Question 408227
here are solutions for these problems...

a)

x^2-7x+12=0

Solution by Factoring using the AC method (Factor by Grouping)
In order to factor x^2-7x+12=0, first multiply the leading coefficient 1 and the last term 12 to get 12. Now we need to ask ourselves: What two numbers multiply to 12 and add to -7? Lets find out by listing all of the possible factors of 12


Factors:

1,2,3,4,6,12,

-1,-2,-3,-4,-6,-12, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 12.

1*12=12

2*6=12

3*4=12

(-1)*(-12)=12

(-2)*(-6)=12

(-3)*(-4)=12

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to -7? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -7

||||||
First Number	|	Second Number	|	Sum
1	|	12	|	1+12=13
2	|	6	|	2+6=8
3	|	4	|	3+4=7
-1	|	-12	|	-1+(-12)=-13
-2	|	-6	|	-2+(-6)=-8
-3	|	-4	|	-3+(-4)=-7


We can see from the table that -3 and -4 add to -7. So the two numbers that multiply to 12 and add to -7 are: -3 and -4

So the original quadratic


x^2-7x+12=0


breaks down to this (just replace -7x with the two numbers that multiply to 12 and add to -7, which are: -3 and -4)


x^2-3x-4x + 12 Replace -7x with -3x-4x

Group the first two terms together and the last two terms together like this:

(x^2-3x) +(-4x + 12)

Factor a 1x out of the first group and factor a -4 out of the second group.


x(x-3) -4(-x - 3)


Now since we have a common term x-3 we can combine the two terms.


(x -4)(-x - 3) Combine like terms.
==============================================================================

Answer:


So the quadratic x^2-7x+12=0 factors to (x -4)(-x - 3) 




Notice how  (x -4)(-x - 3)  foils back to our original problem x^2-7x+12=0. This verifies our answer. 



b)
x^2-10x+16=0
Solution by Factoring using the AC method (Factor by Grouping)
In order to factor x^2-10x+16=0, first multiply the leading coefficient 1 and the last term 16 to get 16. Now we need to ask ourselves: What two numbers multiply to 16 and add to -10? Lets find out by listing all of the possible factors of 16


Factors:

1,2,4,8,16,

-1,-2,-4,-8,-16, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 16.

1*16=16

2*8=16

4*4=16

(-1)*(-16)=16

(-2)*(-8)=16

(-4)*(-4)=16

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to -10? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -10

||||||
First Number	|	Second Number	|	Sum
1	|	16	|	1+16=17
2	|	8	|	2+8=10
4	|	4	|	4+4=8
-1	|	-16	|	-1+(-16)=-17
-2	|	-8	|	-2+(-8)=-10
-4	|	-4	|	-4+(-4)=-8


We can see from the table that -2 and -8 add to -10. So the two numbers that multiply to 16 and add to -10 are: -2 and -8

So the original quadratic


x^2-10x+16=0


breaks down to this (just replace -10%2Ax with the two numbers that multiply to 16 and add to -10, which are: -2 and -8)


x^2-2x-8x+16=0 Replace -10x with -2x-8x

Group the first two terms together and the last two terms together like this:

(x^2-2x)+(-8x+16)=0

Factor a 1x out of the first group and factor a -8 out of the second group.


x(x-2)-8(x-2)=0


Now since we have a common term x-2 we can combine the two terms.


(x-8)(x-2)=0 Combine like terms.
==============================================================================

Answer:


So the quadratic x^2-2x-8x+16=0 factors to (x-8)(x-2)




Notice how (x-8)(x-2) foils back to our original problem x^2-2x-8x+16=0. This verifies our answer. 

c)

x^2+2x-15=0

Solution by Factoring using the AC method (Factor by Grouping)
In order to factor x^2+2x-15=0 , first multiply the leading coefficient 1 and the last term -15 to get -15. Now we need to ask ourselves: What two numbers multiply to -15 and add to 2? Lets find out by listing all of the possible factors of -15


Factors:

1,3,5,15,

-1,-3,-5,-15, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -15.

(-1)*(15)=-15

(-3)*(5)=-15

Now which of these pairs add to 2? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 2

||||
First Number	|	Second Number	|	Sum
1	|	-15	|	1+(-15)=-14
3	|	-5	|	3+(-5)=-2
-1	|	15	|	(-1)+15=14
-3	|	5	|	(-3)+5=2


We can see from the table that -3 and 5 add to 2. So the two numbers that multiply to -15 and add to 2 are: -3 and 5

So the original quadratic


x^2+2x-15=0


breaks down to this (just replace 2x with the two numbers that multiply to -15 and add to 2, which are: -3 and 5)


x^2-3x+5x-15=0 Replace 2x with -3x+5x

Group the first two terms together and the last two terms together like this:

(x^2-3x)+(5x-15)=0 

Factor a 1x out of the first group and factor a 5 out of the second group.


x(x-3)+5(x-3)=0 


Now since we have a common term x-3 we can combine the two terms.


(x+5)(x-3)=0  Combine like terms.
==============================================================================

Answer:


So the quadratic x^2+2x-15=0 factors to (x+5)(x-3)




Notice how (x+5)(x-3) foils back to our original problem x^2+2x-15=0. This verifies our answer. 

d)

x^2- 4x-21=0
Solution by Factoring using the AC method (Factor by Grouping)
In order to factor x^2- 4x-21=0, first multiply the leading coefficient 1 and the last term -21 to get -21. Now we need to ask ourselves: What two numbers multiply to -21 and add to -4? Lets find out by listing all of the possible factors of -21


Factors:

1,3,7,21,

-1,-3,-7,-21, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -21.

(-1)*(21)=-21

(-3)*(7)=-21

Now which of these pairs add to -4? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -4

||||
First Number	|	Second Number	|	Sum
1	|	-21	|	1+(-21)=-20
3	|	-7	|	3+(-7)=-4
-1	|	21	|	(-1)+21=20
-3	|	7	|	(-3)+7=4


We can see from the table that 3 and -7 add to -4. So the two numbers that multiply to -21 and add to -4 are: 3 and -7

So the original quadratic


x^2- 4x-21=0


breaks down to this (just replace -4x with the two numbers that multiply to -21 and add to -4, which are: 3 and -7)


x^2+3x-7x--21=0 .... Replace -4x with 3x-7x

Group the first two terms together and the last two terms together like this:

(x^2+3x)+(-7x-21)

Factor a 1x out of the first group and factor a -7 out of the second group.

x(x-3)-7(x+3)


Now since we have a common term x+3 we can combine the two terms.

(x-7)(x+3)........ Combine like terms.
==============================================================================

Answer:


So the quadratic x^2- 4x-21=0 factors to (x-7)(x+3)




Notice how (x-7)(x+3) foils back to our original problem x^2- 4x-21=0. This verifies our answer. 


e)

x^2-5x+6=0

Solution by Factoring using the AC method (Factor by Grouping)
In order to factor x^2-5x+6=0, first multiply the leading coefficient 1 and the last term 6 to get 6. Now we need to ask ourselves: What two numbers multiply to 6 and add to -5? Lets find out by listing all of the possible factors of 6


Factors:

1,2,3,6,

-1,-2,-3,-6, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 6.

1*6=6

2*3=6

(-1)*(-6)=6

(-2)*(-3)=6

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to -5? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -5

||||
First Number	|	Second Number	|	Sum
1	|	6	|	1+6=7
2	|	3	|	2+3=5
-1	|	-6	|	-1+(-6)=-7
-2	|	-3	|	-2+(-3)=-5


We can see from the table that -2 and -3 add to -5. So the two numbers that multiply to 6 and add to -5 are: -2 and -3

So the original quadratic


x^2-5x+6=0


breaks down to this (just replace -5x with the two numbers that multiply to 6 and add to -5, which are: -2 and -3)


x^2-2x-3x+6=0 Replace -5x with -2x-3x

Group the first two terms together and the last two terms together like this:
(x^2-2x)+(-3x+6)=0

Factor a 1x out of the first group and factor a -3 out of the second group.


x(x-2)+3(-x+2)=0


Now since we have a common term x-2 we can combine the two terms.


(x+3)(-x+2)=0 Combine like terms.
==============================================================================

Answer:


So the quadratic x^2-5x+6=0 factors (x+3)(-x+2)=0 



f)
x^2+19x+18=0

Solution by Factoring using the AC method (Factor by Grouping)
In order to factor x^2+19x+18=0, first multiply the leading coefficient 1 and the last term 18 to get 18. Now we need to ask ourselves: What two numbers multiply to 18 and add to 19? Lets find out by listing all of the possible factors of 18


Factors:

1,2,3,6,9,18,

-1,-2,-3,-6,-9,-18, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 18.

1*18=18

2*9=18

3*6=18

(-1)*(-18)=18

(-2)*(-9)=18

(-3)*(-6)=18

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to 19? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 19

||||||
First Number	|	Second Number	|	Sum
1	|	18	|	1+18=19
2	|	9	|	2+9=11
3	|	6	|	3+6=9
-1	|	-18	|	-1+(-18)=-19
-2	|	-9	|	-2+(-9)=-11
-3	|	-6	|	-3+(-6)=-9


We can see from the table that 1 and 18 add to 19. So the two numbers that multiply to 18 and add to 19 are: 1 and 18

So the original quadratic


x^2+19x+18=0


breaks down to this (just replace 19%2Ax with the two numbers that multiply to 18 and add to 19, which are: 1 and 18)

x^2+1x +18x+18=0 Replace 19x with 1x+ 18x

Group the first two terms together and the last two terms together like this:

x^2+1x +18x+18=0

Factor a 1x out of the first group and factor a 18 out of the second group.

x(x+1) +18(x+1)=0


Now since we have a common term x+1 we can combine the two terms.


(x +18)(x+1)=0Combine like terms.
==============================================================================

Answer:


So the quadratic x^2+19x+18=0 factors to (x +18)(x+1)=0


g)
x^2-17x+72=0

Solution by Factoring using the AC method (Factor by Grouping)
In order to factor x^2-17x+72=0, first multiply the leading coefficient 1 and the last term 72 to get 72. Now we need to ask ourselves: What two numbers multiply to 72 and add to -17? Lets find out by listing all of the possible factors of 72


Factors:

1,2,3,4,6,8,9,12,18,24,36,72,

-1,-2,-3,-4,-6,-8,-9,-12,-18,-24,-36,-72, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 72.

1*72=72

2*36=72

3*24=72

4*18=72

6*12=72

8*9=72

(-1)*(-72)=72

(-2)*(-36)=72

(-3)*(-24)=72

(-4)*(-18)=72

(-6)*(-12)=72

(-8)*(-9)=72

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to -17? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -17

||||||||||||
First Number	|	Second Number	|	Sum
1	|	72	|	1+72=73
2	|	36	|	2+36=38
3	|	24	|	3+24=27
4	|	18	|	4+18=22
6	|	12	|	6+12=18
8	|	9	|	8+9=17
-1	|	-72	|	-1+(-72)=-73
-2	|	-36	|	-2+(-36)=-38
-3	|	-24	|	-3+(-24)=-27
-4	|	-18	|	-4+(-18)=-22
-6	|	-12	|	-6+(-12)=-18
-8	|	-9	|	-8+(-9)=-17


We can see from the table that -8 and -9 add to -17. So the two numbers that multiply to 72 and add to -17 are: -8 and -9

So the original quadratic


x^2-17x+72=0


breaks down to this (just replace -17x with the two numbers that multiply to 72 and add to -17, which are: -8 and -9)


x^2-8x - 9x+72=0 Replace -17x with -8x-9x

Group the first two terms together and the last two terms together like this:

(x^2-8x)+( - 9x+72)=0 

Factor a 1x out of the first group and factor a -9 out of the second group.


x(x-8)-9( x-8)=0 


Now since we have a common term x-8 we can combine the two terms.

(x-9)( x-8)=0  Combine like terms.
==============================================================================

Answer:


So the quadratic x^2-17x+72=0 factors to (x-9)( x-8)=0



h)

x^2+5x=0

Solution by Factoring using the AC method (Factor by Grouping)
Notice how each term in the expression x^2+5x=0

has a common factor of x. We can simply factor out an x like this:
x(x+5)=0


i)

x^2+8x+7=0

Solution by Factoring using the AC method (Factor by Grouping)
In order to factor x^2+8x+7=0, first multiply the leading coefficient 1 and the last term 7 to get 7. Now we need to ask ourselves: What two numbers multiply to 7 and add to 8? Lets find out by listing all of the possible factors of 7


Factors:

1,7,

-1,-7, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 7.

1*7=7

(-1)*(-7)=7

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to 8? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 8

||
First Number	|	Second Number	|	Sum
1	|	7	|	1+7=8
-1	|	-7	|	-1+(-7)=-8


We can see from the table that 1 and 7 add to 8. So the two numbers that multiply to 7 and add to 8 are: 1 and 7

So the original quadratic


x^2+8x+7=0


breaks down to this (just replace 8x with the two numbers that multiply to 7 and add to 8, which are: 1 and 7)


x^2+x+7x+7=0 Replace 8x with x+7x

Group the first two terms together and the last two terms together like this:

(x^2+x)+(7x+7)=0 

Factor a 1x out of the first group and factor a 7 out of the second group.


x(x+1)+7(x+1)=0


Now since we have a common term x+1 we can combine the two terms.


(x+7)(x+1)=0Combine like terms.
==============================================================================

Answer:


So the quadratic x^2+8x+7=0 factors to (x+7)(x+1)=0

j)

3x^2+6x=0

Solution by Factoring using the AC method (Factor by Grouping)
Notice how each term in the expression 3x^2+6x=0

has a common factor of 3x. We can simply factor out an 3x like this:
3x(x+2)=0


k)

2x^2=32-12x.....move all terms to the left

2x^2+12x-32=0.........divide each term by 2

x^2+6x-16=0

Solution by Factoring using the AC method (Factor by Grouping)
In order to factor x^2+6x-16=0, first multiply the leading coefficient 1 and the last term 16 to get 16. Now we need to ask ourselves: What two numbers multiply to 16 and add to 6? Lets find out by listing all of the possible factors of 16


Factors:

1,2,4,8,16,

-1,-2,-4,-8,-16, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 16.

1*16=16

2*8=16

4*4=16

(-1)*(-16)=16

(-2)*(-8)=16

(-4)*(-4)=16

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to 6? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 6

||||||
First Number	|	Second Number	|	Sum
1	|	16	|	1+16=17
2	|	8	|	2+8=10
4	|	4	|	4+4=8
-1	|	-16	|	-1+(-16)=-17
-2	|	-8	|	-2+(-8)=-10
-4	|	-4	|	-4+(-4)=-8


None of these factors add to 6. So the quadratic x^2+6x-16=0 cannot be factored. 


l)

3x^2+7x-6=0
Solution by Factoring using the AC method (Factor by Grouping)
In order to factor 3x^2+7x-6=0, first multiply the leading coefficient 3 and the last term -6 to get -18. Now we need to ask ourselves: What two numbers multiply to -18 and add to 7? Lets find out by listing all of the possible factors of -18


Factors:

1,2,3,6,9,18,

-1,-2,-3,-6,-9,-18, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -18.

(-1)*(18)=-18

(-2)*(9)=-18

(-3)*(6)=-18

Now which of these pairs add to 7? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 7

||||||
First Number	|	Second Number	|	Sum
1	|	-18	|	1+(-18)=-17
2	|	-9	|	2+(-9)=-7
3	|	-6	|	3+(-6)=-3
-1	|	18	|	(-1)+18=17
-2	|	9	|	(-2)+9=7
-3	|	6	|	(-3)+6=3


We can see from the table that -2 and 9 add to 7. So the two numbers that multiply to -18 and add to 7 are: -2 and 9

So the original quadratic


3x^2+7x-6=0


breaks down to this (just replace 7x with the two numbers that multiply to -18 and add to 7, which are: -2 and 9)


3x^2-2x+9x-6=0... Replace 7x with -2x+9x

Group the first two terms together and the last two terms together like this:

(3x^2-2x)+(9x-6)=0

Factor a 1x out of the first group and factor a 3 out of the second group.


x(3x-2)+3(3x-2)=0


Now since we have a common term 3x-2 we can combine the two terms.


(x+3)(3x-2)=0.......... Combine like terms.
==============================================================================

Answer:


So the quadratic 3x^2+7x-6=0 factors to (x+3)(3x-2)=0.




m)

x^2=10x-21............move all terms to the left

x^2-10x+21=0

Solution by Factoring using the AC method (Factor by Grouping)
In order to factor x^2-10x+21=0, first multiply the leading coefficient 1 and the last term 21 to get 21. Now we need to ask ourselves: What two numbers multiply to 21 and add to -10? Lets find out by listing all of the possible factors of 21


Factors:

1,3,7,21,

-1,-3,-7,-21, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 21.

1*21=21

3*7=21

(-1)*(-21)=21

(-3)*(-7)=21

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to -10? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -10

||||
First Number	|	Second Number	|	Sum
1	|	21	|	1+21=22
3	|	7	|	3+7=10
-1	|	-21	|	-1+(-21)=-22
-3	|	-7	|	-3+(-7)=-10


We can see from the table that -3 and -7 add to -10. So the two numbers that multiply to 21 and add to -10 are: -3 and -7

So the original quadratic


x^2-10x+21=0


breaks down to this (just replace -10%2Ax with the two numbers that multiply to 21 and add to -10, which are: -3 and -7)


x^2-3x-7x+21=0 Replace -10x with -3x-7x

Group the first two terms together and the last two terms together like this:

(x^2-3x)+(-7x+21)=0

Factor a 1x out of the first group and factor a -7 out of the second group.


x(x-3)-7(x-3)=0


Now since we have a common term x-3 we can combine the two terms.


(x-7)(x-3)=0.......Combine like terms.
==============================================================================

Answer:


So the quadratic x^2-10x+21=0 factors to (x-7)(x-3)=0

n)

x^2-x-56=0

Solution by Factoring using the AC method (Factor by Grouping)
In order to factor x^2-x-56=0, first multiply the leading coefficient 1 and the last term -56 to get -56. Now we need to ask ourselves: What two numbers multiply to -56 and add to -1? Lets find out by listing all of the possible factors of -56


Factors:

1,2,4,7,8,14,28,56,

-1,-2,-4,-7,-8,-14,-28,-56, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -56.

(-1)*(56)=-56

(-2)*(28)=-56

(-4)*(14)=-56

(-7)*(8)=-56

Now which of these pairs add to -1? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -1

||||||||
First Number	|	Second Number	|	Sum
1	|	-56	|	1+(-56)=-55
2	|	-28	|	2+(-28)=-26
4	|	-14	|	4+(-14)=-10
7	|	-8	|	7+(-8)=-1
-1	|	56	|	(-1)+56=55
-2	|	28	|	(-2)+28=26
-4	|	14	|	(-4)+14=10
-7	|	8	|	(-7)+8=1


We can see from the table that 7 and -8 add to -1. So the two numbers that multiply to -56 and add to -1 are: 7 and -8

So the original quadratic


x^2-x-56=0


breaks down to this (just replace -1x with the two numbers that multiply to -56 and add to -1, which are: 7 and -8)


x^2+7x-8x-56=0 Replace -1x with 7x-8x

Group the first two terms together and the last two terms together like this:

(x^2+7x)+(-8x-56)=0

Factor a 1x out of the first group and factor a -8 out of the second group.


x(x+7)-8(x+7)=0


Now since we have a common term x+7 we can combine the two terms.


(x-8)(x+7)=0..... Combine like terms.
==============================================================================

Answer:


So the quadratic x^2-x-56=0 factors to (x-8)(x+7)=0


o)

3x^2+2x-5=0

Solution by Factoring using the AC method (Factor by Grouping)
In order to factor 3x^2+2x-5=0, first multiply the leading coefficient 3 and the last term -5 to get -15. Now we need to ask ourselves: What two numbers multiply to -15 and add to 2? Lets find out by listing all of the possible factors of -15


Factors:

1,3,5,15,

-1,-3,-5,-15, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -15.

(-1)*(15)=-15

(-3)*(5)=-15

Now which of these pairs add to 2? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 2

||||
First Number	|	Second Number	|	Sum
1	|	-15	|	1+(-15)=-14
3	|	-5	|	3+(-5)=-2
-1	|	15	|	(-1)+15=14
-3	|	5	|	(-3)+5=2


We can see from the table that -3 and 5 add to 2. So the two numbers that multiply to -15 and add to 2 are: -3 and 5

So the original quadratic

3x^2+2x-5=0


breaks down to this (just replace 2x with the two numbers that multiply to -15 and add to 2, which are: -3 and 5)


3x^2-3x+5x-5=0.... Replace 2x with -3x+5x

Group the first two terms together and the last two terms together like this:
(3x^2-3x)+(5x-5)=0.

Factor a 3x out of the first group and factor a 5 out of the second group.


3x(x-1)+5(x-1)=0.


Now since we have a common term x-1 we can combine the two terms.


(3x+5)(x-1)=0. ..Combine like terms.
==============================================================================

Answer:


So the quadratic 3x^2+2x-5=0 factors to (3x+5)(x-1)=0


p)

6x^2=6-5x...move all terms to the left

6x^2+5x-6=0
Solution by Factoring using the AC method (Factor by Grouping)
In order to factor 6x^2+5x-6=0, first multiply the leading coefficient 6 and the last term -6 to get -36. Now we need to ask ourselves: What two numbers multiply to -36 and add to 5? Lets find out by listing all of the possible factors of -36


Factors:

1,2,3,4,6,9,12,18,36,

-1,-2,-3,-4,-6,-9,-12,-18,-36, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -36.

(-1)*(36)=-36

(-2)*(18)=-36

(-3)*(12)=-36

(-4)*(9)=-36

(-6)*(6)=-36

Now which of these pairs add to 5? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 5

||||||||||
First Number	|	Second Number	|	Sum
1	|	-36	|	1+(-36)=-35
2	|	-18	|	2+(-18)=-16
3	|	-12	|	3+(-12)=-9
4	|	-9	|	4+(-9)=-5
6	|	-6	|	6+(-6)=0
-1	|	36	|	(-1)+36=35
-2	|	18	|	(-2)+18=16
-3	|	12	|	(-3)+12=9
-4	|	9	|	(-4)+9=5
-6	|	6	|	(-6)+6=0


We can see from the table that -4 and 9 add to 5. So the two numbers that multiply to -36 and add to 5 are: -4 and 9

So the original quadratic

6x^2+5x-6=0


breaks down to this (just replace 5%2Ax with the two numbers that multiply to -36 and add to 5, which are: -4 and 9)


6x^2-4x+9x-6=0 Replace 5x with -4x+9x

Group the first two terms together and the last two terms together like this:

(6x^2-4x)+(9x-6)=0

Factor a 2x out of the first group and factor a 3 out of the second group.

2x(3x-2)+3(3x-2)=0


Now since we have a common term 3x-2 we can combine the two terms.

(2x+3)(3x-2)=0..........Combine like terms.
==============================================================================

Answer:


So the quadratic 6x^2+5x-6=0 factors to (2x+3)(3x-2)=0


q)

4x^2+19x=5
4x^2+19x-5=0

Solution by Factoring using the AC method (Factor by Grouping)
In order to factor 4x^2+19x-5=0, first multiply the leading coefficient 4 and the last term -5 to get -20. Now we need to ask ourselves: What two numbers multiply to -20 and add to 19? Lets find out by listing all of the possible factors of -20


Factors:

1,2,4,5,10,20,

-1,-2,-4,-5,-10,-20, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -20.

(-1)*(20)=-20

(-2)*(10)=-20

(-4)*(5)=-20

Now which of these pairs add to 19? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 19

||||||
First Number	|	Second Number	|	Sum
1	|	-20	|	1+(-20)=-19
2	|	-10	|	2+(-10)=-8
4	|	-5	|	4+(-5)=-1
-1	|	20	|	(-1)+20=19
-2	|	10	|	(-2)+10=8
-4	|	5	|	(-4)+5=1


We can see from the table that -1 and 20 add to 19. So the two numbers that multiply to -20 and add to 19 are: -1 and 20

So the original quadratic


4x^2+19x-5=0


breaks down to this (just replace 19x with the two numbers that multiply to -20 and add to 19, which are: -1 and 20)


4x^2-x+20x-5=0 ..........Replace 19x with -x+20x

Group the first two terms together and the last two terms together like this:

(4x^2-x)+(20x-5)=0 

Factor a 1x out of the first group and factor a 5 out of the second group.


x(4x-1)+5(4x-1)=0 


Now since we have a common term 4x-1 we can combine the two terms.


(x+5)(4x-1)=0 .......... Combine like terms.
==============================================================================

Answer:


So the quadratic 4x^2+19x-5=0 factors to (x+5)(4x-1)=0 

r)

6x^2-24=0

6(x^2-4)=0............6 is not equal to zero, so (x^2-4)=0

6(x^2-2^2)=0


6(x-2)(x+2)=0

So the quadratic 6x^2-24=0 factors to 6(x-2)(x+2)=0



s)

4x^2-9=0

Solution by Factoring using the AC method (Factor by Grouping)
In order to factor 4x^2-9=0, first multiply the leading coefficient 4 and the last term -9 to get -36. Now we need to ask ourselves: What two numbers multiply to -36 and add to 0? Lets find out by listing all of the possible factors of -36


Factors:

1,2,3,4,6,9,12,18,36,

-1,-2,-3,-4,-6,-9,-12,-18,-36, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -36.

(-1)*(36)=-36

(-2)*(18)=-36

(-3)*(12)=-36

(-4)*(9)=-36

(-6)*(6)=-36

Now which of these pairs add to 0? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 0

||||||||||
First Number	|	Second Number	|	Sum
1	|	-36	|	1+(-36)=-35
2	|	-18	|	2+(-18)=-16
3	|	-12	|	3+(-12)=-9
4	|	-9	|	4+(-9)=-5
6	|	-6	|	6+(-6)=0
-1	|	36	|	(-1)+36=35
-2	|	18	|	(-2)+18=16
-3	|	12	|	(-3)+12=9
-4	|	9	|	(-4)+9=5
-6	|	6	|	(-6)+6=0


We can see from the table that 6 and -6 add to 0. So the two numbers that multiply to -36 and add to 0 are: 6 and -6

So the original quadratic


4x^2-9=0


breaks down to this (just replace 0x with the two numbers that multiply to -36 and add to 0, which are: 6 and -6)


4x^2+ 6x-6x -9=0 ............Replace 0x with 6x-6x

Group the first two terms together and the last two terms together like this:

(4x^2+ 6x)+(-6x -9)=0 

Factor a 2x out of the first group and factor a -3 out of the second group.


2x(2x+ 3)-3(2x +3)=0 


Now since we have a common term 2x+3 we can combine the two terms.


(2x-3)(2x +3)=0 ... Combine like terms.
==============================================================================

Answer:


So the quadratic 4x^2-9=0 factors to (2x-3)(2x +3)=0 


t)

2x^2=13x-20

2x^2-13x+20=0

Solution by Factoring using the AC method (Factor by Grouping)
In order to factor 2x^2-13x+20=0, first multiply the leading coefficient 2 and the last term 20 to get 40. Now we need to ask ourselves: What two numbers multiply to 40 and add to -13? Lets find out by listing all of the possible factors of 40


Factors:

1,2,4,5,8,10,20,40,

-1,-2,-4,-5,-8,-10,-20,-40, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 40.

1*40=40

2*20=40

4*10=40

5*8=40

(-1)*(-40)=40

(-2)*(-20)=40

(-4)*(-10)=40

(-5)*(-8)=40

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to -13? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -13

||||||||
First Number	|	Second Number	|	Sum
1	|	40	|	1+40=41
2	|	20	|	2+20=22
4	|	10	|	4+10=14
5	|	8	|	5+8=13
-1	|	-40	|	-1+(-40)=-41
-2	|	-20	|	-2+(-20)=-22
-4	|	-10	|	-4+(-10)=-14
-5	|	-8	|	-5+(-8)=-13


We can see from the table that -5 and -8 add to -13. So the two numbers that multiply to 40 and add to -13 are: -5 and -8

So the original quadratic


2x^2-13x+20=0


breaks down to this (just replace -13x with the two numbers that multiply to 40 and add to -13, which are: -5 and -8)


2x^2-5x-8x+20=0........... Replace -13x with -5x-8x

Group the first two terms together and the last two terms together like this:

(2x^2-5x)+(-8x+20)=0

Factor a 1x out of the first group and factor a -4 out of the second group.


x(2x-5)-4(2x+5)=0


Now since we have a common term 2x-5 we can combine the two terms.


(x-4)(2x+5)=0.... Combine like terms.
==============================================================================

Answer:


So the quadratic 2x^2-13x+20=0 factors to (x-4)(2x+5)=0


u)

12x^2+7x=12
12x^2+7x-12=0


Solution by Factoring using the AC method (Factor by Grouping)
In order to factor 12x^2+7x-12=0, first multiply the leading coefficient 12 and the last term -12 to get -144. Now we need to ask ourselves: What two numbers multiply to -144 and add to 7? Lets find out by listing all of the possible factors of -144


Factors:

1,2,3,4,6,8,9,12,16,18,24,36,48,72,144,

-1,-2,-3,-4,-6,-8,-9,-12,-16,-18,-24,-36,-48,-72,-144, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -144.

(-1)*(144)=-144

(-2)*(72)=-144

(-3)*(48)=-144

(-4)*(36)=-144

(-6)*(24)=-144

(-8)*(18)=-144

(-9)*(16)=-144

(-12)*(12)=-144

Now which of these pairs add to 7? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 7

||||||||||||||||
First Number	|	Second Number	|	Sum
1	|	-144	|	1+(-144)=-143
2	|	-72	|	2+(-72)=-70
3	|	-48	|	3+(-48)=-45
4	|	-36	|	4+(-36)=-32
6	|	-24	|	6+(-24)=-18
8	|	-18	|	8+(-18)=-10
9	|	-16	|	9+(-16)=-7
12	|	-12	|	12+(-12)=0
-1	|	144	|	(-1)+144=143
-2	|	72	|	(-2)+72=70
-3	|	48	|	(-3)+48=45
-4	|	36	|	(-4)+36=32
-6	|	24	|	(-6)+24=18
-8	|	18	|	(-8)+18=10
-9	|	16	|	(-9)+16=7
-12	|	12	|	(-12)+12=0


We can see from the table that -9 and 16 add to 7. So the two numbers that multiply to -144 and add to 7 are: -9 and 16

So the original quadratic

12x^2+7x-12=0


breaks down to this (just replace 7%2Ax with the two numbers that multiply to -144 and add to 7, which are: -9 and 16)


12x^2-9x+16x-12=0... Replace 7x with -9x+16x

Group the first two terms together and the last two terms together like this:

(12x^2-9x)+(16x-12)=0

Factor a 3x out of the first group and factor a 4 out of the second group.


3x(4x-3)+4(4x-3)=0


Now since we have a common term 4x-3 we can combine the two terms.

(3x+4)(4x-3)=0........ Combine like terms.
==============================================================================

Answer:


So the quadratic 12x^2+7x-12=0 factors to (3x+4)(4x-3)=0




v)

6x^2+29x-5=0

In order to factor 6x^2+29x-5=0, first multiply the leading coefficient 6 and the last term -5 to get -30. Now we need to ask ourselves: What two numbers multiply to -30 and add to 29? Lets find out by listing all of the possible factors of -30


Factors:

1,2,3,5,6,10,15,30,

-1,-2,-3,-5,-6,-10,-15,-30, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -30.

(-1)*(30)=-30

(-2)*(15)=-30

(-3)*(10)=-30

(-5)*(6)=-30

Now which of these pairs add to 29? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 29

||||||||
First Number	|	Second Number	|	Sum
1	|	-30	|	1+(-30)=-29
2	|	-15	|	2+(-15)=-13
3	|	-10	|	3+(-10)=-7
5	|	-6	|	5+(-6)=-1
-1	|	30	|	(-1)+30=29
-2	|	15	|	(-2)+15=13
-3	|	10	|	(-3)+10=7
-5	|	6	|	(-5)+6=1


We can see from the table that -1 and 30 add to 29. So the two numbers that multiply to -30 and add to 29 are: -1 and 30

So the original quadratic


6x^2+29x-5=0


breaks down to this (just replace 29x with the two numbers that multiply to -30 and add to 29, which are: -1 and 30)


6x^2-x+30x-5=0..... Replace 29x with -x+30x

Group the first two terms together and the last two terms together like this:

(6x^2-x)+(30x-5)=0

Factor a 1x out of the first group and factor a 5 out of the second group.


x(6x-1)+5(6x-1)=0


Now since we have a common term 6x-1 we can combine the two terms.


(x+5)(6x-1)=0.........Combine like terms.
==============================================================================

Answer:


So the quadratic 6x^2+29x-5=0 factors to (x+5)(6x-1)=0



w)

(x-5)^2=36..........take a square root

sqrt((x-5)^2)=sqrt(36)

x-5=6

x-5-6=0

x-11=0.........



x)

(x+2)(x+3)=2

(x+2)(x+3)-2=0

x^2+3x+2x+6-2=0

x^2+5x+4=0

In order to factor x^2+5x+4=0, first multiply the leading coefficient 1 and the last term 4 to get 4. Now we need to ask ourselves: What two numbers multiply to 4 and add to 5? Lets find out by listing all of the possible factors of 4


Factors:

1,2,4,

-1,-2,-4, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 4.

1*4=4

2*2=4

(-1)*(-4)=4

(-2)*(-2)=4

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to 5? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 5

||||
First Number	|	Second Number	|	Sum
1	|	4	|	1+4=5
2	|	2	|	2+2=4
-1	|	-4	|	-1+(-4)=-5
-2	|	-2	|	-2+(-2)=-4


We can see from the table that 1 and 4 add to 5. So the two numbers that multiply to 4 and add to 5 are: 1 and 4

So the original quadratic


x^2+5x+4=0


breaks down to this (just replace 5x with the two numbers that multiply to 4 and add to 5, which are: 1 and 4)

x^2+x+4x+4=0 ..........Replace 5x with x+4x

Group the first two terms together and the last two terms together like this:

(x^2+x)+(4x+4)=0

Factor a 1x out of the first group and factor a 4 out of the second group.


x(x+1)+4(x+1)=0


Now since we have a common term x+1 we can combine the two terms.


(x+4)(x+1)=0 ..........Combine like terms.
==============================================================================

Answer:


So the quadratic x^2+5x+4=0 factors to (x+4)(x+1)=0