p2-5p-50= (p-10) (p+5)
x(x+5)
25xy+5yz is 5y(5x+z) when factored
11
Factorise: 5x2 + 2x - 192 = (x - 6)(5x + 32) = 0; whence, x = 6 or -6.4.
p2-5p-50= (p-10) (p+5)
qwertyuiopasdfjk
(5x + 2)(x + 3)
you factor out the common 5so it would be: 5(x + 2)so that when you multiply back out 5 by x you get 5x and 5 by 2 you get 10
x(x+5)
6x^2 + 5x + 1 = (3x +1) (2x + 1)
1
(4x-3)(x+2)
Yes and it is 4(5x-7) when factored
25xy+5yz is 5y(5x+z) when factored
5x2 + 9x + 13 has no real factors.
11