A real coefficients polynomial function f(x) is of degree 4. It has a complex root x=square root 2i and a simple root at 2. If f(-1)=0 and its y-intercept is 8

Question

Find this function. Write your answer in standard form , in descending order

 

Answer 1

Begin by writing down what you know. The polynomial is to the fourth degree, which means you have a x⁴. You know two of the roots are x=sqrt(2)i and x=2, so you know two of the factors right off the bat. You know that the y-intercept is 8, and that when x=-1, the function is equal to zero.

 

If the roots are x=sqrt(2)i and x=2, that means that your factors are (x²+2)=0 and (x-2)=0, or (x²+2)(x-2). But those can't be the only factors, because your function is to the fourth degree, and when x=-1 they don't equal zero. To make it so the function is equal to zero when x=-1, you can add the factor (x+1), to give (x²+2)(x-2)(x+1). Since -1+1 is equal to zero, this would make your whole equation 0 at f(-1).

 

Now you have a fourth degree function with both of the roots that were given, but you have to make sure the y-intercept is 8 before finishing. Check the y-intercept by solving at f(0). 

 

At f(0) the equation is equal to -4. Which is not correct. However, we can multiply -4 by -2 to get 8. 

 

If you do this, you will have the function f(x)=-2(x²+2)(x-2)(x+1). This function meets all of the above requirements.