help with some maths?: O?

1)
p(p^2 - 3p)
= p^3 - 3p^2

2)
y^2 + 5y
= y(y + 5)

3)
x^2 - 2x - 15 = 0
x^2 + 3x - 5x - 15 = 0
(x + 3)(x - 5) = 0

x + 3 = 0
x = -3

x - 5 = 0
x = 5

∴ x = -3 , 5

p^3-3p^2

y( y+5)

2x( x+3y)

X^2-2x-15= 0
(x-5)(x+3)=0
x= 5, x= -3

The first one's wrong. It should be:
p^3 - 39^2
*Note, when you multiply p by p, you get p^2

To factorise, take out a common factor. In your case, it's y because both components have y in it:
y(y + 5)

For the next one, it's just like the one above. Find a common factor. Notice both components can be divided by 2 AND also by x so:
2x(x + 6y)

To solve the quadratic, you must know your sum and product rule. You need two numbers that add up to -2 AND also their product must be -15.

The answer is -5 and 3

3 + (-5) = -2
3 * -5 = -15

So the answer is:
(x+5)(x-3) = 0

It's the opposite sign on front of those two numbers that go in the brackets but your answer is still -5 and 3.

1
wrong p^3-3p^2
2
y*(2y+5)
3
2x*(x+3y)
4
3 or -5

yes, thats correct

y^2 + 5y = y(y + 5)

2x^2 + 6xy = 2x(x + 6y)

x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0
x - 5 = 0 and x + 3 = 0
x = 5 and x = -3

It should be p^3 - 3p^2.
y(y+5) is the second.
2x(x+3y) is the third
(x-5)(x+3)=0 So x = 5 or - 3 is the next.

p^3-3p^2


y(y+5)

2x(x+3y)

(x-5)(x+3)=0 x=5 x=-3