# How do you find all zeros of the function  f(x) = 4(x + 7)^2(x - 7)^3?

Zeros of $f \left(x\right) = 4 {\left(x + 7\right)}^{2} {\left(x - 7\right)}^{2}$ are $- 7$ and $7$.
If $f \left(x\right) = u {\left(x - a\right)}^{m} {\left(x - b\right)}^{n} {\left(x - c\right)}^{p} {\left(x - d\right)}^{q}$, a multiplication of number of binomials of degree one, then zeros of polynomials are $a$, $b$, $c$ and $d$, as any of them when put in place of $x$ will make $f \left(x\right) = 0$. Here $u$ is just a constant.
Hence, zeros of $f \left(x\right) = 4 {\left(x + 7\right)}^{2} {\left(x - 7\right)}^{2}$ are $- 7$ and $7$.