If we start with x^2-7x+10x2−7x+10, we can factor this by considering what multiplys to 1010 (1*101⋅10) and adds to -7−7. I like to do this via table, like this:
color(white)(.).x1010color(white)(.......) add to:
..................................................
1*10color(white)(00....)=10
-1*-10color(white)()=-10
1*-10color(white)(00)=-9
10*-1color(white)((.))=9
2*5color(white)(.......0.)=7
-2*-5color(white)(ct)=-7
2*-5color(white)(......)=3
5*-2color(white)(......)=3
This might look like a mess (and I admit, it kinda is), but we now we know that the only factors of x^2-7x+10 that multiply to 10 and add to -7 are: -2 and -5. That means that we can factor x^2-7x+10 to (x-2)(x-5). Now, to find the zeroes, we just set each parentheses equal to zero and solve for x, like this:
x-2=0color(white)(..........)x-5=0
x=2color(white)(................)x=5
Now we've got it. x=2 and 5. Nice job!
