ab + 2a + 3b + 6
Answer = 20 Let the number be ab where a is the "tens" value and b is the "units" value. Then a + 4 = 3(a + b) a + 4 = 3a + 3b 4 = 2a + 3b The only condition that satisfies this is when b = 0 then 4 = 2a : a = 2
ab + 4 + a + 4b factors to (a + 4)(b + 1)
(a + b)(b - 2c)
(b - x)(ab - xy)
ab + 2a + 3b + 6
so let's group and then factor ab+2a+3b+6= (ab+2a)+(3b+6)= a(b+2)+3(b+2)= (a+3)(b+2) and that is our final answer! Doctor Chuck
a(b+3)+b(b+3)
This expression can be factored. ab + 3a + b2 + 3b = a(b + 3) + b(b + 3) = (a + b)(b + 3)
Answer = 20 Let the number be ab where a is the "tens" value and b is the "units" value. Then a + 4 = 3(a + b) a + 4 = 3a + 3b 4 = 2a + 3b The only condition that satisfies this is when b = 0 then 4 = 2a : a = 2
2a x 3b = 6ab
ab-2ac+b^2-2bc
You want: abc + ab Factor out the common terms which are "a" and "b" ab ( c + 1 )
(a + b)(b + c)
x2y + axy + abx + a2b Factor by grouping. xy(x + a) + ab(x + a) (xy + ab)(x + a)
ab + 4 + a + 4b factors to (a + 4)(b + 1)
The GCF is a.