Applying an algorithm on $3x^2-2x+12x-8$ [duplicate]

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This question is an exact duplicate of:

• 
  How do you apply this algorithm to $x^2+5x+6$?
  
  1 answer
  
  

There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$)

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$


2. Find the product $ac$, including sign.


3. Find the prime factorization of $ac$ using the factor tree.


4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.


5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.


6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$


7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

I'm currently trying to apply this method on $3x^2-2x+12x-8$. However, this equation does not seem to be similar to the form $ax^2+bx+c$.

polynomials factoring

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asked Aug 2 at 13:50

Maxwell

296

marked as duplicate by Community♦ Aug 2 at 14:41

This question was marked as an exact duplicate of an existing question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not sure but add $-2x$ and $12x$ up to $10x$? Form there you have your wanted form $ax^2+bx+c$.
  – mrtaurho
  Aug 2 at 13:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  They're trying to tell you that $s=-2$ and $t=12$.
  – steven gregory
  Aug 2 at 13:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @stevengregory I already tried to simpifly that.
  – Maxwell
  Aug 2 at 14:00
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

This question is an exact duplicate of:

• 
  How do you apply this algorithm to $x^2+5x+6$?
  
  1 answer
  
  

There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$)

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$


2. Find the product $ac$, including sign.


3. Find the prime factorization of $ac$ using the factor tree.


4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.


5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.


6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$


7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

I'm currently trying to apply this method on $3x^2-2x+12x-8$. However, this equation does not seem to be similar to the form $ax^2+bx+c$.

polynomials factoring

share|cite|improve this question

asked Aug 2 at 13:50

Maxwell

296

marked as duplicate by Community♦ Aug 2 at 14:41

This question was marked as an exact duplicate of an existing question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not sure but add $-2x$ and $12x$ up to $10x$? Form there you have your wanted form $ax^2+bx+c$.
  – mrtaurho
  Aug 2 at 13:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  They're trying to tell you that $s=-2$ and $t=12$.
  – steven gregory
  Aug 2 at 13:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @stevengregory I already tried to simpifly that.
  – Maxwell
  Aug 2 at 14:00
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

This question is an exact duplicate of:

• 
  How do you apply this algorithm to $x^2+5x+6$?
  
  1 answer
  
  

There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$)

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$


2. Find the product $ac$, including sign.


3. Find the prime factorization of $ac$ using the factor tree.


4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.


5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.


6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$


7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

I'm currently trying to apply this method on $3x^2-2x+12x-8$. However, this equation does not seem to be similar to the form $ax^2+bx+c$.

polynomials factoring

share|cite|improve this question

asked Aug 2 at 13:50

Maxwell

296

This question is an exact duplicate of:

• 
  How do you apply this algorithm to $x^2+5x+6$?
  
  1 answer
  
  

There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$)

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$


2. Find the product $ac$, including sign.


3. Find the prime factorization of $ac$ using the factor tree.


4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.


5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.


6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$


7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

I'm currently trying to apply this method on $3x^2-2x+12x-8$. However, this equation does not seem to be similar to the form $ax^2+bx+c$.

This question is an exact duplicate of:

• 
  How do you apply this algorithm to $x^2+5x+6$?
  
  1 answer
  
  

polynomials factoring

share|cite|improve this question

asked Aug 2 at 13:50

Maxwell

296

share|cite|improve this question

share|cite|improve this question

asked Aug 2 at 13:50

Maxwell

296

asked Aug 2 at 13:50

Maxwell

296

asked Aug 2 at 13:50

Maxwell

296

296

marked as duplicate by Community♦ Aug 2 at 14:41

This question was marked as an exact duplicate of an existing question.

marked as duplicate by Community♦ Aug 2 at 14:41

This question was marked as an exact duplicate of an existing question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not sure but add $-2x$ and $12x$ up to $10x$? Form there you have your wanted form $ax^2+bx+c$.
  – mrtaurho
  Aug 2 at 13:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  They're trying to tell you that $s=-2$ and $t=12$.
  – steven gregory
  Aug 2 at 13:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @stevengregory I already tried to simpifly that.
  – Maxwell
  Aug 2 at 14:00
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not sure but add $-2x$ and $12x$ up to $10x$? Form there you have your wanted form $ax^2+bx+c$.
  – mrtaurho
  Aug 2 at 13:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  They're trying to tell you that $s=-2$ and $t=12$.
  – steven gregory
  Aug 2 at 13:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @stevengregory I already tried to simpifly that.
  – Maxwell
  Aug 2 at 14:00
  

  
  
  
  

I am not sure but add $-2x$ and $12x$ up to $10x$? Form there you have your wanted form $ax^2+bx+c$.
– mrtaurho
Aug 2 at 13:51

I am not sure but add $-2x$ and $12x$ up to $10x$? Form there you have your wanted form $ax^2+bx+c$.
– mrtaurho
Aug 2 at 13:51

They're trying to tell you that $s=-2$ and $t=12$.
– steven gregory
Aug 2 at 13:58

They're trying to tell you that $s=-2$ and $t=12$.
– steven gregory
Aug 2 at 13:58

@stevengregory I already tried to simpifly that.
– Maxwell
Aug 2 at 14:00

@stevengregory I already tried to simpifly that.
– Maxwell
Aug 2 at 14:00

add a comment | 

4 Answers
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up vote
0
down vote



accepted

We have

$$3x^2+10x+8$$

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$

$$left(dfrac3xphantom+4phantom5right)
left(dfrac3xphantom+4phantom5right)$$

2. Find the product $ac$, including sign.

$$ac=24$$

3. Find the prime factorization of $ac$ using the factor tree.

$$ac=24=8cdot 3$$

4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.

$$(1,24),(2,12),(3,8),(4,6),...$$

5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.

$$(s,t)=(4,6)$$

6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$

$$left(dfrac3x+4phantom5right)left(dfrac3x+6phantom5right)$$

7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

$$left(dfrac3x+41right)left(dfrac3x+63right)=left(3x+4right)left(x+2right)$$

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answered Aug 2 at 14:15

gimusi

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up vote
1
down vote

a=3, b=10 and c=-8 for this case.
The '-2x+12x' term simplifies to 10x, giving you b.

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answered Aug 2 at 13:55

Meeta Jo

1418

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Given the quadratic equation $3x^2+10x-8=0$,
$$beginalign &2) ac=3cdot (-8)=-24;\
&3) pm(1,2,3,4,6,8,12,24)\
&5) -12;2 (textsince ac=-24<0) textand -12+2=10=b;\
&7) left(frac3x-21right)left(frac3x+123right)=(3x-2)(x+4).endalign$$

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answered Aug 2 at 14:21

farruhota

13.5k2632

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up vote
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down vote

$$3x^2+10x-8$$

There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$)

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$

$$left(dfrac3xphantom+4phantom5right)
left(dfrac3xphantom+4phantom5right).$$

2. Find the product $ac$, including sign.

$$ac=-24$$

3. Find the prime factorization of $ac$ using the factor tree.

$$24 = 2^3 cdot 3$$

4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.

$$beginarrayrrr
1 & 24 \
colorred2 & colorred12 \
3 & 8 \
4 & 6
endarray$$

5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.

$$ac = -24 < 0$$
$$-2 + 12 = 10$$

6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$

$$left(dfrac3x-2phantom5right)left(dfrac3x+12phantom5right).$$

7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

$$left(dfrac3x-21right)left(dfrac3x+123right).$$
$$1 cdot 3 = 3 checkmark$$

$$(3x-2)(x+4) = 3x^2-2x+12x - 8 = 3x^2+10x-8$$

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edited Aug 2 at 14:26

answered Aug 2 at 14:21

steven gregory

16.3k22055

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

We have

$$3x^2+10x+8$$

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$

$$left(dfrac3xphantom+4phantom5right)
left(dfrac3xphantom+4phantom5right)$$

2. Find the product $ac$, including sign.

$$ac=24$$

3. Find the prime factorization of $ac$ using the factor tree.

$$ac=24=8cdot 3$$

4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.

$$(1,24),(2,12),(3,8),(4,6),...$$

5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.

$$(s,t)=(4,6)$$

6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$

$$left(dfrac3x+4phantom5right)left(dfrac3x+6phantom5right)$$

7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

$$left(dfrac3x+41right)left(dfrac3x+63right)=left(3x+4right)left(x+2right)$$

share|cite|improve this answer

answered Aug 2 at 14:15

gimusi

63.8k73480

add a comment | 

up vote
0
down vote



accepted

We have

$$3x^2+10x+8$$

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$

$$left(dfrac3xphantom+4phantom5right)
left(dfrac3xphantom+4phantom5right)$$

2. Find the product $ac$, including sign.

$$ac=24$$

3. Find the prime factorization of $ac$ using the factor tree.

$$ac=24=8cdot 3$$

4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.

$$(1,24),(2,12),(3,8),(4,6),...$$

5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.

$$(s,t)=(4,6)$$

6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$

$$left(dfrac3x+4phantom5right)left(dfrac3x+6phantom5right)$$

7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

$$left(dfrac3x+41right)left(dfrac3x+63right)=left(3x+4right)left(x+2right)$$

share|cite|improve this answer

answered Aug 2 at 14:15

gimusi

63.8k73480

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

We have

$$3x^2+10x+8$$

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$

$$left(dfrac3xphantom+4phantom5right)
left(dfrac3xphantom+4phantom5right)$$

2. Find the product $ac$, including sign.

$$ac=24$$

3. Find the prime factorization of $ac$ using the factor tree.

$$ac=24=8cdot 3$$

4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.

$$(1,24),(2,12),(3,8),(4,6),...$$

5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.

$$(s,t)=(4,6)$$

6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$

$$left(dfrac3x+4phantom5right)left(dfrac3x+6phantom5right)$$

7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

$$left(dfrac3x+41right)left(dfrac3x+63right)=left(3x+4right)left(x+2right)$$

share|cite|improve this answer

answered Aug 2 at 14:15

gimusi

63.8k73480

We have

$$3x^2+10x+8$$

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$

$$left(dfrac3xphantom+4phantom5right)
left(dfrac3xphantom+4phantom5right)$$

2. Find the product $ac$, including sign.

$$ac=24$$

3. Find the prime factorization of $ac$ using the factor tree.

$$ac=24=8cdot 3$$

4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.

$$(1,24),(2,12),(3,8),(4,6),...$$

5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.

$$(s,t)=(4,6)$$

6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$

$$left(dfrac3x+4phantom5right)left(dfrac3x+6phantom5right)$$

7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

$$left(dfrac3x+41right)left(dfrac3x+63right)=left(3x+4right)left(x+2right)$$

share|cite|improve this answer

answered Aug 2 at 14:15

gimusi

63.8k73480

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 14:15

gimusi

63.8k73480

answered Aug 2 at 14:15

gimusi

63.8k73480

answered Aug 2 at 14:15

gimusi

63.8k73480

63.8k73480

add a comment | 

add a comment | 

up vote
1
down vote

a=3, b=10 and c=-8 for this case.
The '-2x+12x' term simplifies to 10x, giving you b.

share|cite|improve this answer

answered Aug 2 at 13:55

Meeta Jo

1418

add a comment | 

up vote
1
down vote

a=3, b=10 and c=-8 for this case.
The '-2x+12x' term simplifies to 10x, giving you b.

share|cite|improve this answer

answered Aug 2 at 13:55

Meeta Jo

1418

add a comment | 

up vote
1
down vote

up vote
1
down vote

a=3, b=10 and c=-8 for this case.
The '-2x+12x' term simplifies to 10x, giving you b.

share|cite|improve this answer

answered Aug 2 at 13:55

Meeta Jo

1418

a=3, b=10 and c=-8 for this case.
The '-2x+12x' term simplifies to 10x, giving you b.

share|cite|improve this answer

answered Aug 2 at 13:55

Meeta Jo

1418

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 13:55

Meeta Jo

1418

answered Aug 2 at 13:55

Meeta Jo

1418

answered Aug 2 at 13:55

Meeta Jo

1418

1418

add a comment | 

add a comment | 

up vote
0
down vote

Given the quadratic equation $3x^2+10x-8=0$,
$$beginalign &2) ac=3cdot (-8)=-24;\
&3) pm(1,2,3,4,6,8,12,24)\
&5) -12;2 (textsince ac=-24<0) textand -12+2=10=b;\
&7) left(frac3x-21right)left(frac3x+123right)=(3x-2)(x+4).endalign$$

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answered Aug 2 at 14:21

farruhota

13.5k2632

add a comment | 

up vote
0
down vote

Given the quadratic equation $3x^2+10x-8=0$,
$$beginalign &2) ac=3cdot (-8)=-24;\
&3) pm(1,2,3,4,6,8,12,24)\
&5) -12;2 (textsince ac=-24<0) textand -12+2=10=b;\
&7) left(frac3x-21right)left(frac3x+123right)=(3x-2)(x+4).endalign$$

share|cite|improve this answer

answered Aug 2 at 14:21

farruhota

13.5k2632

add a comment | 

up vote
0
down vote

up vote
0
down vote

Given the quadratic equation $3x^2+10x-8=0$,
$$beginalign &2) ac=3cdot (-8)=-24;\
&3) pm(1,2,3,4,6,8,12,24)\
&5) -12;2 (textsince ac=-24<0) textand -12+2=10=b;\
&7) left(frac3x-21right)left(frac3x+123right)=(3x-2)(x+4).endalign$$

share|cite|improve this answer

answered Aug 2 at 14:21

farruhota

13.5k2632

Given the quadratic equation $3x^2+10x-8=0$,
$$beginalign &2) ac=3cdot (-8)=-24;\
&3) pm(1,2,3,4,6,8,12,24)\
&5) -12;2 (textsince ac=-24<0) textand -12+2=10=b;\
&7) left(frac3x-21right)left(frac3x+123right)=(3x-2)(x+4).endalign$$

share|cite|improve this answer

answered Aug 2 at 14:21

farruhota

13.5k2632

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answered Aug 2 at 14:21

farruhota

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answered Aug 2 at 14:21

farruhota

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answered Aug 2 at 14:21

farruhota

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$$3x^2+10x-8$$

There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$)

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$

$$left(dfrac3xphantom+4phantom5right)
left(dfrac3xphantom+4phantom5right).$$

2. Find the product $ac$, including sign.

$$ac=-24$$

3. Find the prime factorization of $ac$ using the factor tree.

$$24 = 2^3 cdot 3$$

4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.

$$beginarrayrrr
1 & 24 \
colorred2 & colorred12 \
3 & 8 \
4 & 6
endarray$$

5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.

$$ac = -24 < 0$$
$$-2 + 12 = 10$$

6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$

$$left(dfrac3x-2phantom5right)left(dfrac3x+12phantom5right).$$

7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

$$left(dfrac3x-21right)left(dfrac3x+123right).$$
$$1 cdot 3 = 3 checkmark$$

$$(3x-2)(x+4) = 3x^2-2x+12x - 8 = 3x^2+10x-8$$

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edited Aug 2 at 14:26

answered Aug 2 at 14:21

steven gregory

16.3k22055

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up vote
0
down vote

$$3x^2+10x-8$$

There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$)

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$

$$left(dfrac3xphantom+4phantom5right)
left(dfrac3xphantom+4phantom5right).$$

2. Find the product $ac$, including sign.

$$ac=-24$$

3. Find the prime factorization of $ac$ using the factor tree.

$$24 = 2^3 cdot 3$$

4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.

$$beginarrayrrr
1 & 24 \
colorred2 & colorred12 \
3 & 8 \
4 & 6
endarray$$

5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.

$$ac = -24 < 0$$
$$-2 + 12 = 10$$

6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$

$$left(dfrac3x-2phantom5right)left(dfrac3x+12phantom5right).$$

7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

$$left(dfrac3x-21right)left(dfrac3x+123right).$$
$$1 cdot 3 = 3 checkmark$$

$$(3x-2)(x+4) = 3x^2-2x+12x - 8 = 3x^2+10x-8$$

share|cite|improve this answer

edited Aug 2 at 14:26

answered Aug 2 at 14:21

steven gregory

16.3k22055

add a comment | 

up vote
0
down vote

up vote
0
down vote

$$3x^2+10x-8$$

There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$)

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$

$$left(dfrac3xphantom+4phantom5right)
left(dfrac3xphantom+4phantom5right).$$

2. Find the product $ac$, including sign.

$$ac=-24$$

3. Find the prime factorization of $ac$ using the factor tree.

$$24 = 2^3 cdot 3$$

4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.

$$beginarrayrrr
1 & 24 \
colorred2 & colorred12 \
3 & 8 \
4 & 6
endarray$$

5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.

$$ac = -24 < 0$$
$$-2 + 12 = 10$$

6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$

$$left(dfrac3x-2phantom5right)left(dfrac3x+12phantom5right).$$

7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

$$left(dfrac3x-21right)left(dfrac3x+123right).$$
$$1 cdot 3 = 3 checkmark$$

$$(3x-2)(x+4) = 3x^2-2x+12x - 8 = 3x^2+10x-8$$

share|cite|improve this answer

edited Aug 2 at 14:26

answered Aug 2 at 14:21

steven gregory

16.3k22055

$$3x^2+10x-8$$

There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$)

1. First write $left(dfracaxphantom+4phantom5right)
   left(dfracaxphantom+4phantom5right).$

$$left(dfrac3xphantom+4phantom5right)
left(dfrac3xphantom+4phantom5right).$$

2. Find the product $ac$, including sign.

$$ac=-24$$

3. Find the prime factorization of $ac$ using the factor tree.

$$24 = 2^3 cdot 3$$

4. Find all factor pairs of $ac$ using the factor tree: begin with $1$,$ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached sqrtac.

$$beginarrayrrr
1 & 24 \
colorred2 & colorred12 \
3 & 8 \
4 & 6
endarray$$

5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.

$$ac = -24 < 0$$
$$-2 + 12 = 10$$

6. Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$

$$left(dfrac3x-2phantom5right)left(dfrac3x+12phantom5right).$$

7. Divide each of these binomials by its own GCF $left(dfracax+soperatornamegcf(a,s)right)
   left(dfracax+toperatornamegcf(a,t)right).$ Check that $$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$

$$left(dfrac3x-21right)left(dfrac3x+123right).$$
$$1 cdot 3 = 3 checkmark$$

$$(3x-2)(x+4) = 3x^2-2x+12x - 8 = 3x^2+10x-8$$

share|cite|improve this answer

edited Aug 2 at 14:26

answered Aug 2 at 14:21

steven gregory

16.3k22055

share|cite|improve this answer

share|cite|improve this answer

edited Aug 2 at 14:26

edited Aug 2 at 14:26

edited Aug 2 at 14:26

answered Aug 2 at 14:21

steven gregory

16.3k22055

answered Aug 2 at 14:21

steven gregory

16.3k22055

answered Aug 2 at 14:21

steven gregory

16.3k22055

16.3k22055

add a comment | 

add a comment |