Question

Question #802bb

Answer 1

`9p^2 + 15p + 6`

Explanation:

`(3p+3)(3p+2)`

1. Multiply the `3p` and the `3p` for the first term
2. Multiply the `3p` and the `2` for the second term
3. Multiply the `3` and the `3p` for the third term
4. Multiply the `3` and the `2` for the fourth term

This is what it should look like in the end:

`9p^2 + 6p + 9p + 6`

Notice how the second and third terms have the same variable of `x`
Add both of the numbers together so there will only by one number with the `x` variable. It should look like this in the end:

`9p^2 + 15p + 6`

Answer 2

you can use distributive property or this formula
`(x+a)(x+b)=x^2+(a+b)x+ab`
Distributive property is
`(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd`
Here, `a,b,c,d` can be anything, they even can be the same number
So, using this
`(3p)^2+(3+2)xx3p+3xx2`
Notice that `(3p)^2 != 3p^2`
`9p^2+5xx3p+6`
Solve
`9p^2+15p+6`