Question

find the zeros of the polynomial x^2 + 8x + 16

Answer 1

hey mate,

$x^2 + 8x + 16 = 0$

<=>

$(x+4)^2 = 0$

As we know that $(a+b)^2 = a^2+b^2+2ab$

then

$(x+4)^2 = 0$

<=> (x+4) = 0

<=> x = -4

so there is one solution x = -4

thanks