Quadratics Question 7

Given constants $c$ , $d$ , $m$ , and $n$ such that $x^2+mx+c$ has factors of $(x+2)$ and $(x+4)$ and $x^2+nx+d$ has factors $(x+3)$ and $(x+7)$ , what is $mn$ ?

16

18

29

60

168

Approach

x^2+mx+c = (x+2)(x+4)

x^2+mx+c = x^2+6x+8

m=6

x^2+nx+d = (x+3)(x+7)

x^2+nx+d = x^2+10x+21

n=10

mn = 6(10) = \boxed{60}