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

Answer:[tex](x-2)(x+3-\sqrt{2})(x+3+\sqrt{2})[/tex]Step-by-step explanation:[tex]f(x) = x^3 + 4x^2 -5x-14\\[/tex]since this is a cubic polynomial (highest power of  x is 3), we know that there are 3 possible values of x for which the f(x) is equal to zero (or there are values of x which satisfy f(x) = 0 ). In other words, there are 3 linear factors of this f(x) as the highest power of x is 3. given that "2 is a zero", this actually means that one of the value of x for which f(x) = 0 is x =2, we can check this out by: [tex]f(2) = (2)^3 + 4(2)^2 -5(2)-14\\f(2) = 0[/tex]hence x = 2, is one of the values that satisfy f(x) = 0, we can write this out in linear factors as (x-2) (think of this as taking 2 from the right hand side of the equation x = 2 to the left making x - 2 = 0)now we need to find the rest of the values of x, for this we first need to reduce this cubic (3rd power) polynomial f(x) into a quadratic (2nd power) polynomial. We can do this by "comparison method" since we already know that one of the linear factors of f(x) is (x-2) we can write[tex]x^3 + 4x^2 -5x-14 == (x-2)(ax^2+bx+c)[/tex]here, '==' is simply implying that the two expressions are the same. All we need to find now is the values of a,b and c. First simplify the right hand side of the equation by multiplying [tex](x-2)[/tex] with [tex](ax^2+bx+c)[/tex]. this will yield[tex]x^3 + 4x^2 -5x-14 == ax^3 + bx^2 - 2ax^2 + cx -2bx-2c\\x^3 + 4x^2 -5x-14 == ax^3 + (b-2a)x^2+(c-2b)x-2c\\[/tex]here is when the comparison takes place. compare all the x^3 terms on both sides, similarly the x^2 terms, the x terms, and finally the terms without any x. By comparing you can easily find the values of a b and c[tex]x^3 = ax^3\\4x^2 = (b-2a)x^2\\-5x = (c -2b)x\\-14 = -2c[/tex]you can see that all the x's cancel out. and the values can be found[tex]a = 1\\[/tex]from the 2nd equation[tex]4=b-2(1)\\b = 6[/tex]from the last equation[tex]c = 7[/tex]now you have your quadratic equation[tex](ax^2 + bx + c)\\(x^2 + 6x + 7)[/tex]now you can solve this equation using the quadratic formula[tex]x = \frac{-b +\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-b -\sqrt{b^2-4ac}}{2a}[/tex][tex]x = \frac{-6 +\sqrt{6^2-4(1)(7)}}{2(1)}\\\\x = \frac{-6 -\sqrt{6^2-4(1)(7)}}{2(1)}[/tex]solving and simplifying will yield[tex]x = -3 -\sqrt{2}\\x = -3 +\sqrt{2} \\[/tex]convert them to linear factors [tex](x+3+\sqrt{2})\\(x+3-\sqrt{2})[/tex]and the final answer is (don't forget (x-2) that is part of the answer) ::[tex](x-2)(x+3-\sqrt{2})(x+3+\sqrt{2})[/tex]hope this helps :)