Question

For the polynomial below, -3 and 1 are zeros. Express f (x) as a product of linear factors.

Answer 1

Step-by-step explanation

Since -3 and 1 are zeros of the functions, it implies that

`(x+3)\text{ }and\text{ }(x-1)`

are factors of the equation.

Therefore we can find the remaining factors below

`(x+3)(x-1)=x^2+2x-3`

By long division

`remaining\text{ expression =}(x^4+6x^3+7x^2-8x-6)/(x^2+2x-3)=x^2+4x+2`

By quadratic formula

`\begin{gathered} x_(1,2)=(-4\pm√(4^2-4*1*2))/(2*1) \\ x_1=(-4+2√(2))/(2),x_2=(-4-2√(2))/(2) \\ x=-2+√(2),x=-2-√(2) \\ therefore \\ (x+2-√(2))(x+2+√(2)) \end{gathered}`

The linear factor are

Answer:

`f(x)=(x+3)(x-1)(x+2-√(2))(x+2+√(2))`