
well two times five is ten
and two plus five is 7
so we have it made
(x+2)(x+5) = 0
that works if x+5 = 0 so x = 5
and it works if x+2 = 0 so x = 2posted by Damon

x² + 7 x + 10 = 0
The first term is, x² its coefficient is 1
The middle term is, +7 x its coefficient is 7
The last term, the constantis +10
Multiply the coefficient of the first term by the constant 1 • 10 = 10
Find two factors of 10 whose sum equals the coefficient of the middle term, which is 7
 10 • (  1 ) = 10
 10 + (  1 ) =  10  1 =  11
 5 • (  2 ) = 10
 5 + (  2 ) =  5  2 =  7
 2 • (  5 ) = 10
 2 + (  5 ) =  2  5 =  7
 1 • (  10 ) = 10
 1 + (  10 ) =  1  10 =  11
1 • 10 = 10
1 + 10 = 11
2 • 5 = 10
2 + 5 = 7
That's it 2 and 5
Rewrite the polynomial splitting the middle term using the factors 2 and 5
x² + 2 x + 5 x + 10 = 0
( x² + 2 x ) + ( 5 x + 10 ) = 0
x • ( x + 2 ) + 5 • ( x + 2 ) = 0
( x + 2 ) • x + ( x + 2 ) • 5 = 0
( x + 2 ) • ( x + 5 ) = 0
x + 2 = 0
x =  2
x + 5 = 0
x =  5
The solutions are:
x =  5 and x =  2posted by Bosnian
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