-10
5(x) - 15
You need to find a common factor in this equation, the common factor in this equation is 5. Therfore you put 5 in front of a set of brackets, like this - 5( ).
You then put 5(x-3), because it is 5 times x and 5 x 3 to get the answer in the brackets.
5 ( X - 3 )
10
32
32
55
5(x-3)
10
6x^2 + 5x + 1 = (3x +1) (2x + 1)
1
1.6667
25x2+30x+9 25*9=225 15*15=225 15+15=3025x2+15x+15x+9[5*5xx 3*5x] [3*5x 3*3]5x(5x+3)+3(5x+3)(5x+3)(5x+3)Answer: (5x+3)2
-5
qwertyuiopasdfjk
(5x + 2)(x + 3)
x(x+5)
x2 + 2x - 15 = x2 - 3x + 5x - 15 = x(x - 3) + 5(x - 3) = (x + 5)(x - 3)
6x^2 + 5x + 1 = (3x +1) (2x + 1)
1
(4x-3)(x+2)
(x - 6)(x + 1)
Yes and it is 4(5x-7) when factored
25xy+5yz is 5y(5x+z) when factored
5x2 + 9x + 13 has no real factors.
p2-5p-50= (p-10) (p+5)