(ax+b)(cx+d)=0 is the format of a factored quadratic.
After "FOIL"ing that expression, you have
(a•c)x^2+(a•d+b•c)x+(b•d)=0
5 x^2 + 9 x -72 =0
Using your equation,
a•c=5, so a=1 and c=5, or a=5 and c=1
b•d=-72, so b={1,2,3,4,6,8,9,12,18,24,36,72} and d=negative{72,36,24,18,12,9,8,6,4,3,2,1}
a•d+b•c=9, so you search for the right combination of those numbers.
The positive value of a•d or b•c will be greater, as 9 is positive.
So I look for a 5•d +(1•c) when d is positive. I could not find any: 5•8 +1•-9=40+-9=31
5•6+1•-12=30+-12=18.
5•4 +1•-18=20+-18=2
Then I look for 5•d +(1•c) when d is negative. 5•-3=-15, 1•24= 9
So I find the answer
(5x+24)(x-3)=5x^2 -15x +24x -72 =0
So now, use both binomials to solve what values of x would give you a 0 in that equation.
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