One zero of the polynomial is given. Use it to express the polynomial as a product of linear factors over the complex numbers. \( f(x)=x^{4}+5 x^{3}+9 x^{2}+15 x+18 \); zero: \( x=-2 \) \( f(x)=\square \)

Given that one zero of the polynomial \( f(x) = x^{4} + 5x^{3} + 9x^{2} + 15x + 18 \) is \( x = -2 \), we can express the polynomial as a product of linear factors over the complex numbers. Since \( x = -2 \) is a zero of the polynomial, we can write the polynomial as: \[ f(x) = (x + 2)(x + a)(x + b)(x + c) \] where \( a \), \( b \), and \( c \) are the other zeros of the polynomial. To find the other zeros, we can use the fact that the sum of the zeros of a polynomial is equal to the negation of the coefficient of the \( x^{n-1} \) term divided by the coefficient of the \( x^n \) term, where \( n \) is the degree of the polynomial. In this case, the sum of the zeros is \( -5 \), so we have: \[ -2 + a + b + c = -5 \] Simplifying, we get: \[ a + b + c = -3 \] Now, we need to find the values of \( a \), \( b \), and \( c \) that satisfy this equation. Let's find the other zeros of the polynomial by factoring the polynomial \( f(x) \) further. Factor the expression by following steps: - step0: Factor: \(x^{4}+5x^{3}+9x^{2}+15x+18\) - step1: Evaluate: \(x^{4}+9x^{2}+5x^{3}+15x+18\) - step2: Rewrite the expression: \(x^{4}+3x^{2}+3x^{3}+9x+2x^{3}+6x+6x^{2}+18\) - step3: Factor the expression: \(x\left(x^{3}+3x+3x^{2}+9\right)+2\left(x^{3}+3x+3x^{2}+9\right)\) - step4: Factor the expression: \(\left(x+2\right)\left(x^{3}+3x+3x^{2}+9\right)\) - step5: Factor the expression: \(\left(x+2\right)\left(x+3\right)\left(x^{2}+3\right)\) The polynomial \( f(x) = x^{4} + 5x^{3} + 9x^{2} + 15x + 18 \) can be factored as: \[ f(x) = (x + 2)(x + 3)(x^{2} + 3) \] Now, we can express the polynomial as a product of linear factors over the complex numbers: \[ f(x) = (x + 2)(x + 3)(x + i\sqrt{3})(x - i\sqrt{3}) \] Therefore, the polynomial \( f(x) \) can be expressed as a product of linear factors over the complex numbers.