7.4 factoring with a leading coefficent greater then one
solve all problems below with work shown

2) 6x²+20x+16

3) 5x²+21x+4

4) -7x²+59x-24

5) 3x²-14x+15

6) 2x²+11x-13

7) 3x²+2x-16

8) 2x²+17x-20

9) 11x²+25x-24

We can only solve it is equal to something
if we assume that it eausl zero, then if we factor it we can assume the factors eqal zero
xy=0, then assume x and y=0

so what we do is if we have
ax^2+bx+c
we factor a and take the factors closest together
then place them with the x
then factor c and chose the closest ones
then try them and see if it works

so if we had
6x^2+20x+16
6=2 times 3
(2x+ )(3x+ )
we know that they are both + because the middle terms is positive and the end terms is positive
16=4 times 4
(2x+4)(3x+4)=6x^2+12x+8x+16=6x^2+20x+16

th easy way to check is we know that teh end terms will be right since we made them that way
to check the middle terms
if we have
(ax+b)(cx+d)
to check what the middle term is we do
bcx+dax
and see if that equals the middle term

so
2. 6x^2+20x+16
6=2 times 3
(2x+ )(3x+ )
we know that they are both + because the middle terms is positive and the end terms is positive
16=4 times 4
(2x+4)(3x+4)=6x^2+12x+8x+16=6x^2+20x+16
facored is (2x+4)(3x+4)=0
2x+4=0
x=-2
3x+4=0
x=-4/3

3. 5x^2+21x+4
5=1 times 5 prime is nice since we only have 1 option
(x+ )(5x+ )
we know they are both pluss since middle and end term is positive
4=2 times 2
(x+2)(5x+2)
(2 times 5x)+(x times 2)=10x+2x=12x, we need 21 so this wrong
change end term
(x+4)(5x+1)
(4 times 5x)+(1 times x)=20x+x=21x correct
factore form
(x+4)(5x+1)=0
x+4=0
x=-4
5x+1=0
x=-1/5

4.
-7x^2+59x-24
-7=-1 and 7 or -7 and 1
we pick -1 and 7 since we want the middle term to be positive
(7x- )(-x+)
signs are differenc since end term is negaive
-24=8 and -3
(7x-3)(-x+8)
(7x times 8)+(-3 times -x)=56x+3x=59x
correct
(7x-3)(-x+8)=0
7x-3=0
x=3/7
-x+8=0
x=8

OK, so I am getting tired of this so I will just do the rest in factored form isnce you probably shoud have figured out how to do it from the previous examples

5. (x-3)(3x-5)
6. (x-1)(2x+13)
7. (x-2)(3x+8)
8. 2x^2+17x-20 cannot be factored but 2x^2+18x-20 can maybe typing mistake
9. (x+3)(11x-8)

answers:
2.(2x+4)(3x+4)
3.(x+4)(5x+1)
4.(7x-3)(-x+8)
5. (x-3)(3x-5)
6. (x-1)(2x+13)
7. (x-2)(3x+8)
8. 2x^2+17x-20 cannot be factored but 2x^2+18x-20 can maybe typing mistake
9. (x+3)(11x-8)

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.