Question

Find all the zeros help please

Answer 1

With f(7) = 0, you know that f(x) is divisible by x-7. So

$f(x) = (x-7)(ax^2+bx+c) \\\\ x^3-4x^2-25x+28 = ax^3+(b-7a)x^2+(c-7b)x-7c \\\\ \implies \begin{cases}a=1\\b-7a=-4\\c-7b=-25\\-7c=28\end{cases} \implies a=1,b=3,c=-4$

Then we have

$f(x)=(x-7)(x^2+3x-4)=(x-7)(x+4)(x-1)$

which makes the zeros 7, -4, and 1.