Question

Help Needed
Let ​f(x)=x2+12x+32.​ What are the zeros of the function? Enter your answers in the boxes.

Answer 1

The zeros of this function is -4 and -8

Answer 2

Answer:

Step-by-step explanation:

The given equation is:

$f(x)=x^2+12x+32$

Simplifying the above expression, we have

$f(x)=x^2+4x+8x+32$

$f(x)=x(x+4)+8(x+4)$

$f(x)=(x+4)(x+8)$

Thus, zeroes of the given function will be:

$f(x)=0$

$(x+4)(x+8)=0$

$x=-4 and x=-8$

Therefore the zeroes of the given function are $x=-4 and x=-8$.