Question

Which are the real zeros of this function? f(x) = x3 – 6x2 – 16x

Answer 1

To find the zeros of the function $f(x)= x^{3} -6x^{2}-16x$ we need to factorize the expression.

$f(x)= x^{3} -6x^{2}-16x =x(x^{2} -6x-16 )$

a nice way to factorize $x^{2} -6x-16$, if possible, is by completing the square as follows:

$x^{2} -6x-16 =x^{2} -2*3x+ (3)^{2}-(3)^{2} -16$

$=(x^{2} -2*3x+ (3)^{2})-9-16= (x-3)^{2}-25= (x-3)^{2}- 5^{2}$

now we use the difference of squares formula $a^{2} - b^{2} =(a+b)(a-b)$:

$(x-3)^{2}- 5^{2}=[(x-3)+5][(x-3)-5]=[x+2][x-8]$


finally, we combine the results:

$f(x)=x(x+2)(x-8)$

the zeros of f, are the values of x which make f(x)=0,

they are x=0, x=-2 and x=8


Answer: {-2, 0, 8}