Questions: The polynomial function f(x)=3x^4+2x^3-12x^2-8x has (x-2) as one of its linear factors. Use long division to find the other linear factors. Separate multiple factors with a comma. If there are any repeated factors, only write them once.

Solution Steps

To find the other linear factors of the polynomial \( f(x) = 3x^4 + 2x^3 - 12x^2 - 8x \), given that \( (x-2) \) is a factor, we can use polynomial long division. We will divide the polynomial by \( (x-2) \) to find the quotient, which will be a cubic polynomial. We will then factor the resulting cubic polynomial to find the remaining linear factors.

We start with the polynomial \( f(x) = 3x^4 + 2x^3 - 12x^2 - 8x \) and divide it by the known factor \( (x - 2) \). This division yields a quotient of \( 3x^3 + 8x^2 + 4x \) and a remainder of \( 0 \), confirming that \( (x - 2) \) is indeed a factor.

Next, we factor the quotient \( 3x^3 + 8x^2 + 4x \). We can factor out \( x \) from the polynomial, resulting in \( x(3x^2 + 8x + 4) \).

Now, we need to factor the quadratic \( 3x^2 + 8x + 4 \). We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3 \), \( b = 8 \), and \( c = 4 \). The discriminant \( b^2 - 4ac = 8^2 - 4 \cdot 3 \cdot 4 = 64 - 48 = 16 \) is a perfect square, allowing us to find the roots.

Calculating the roots: \[ x = \frac{-8 \pm \sqrt{16}}{2 \cdot 3} = \frac{-8 \pm 4}{6} \] This gives us two roots: \[ x_1 = \frac{-4}{6} = -\frac{2}{3}, \quad x_2 = \frac{-12}{6} = -2 \]

Thus, the quadratic factors as \( 3(x + \frac{2}{3})(x + 2) \).

Final Answer