for number 5:

$\displaystyle y^2+23y-24=0$

ok, you first set everything equal to zero, as i did above, then you look at the last term, and think: are there any factors of -24 that will add up to 23. YES there are

these two factors will be 24 and -1. so, to factor this:

$\displaystyle (y-1)(y+24)$

like so.

you just plug in the factors. Note that this will only work when the coefficient of the first variable is 1 (see how $\displaystyle y^2$ is by itself? this will only work if $\displaystyle y^2$ has no coefficient.

Now you set y-1 equal to zero. So, y-1=0, and y=1. you do the same for y+24. Y+24=0 Y=-24. So, Y=1 and Y=-24.

for number 7: you will use the distributive property.

Problem: $\displaystyle 3x(2x+1)=0.$ Note, if you set 2x+1 equal to zero, and solve, you will get -1/2 this is also a solution.

You will distribute the number outside of the parenthesis to the numbers in the parenthesis. This means you will multiply 3x by 2x and by 1. so you get:

$\displaystyle 6x^2+3x=0$

thus, x=0

x=-1/2