + 4 + 35 - 12 = 24 \quad \text{(not a root)} \]
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Testing \(x = 2\): \[ 3(2)^3 + 4(2)^2 - 35(2) - 12 = 24 + 16 - 70 - 12 = -42 \quad \text{(not a root)} \]
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Testing \(x = -2\): \[ 3(-2)^3 + 4(-2)^2 - 35(-2) - 12 = -24 + 16 + 70 - 12 = 50 \quad \text{(not a root)} \]
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Testing \(x = -3\): \[ 3(-3)^3 + 4(-3)^2 - 35(-3) - 12 = -81 + 36 + 105 - 12 = 48 \quad \text{(not a root)} \]
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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: \[ \boxed{(x + 4)(3x + 1)(x - 3)} \]