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)

Idk idk idk idk idk idk idk idk

Aand b are correctly balanced reactions. they have the same number of atoms in both sides of the reaction