The function f is defined by f(x)=x^2+3x-10 if f(x+5)=x^2+kx+30, k= find the smallest zero of f(x+5), x=

F(x+5)=(x+5)^2+3(x+5)-10

f(x+5)=x^2+10x+25+3x+15-10

f(x+5)=x^2+13x+30  and f(x+5)=x^2+kx+30 so

k=13

Now factor x^2+13x+30

Find j and k such that jk=ac=30 and j+k+b=13 so j and k are 10 and 3 so

(x+3)(x+10)

So the two zeros occur when x=-3 and -10 the smallest of which is:

x=-10

I’m not absolutely sure but i think it’s 7x6x5x4x3x2=5040 because as the first character you can take any of the 7 given then you are left with 5 then 6 then 4 etc.