For the polynomial below, 1 is a zero. $$ f(x)=x^{3}+x^{2}-4 x+2 $$ Express $$ f(x) $$ as a product of linear factors. $$ f(x)=\square $$

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We begin with the polynomial
$$
f(x)=x^{3}+x^{2}-4x+2,
$$
and we are given that $$x=1$$ is a zero, so $$(x-1)$$ is a factor.

**Step 1. Divide by $$(x-1)$$:**

Using synthetic division with $$1$$ and the coefficients $$1,\,1,\,-4,\,2$$:

- Bring down the $$1$$.
- Multiply $$1 \times 1 = 1$$; add to the next coefficient: $$1+1=2$$.
- Multiply $$2 \times 1 = 2$$; add to the next coefficient: $$-4+2=-2$$.
- Multiply $$-2 \times 1 = -2$$; add to the constant: $$2+(-2)=0$$.

The quotient polynomial is
$$
x^2+2x-2.
$$

**Step 2. Factor the quadratic $$x^2+2x-2$$:**

We use the quadratic formula
$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
$$
with $$a=1$$, $$b=2$$, $$c=-2$$. Then,
$$
x=\frac{-2\pm\sqrt{2^2-4(1)(-2)}}{2(1)}=\frac{-2\pm\sqrt{4+8}}{2}=\frac{-2\pm\sqrt{12}}{2}.
$$
Simplify $$\sqrt{12} = 2\sqrt{3}$$:
$$
x=\frac{-2\pm 2\sqrt{3}}{2}=-1\pm\sqrt{3}.
$$
Thus, the zeros are $$x=-1+\sqrt{3}$$ and $$x=-1-\sqrt{3}$$.

**Step 3. Write $$ f(x) $$ as a product of linear factors:**

The zeros give us the factors
$$
(x-1),\quad \left(x-(-1+\sqrt{3})\right)=x+1-\sqrt{3},\quad \text{and} \quad \left(x-(-1-\sqrt{3})\right)=x+1+\sqrt{3}.
$$
Thus, the fully factored form is
$$
f(x)=(x-1)(x+1-\sqrt{3})(x+1+\sqrt{3}).
$$

Therefore,
$$
f(x)=(x-1)(x+1-\sqrt{3})(x+1+\sqrt{3}).
$$
$$ f(x) = (x - 1)(x + 1 - \sqrt{3})(x + 1 + \sqrt{3}) $$

For the polynomial below, 1 is a zero.

Express as a product of linear factors.

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For the polynomial below, 1 is a zero. Express as a product of linear factors.