Find an equation for fx the polynomial of smallest degree wi

Find an equation for f(x), the polynomial of smallest degree with real coefficients such that f(x) breaks through the x-axis at -5, bounces off of the x- axis at -1, has complex roots of 4+2i and -4+4i and passes through the point (0,-4). f(x)=

Solution

f (x) has zeros at - 5, - 1, 4 +2i and - 4 + 4i . Let f (x) = a( x + 5) (x +1)( x - 4 - 2i) (x + 4 - 4i) . This function has zeros at - 5, - 1, 4 +2i and - 4 + 4i . Since the function passes through ( 0, - 4) , we have f ( 0 ) = - 4. Therefore, - 4 = a( 0 + 5) (0 +1)( 0 - 4 - 2i) (0 + 4 -4i) = 5a(2i +4)(4i -4) = 40a(i + 2) (i -1) Therefore, a = (-1/10)* 1/ ( i + 2)( i - 1) . Onmultiplying both the numerator and the denominator of the RHS by ( i - 2)( i + 1), we have a = (-1/10)* ( i - 2)( i + 1) / ( - 5)( - 2) = ( - 1/100)( i - 2)( i + 1) ( as i² = -1) Thus, the required function is f (x) = ( - 1/100)( i - 2)( i + 1)( x + 5) (x +1)( x - 4 - 2i) (x + 4 - 4i) .