Question

Factor the polynomial function over the complex numbers. f(x)=x4+9x3+15x2+9x+14 Enter your answer in the box. f(x)=

Answer 1

The polynomial function

$f(x)=x^4+9x^3+15x^2+9x+14$

can be factored over the complex numbers as follows

$f(x)=(x+7)(x+2)(x-i)(x+i)$

The constant term of the fourth degree polynomial

$x^4+9x^3+15x^2+9x+14$

is positive, and has the factors $+7$ and $+2$. So, we will try out polynomial division with $(x+7)$

$\dfrac{x^4+9x^3+15x^2+9x+14}{x+7}=x^3+2x^2+x+2$

so $(x+7)$ is a factor.

Try out the polynomial division again, this time dividing the quotient with $(x+2)$

$\dfrac{x^3+2x^2+x+2}{x+2}=x^2+1$

so $(x+2)$ is also a factor.

So far, we have factored the fourth degree polynomial as follows

$x^4+9x^3+15x^2+9x+14=(x+7)(x+2)(x^2+1)$

we can still factor $(x^2+1)$ further by arranging it as a difference of two squares

$x^2+1=x^2-(-1)\\\\=x^2-i^2\\\\=(x-i)(x+i)$

So, our final factorization of the polynomial function is

$f(x)=x^4+9x^3+15x^2+9x+14\\\\=(x+7)(x+2)(x-i)(x+i)$

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