Given $$ f(x)=6 x^{3}+31 x^{2}+3 x-10 $$, answer the following. Part: $$ \mathbf{0} / 2 $$ Part 1 of 2 Factor $$ f(x) $$, given that $$ \frac{1}{2} $$ is a zero. $$ f(x)= $$

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1. Since $$ \frac{1}{2} $$ is a zero of $$ f(x)=6x^3+31x^2+3x-10 $$, it follows that $$ x-\frac{1}{2} $$ is a factor. For convenience, we can factor instead by using $$ 2x-1 $$ (which yields the same zero).

2. Assume
$$
f(x) = (2x-1)(Ax^2+Bx+C).
$$

3. Multiply out the factors:
$$
(2x-1)(Ax^2+Bx+C) = 2A\,x^3 + 2B\,x^2 + 2C\,x - A\,x^2 - B\,x - C.
$$
Combine like terms:
$$
= 2A\,x^3 + (2B-A)x^2 + (2C-B)x - C.
$$

4. Equate the coefficients to those in $$ f(x)=6x^3+31x^2+3x-10 $$:
   - For $$ x^3 $$: 
     $$
     2A = 6 \quad \Rightarrow \quad A = 3.
     $$
   - For $$ x^2 $$:
     $$
     2B - A = 31 \quad \Rightarrow \quad 2B - 3 = 31 \quad \Rightarrow \quad 2B = 34 \quad \Rightarrow \quad B = 17.
     $$
   - For $$ x $$:
     $$
     2C - B = 3 \quad \Rightarrow \quad 2C - 17 = 3 \quad \Rightarrow \quad 2C = 20 \quad \Rightarrow \quad C = 10.
     $$
   - For the constant term:
     $$
     -C = -10 \quad \Rightarrow \quad C = 10,
     $$
     which is consistent.

5. Thus,
$$
f(x)= (2x-1)(3x^2+17x+10).
$$

6. Next, factor the quadratic $$ 3x^2+17x+10 $$. Look for two numbers whose product is $$ 3 \times 10 = 30 $$ and whose sum is $$ 17 $$. The numbers $$ 15 $$ and $$ 2 $$ work because $$ 15+2=17 $$ and $$ 15 \times 2 = 30 $$.

7. Rewrite and factor by grouping:
$$
3x^2+17x+10 = 3x^2+15x+2x+10.
$$
Group the terms:
$$
= (3x^2+15x) + (2x+10) = 3x(x+5) + 2(x+5).
$$
Factor out the common factor $$ (x+5) $$:
$$
= (x+5)(3x+2).
$$

8. Therefore, the full factorization of $$ f(x) $$ is:
$$
f(x) = (2x-1)(x+5)(3x+2).
$$
$$ f(x) = (2x - 1)(x + 5)(3x + 2) $$

Given \( f(x)=6 x^{3}+31 x^{2}+3 x-10 \), answer the following. Part: \( \mathbf{0} / 2 \) Part 1 of 2 Factor \( f(x) \), given that \( \frac{1}{2} \) is a zero. \( f(x)= \)

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