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) breaks through the x-axis at -5,-----> zerp at x=--5

bounces off of the x-axis at -1-----> zero at x =-1

has complex roots of 4+2i and -4+4i   

now if (4 +2i) is a root of f(x) ---4-2i would also be a root

now if (-4 +4i) is a root of f(x) --- -4+ i would also be a root

So, f(x) = a( x+1)(x+5)(x -4 -2i)(x -4 +2i)( x +4 -4i)(x +4 +4i)

= a(x+1)(x+5) { ( x^2 -4x +2ix -4x +16 -8i -2ix +8i +4)( x^2 +4x+4ix+4x +16 +16i - 4ix -16i +16)

=a(x+1)(x+5)(x^2 - 8x+20)(x^2+8x +32)

Find \'a\' using (0, 4)

So, 4 = a( 1*5*20*32)

1 = a(5*5*32)

a = 1/800

f(x) = (x+1)(x+5)(x^2 - 8x+20)(x^2+8x +32)/800