99 !
factor each expression completely.
1. x^2-5x-24
2. 13x^2-52
3. find the equation of a line through (-6, 3) and parallel to 3x-9y=14
4. solve (5x+4)(3x-7)=0; {rational numbers}

Try to factor the greatest common factor out of each then if you have
1x²+bx+c, then find what 2 numbers (let's call them r and w) mutliply to get c and add to get b
it will factor to (x+w)(x+r)

and differnce of 2 perfect squares: a²-b²=(a-b)(a+b)

and paralell lines have same slope

and if xy=0 then assume x and/or y=0

so

1.
what 2 numbers multiply to get -24 and add to get -5
-8 and 3
(x-8)(x+3)

2.
factor out 13 from each
13(x²-4)
differnce of 2 perfect squares
13(x²-2²)=13(x-2)(x+2)

3.
a nice hack is for ax+by=c, a paralell line will be ax+by=z where the values of a and b are the same for both equations

so 3x-9y=14
a paralell line will be
3x-9y=c
find c
subsitute the point (-6,3)
x=-6 and y=3
subsitute
3(-6)-9(3)=c
-18-27=c
-45=c
the equation is 3x-9y=-45

4.
for so since those  2 things multiply to get 0, set them both equal to 0 and solve

5x+4=0
minus 4 both sides
5x=-4
divide by 5 both sides
x=-4/5

3x-7=0
add 7 both sides
3x=7
divide by 3 both sides
x=7/3

solutions are x=-4/5 and 7/3

1. (x-8)(x+3)
2. 13(x-2)(x+2)
3. 3x-9y=-45 (or simplified would be x-3y=-15)
4. x=-4/5 and 7/3

1. (x-8)(x+3)
2. 13(x^2-4)
3. (3/9)x+5=y
4. -0.8 and 2 1/3

Zero is the mid point of the line segment