For the polynomial below, 2 is a zero. $$ f(x)=x^{3}+4 x^{2}-5 x-14 $$ Express $$ f(x) $$ as a product of linear factors. $$ f(x)= $$ (

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Since 2 is a zero of the polynomial $$ f(x) = x^{3} + 4x^{2} - 5x - 14 $$, we can use synthetic division to divide the polynomial by $$ x - 2 $$.
Factor the expression by following steps:
- step0: Factor:
  $$x^{3}+4x^{2}-5x-14$$
- step1: Calculate:
  $$x^{3}+6x^{2}+7x-2x^{2}-12x-14$$
- step2: Rewrite the expression:
  $$x	imes x^{2}+x	imes 6x+x	imes 7-2x^{2}-2	imes 6x-2	imes 7$$
- step3: Factor the expression:
  $$x\left(x^{2}+6x+7ight)-2\left(x^{2}+6x+7ight)$$
- step4: Factor the expression:
  $$\left(x-2ight)\left(x^{2}+6x+7ight)$$
After performing synthetic division, we have $$ f(x) = (x - 2)(x^{2} + 6x + 7) $$. Now, we need to factor the quadratic $$ x^{2} + 6x + 7 $$.
The quadratic $$ x^{2} + 6x + 7 $$ does not factor nicely into real linear factors, as its discriminant is negative. Therefore, we will express it in its current form.
The polynomial can be expressed as $$ f(x) = (x - 2)(x^{2} + 6x + 7) $$.

For the polynomial below, 2 is a zero.

Express as a product of linear factors.
(

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For the polynomial below, 2 is a zero. Express as a product of linear factors. (