Let ​f(x)=x2+14x+48​. What are the zeros of the function? Enter your answers in the boxes.

Remember that to find the zeroes of a quadratic expression of the form [tex]a x^{2} +bx+c[/tex] we could either factor it or use the quadratic formula. When [tex]a=1[/tex], like in our case, is often way easier and faster factor the expression than using the quadratic formula. The only thing we need to do to factor the expression is find tow number whose product is 48 and its sum is 14; those numbers are 6 and 8. [tex](6)(8)=48[/tex] and [tex]6+8=14[/tex]. Now we can factor our quadratic like follows:
[tex] x^{2} +14x+48=(x+6)(x+8)[/tex]
Since we are trying to find the zeroes of the function, we are going to set each one of our factored binomial equal to zero and solve for x:
[tex]x+6=0[/tex]
[tex]x=-6[/tex]
and
[tex]x+8=0[/tex]
[tex]x=-8[/tex]

We can conclude that the zeroes of the function [tex]f(x)= x^{2} +14x+48[/tex] are [tex]x=-6[/tex] and [tex]x=-8[/tex].