Question 408227:
Found 2 solutions by MathLover1, richard1234:
Answer by MathLover1(20849)   (Show Source):
You can put this solution on YOUR website! 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
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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.
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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
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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.
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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
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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.
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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
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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.
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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
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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.
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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
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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.
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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
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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.
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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
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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.
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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
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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
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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.
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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
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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.
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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
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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.
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! I recommend not to post this many questions into one post....
All of the questions can be solved using either factoring, directly taking the square root of both sides, or by the quadratic formula. Look in your textbook or online for more examples on how to solve the problem. Problems like these take practice and may take a while to get used to.
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