How can I prove this difference satisfies this condition? [closed]

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Consider the following polynomial difference $A(x)-B(x)$:
$$A(x)=x^4-6x^3+8x^2+6x-9$$
$$B(x)=x^4-6x^2-8x-3$$

How can I prove/know that this difference satisfies this condition:
$$A(x)-B(x)<0 iff x:in :]-1,:frac13:[;;cup;;]:3,:+infty[ $$ ?

elementary-set-theory polynomials

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edited Jul 14 at 15:40

amWhy

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asked Jul 14 at 15:04

Daniel Oscar

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closed as off-topic by Alex Francisco, amWhy, Jyrki Lahtonen, Cesareo, Parcly Taxel Jul 15 at 1:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, amWhy, Jyrki Lahtonen, Cesareo, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
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  Use cup to make the union symbol ($cup$) and iff to make the if and only if symbol ($iff$).
  – parsiad
  Jul 14 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. Already edited.
  – Daniel Oscar
  Jul 14 at 15:13
  

  
  
  
  

add a comment | 

up vote
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Consider the following polynomial difference $A(x)-B(x)$:
$$A(x)=x^4-6x^3+8x^2+6x-9$$
$$B(x)=x^4-6x^2-8x-3$$

How can I prove/know that this difference satisfies this condition:
$$A(x)-B(x)<0 iff x:in :]-1,:frac13:[;;cup;;]:3,:+infty[ $$ ?

elementary-set-theory polynomials

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edited Jul 14 at 15:40

amWhy

189k25219431

asked Jul 14 at 15:04

Daniel Oscar

494

closed as off-topic by Alex Francisco, amWhy, Jyrki Lahtonen, Cesareo, Parcly Taxel Jul 15 at 1:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, amWhy, Jyrki Lahtonen, Cesareo, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Use cup to make the union symbol ($cup$) and iff to make the if and only if symbol ($iff$).
  – parsiad
  Jul 14 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. Already edited.
  – Daniel Oscar
  Jul 14 at 15:13
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Consider the following polynomial difference $A(x)-B(x)$:
$$A(x)=x^4-6x^3+8x^2+6x-9$$
$$B(x)=x^4-6x^2-8x-3$$

How can I prove/know that this difference satisfies this condition:
$$A(x)-B(x)<0 iff x:in :]-1,:frac13:[;;cup;;]:3,:+infty[ $$ ?

elementary-set-theory polynomials

share|cite|improve this question

edited Jul 14 at 15:40

amWhy

189k25219431

asked Jul 14 at 15:04

Daniel Oscar

494

Consider the following polynomial difference $A(x)-B(x)$:
$$A(x)=x^4-6x^3+8x^2+6x-9$$
$$B(x)=x^4-6x^2-8x-3$$

How can I prove/know that this difference satisfies this condition:
$$A(x)-B(x)<0 iff x:in :]-1,:frac13:[;;cup;;]:3,:+infty[ $$ ?

elementary-set-theory polynomials

share|cite|improve this question

edited Jul 14 at 15:40

amWhy

189k25219431

asked Jul 14 at 15:04

Daniel Oscar

494

share|cite|improve this question

share|cite|improve this question

edited Jul 14 at 15:40

amWhy

189k25219431

edited Jul 14 at 15:40

amWhy

189k25219431

edited Jul 14 at 15:40

amWhy

189k25219431

189k25219431

asked Jul 14 at 15:04

Daniel Oscar

494

asked Jul 14 at 15:04

Daniel Oscar

494

asked Jul 14 at 15:04

Daniel Oscar

494

494

closed as off-topic by Alex Francisco, amWhy, Jyrki Lahtonen, Cesareo, Parcly Taxel Jul 15 at 1:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, amWhy, Jyrki Lahtonen, Cesareo, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Alex Francisco, amWhy, Jyrki Lahtonen, Cesareo, Parcly Taxel Jul 15 at 1:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, amWhy, Jyrki Lahtonen, Cesareo, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Use cup to make the union symbol ($cup$) and iff to make the if and only if symbol ($iff$).
  – parsiad
  Jul 14 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. Already edited.
  – Daniel Oscar
  Jul 14 at 15:13
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Use cup to make the union symbol ($cup$) and iff to make the if and only if symbol ($iff$).
  – parsiad
  Jul 14 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. Already edited.
  – Daniel Oscar
  Jul 14 at 15:13
  

  
  
  
  

1

1

Use cup to make the union symbol ($cup$) and iff to make the if and only if symbol ($iff$).
– parsiad
Jul 14 at 15:06

Use cup to make the union symbol ($cup$) and iff to make the if and only if symbol ($iff$).
– parsiad
Jul 14 at 15:06

Thank you. Already edited.
– Daniel Oscar
Jul 14 at 15:13

Thank you. Already edited.
– Daniel Oscar
Jul 14 at 15:13

add a comment | 

3 Answers
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HINT:

Note that $$A(x)-B(x)=-6x^3+14x^2+14x-6=-2(3x^3-7x^2-7x+3)$$

The polynomial in brackets is palindromic so $x=-1$ is clearly a root. Factoring gives $$3x^3-7x^2-7x+3=(x+1)(x-3)(3x-1)$$ so $$A(x)-B(x)=-6(x+1)(x-3)left(x-frac13right)$$

This is negative only when $$(x+1)(x-3)left(x-frac13right)$$ is positive.

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answered Jul 14 at 15:16

TheSimpliFire

9,69261952

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1) $x:in :]-1,:frac13:[cup ]:3,:+infty[$

Means that if we divide the real numbers into four regions:

$A = (-infty, -1]; B= (-1,frac 13);C=[frac 13, 3]; D = (3, infty)$

If $x in A$ then $(x + 1) le 0; (x - frac 13) < 0; (x - 3) < 0$ and $(x+1)(x-frac 13)(x-3) le 0$ and $-6(x+1)(x-frac 13)(x-3) ge 0$.

If $x in B$ then $(x + 1) > 0; (x -frac 13) < 0; (x-3)< 0$ and $(x+1)(x-frac 13)(x-3) > 0$ and $-6(x+1)(x-frac 13)(x-3) < 0$.

If $x in C$ then $(x + 1) > 0; (x - frac 13) > 0; (x - 3)le 0$ and $(x+1)(x-frac 13)(x-3) le 0$ and $-6(x+1)(x-frac 13)(x-3) ge 0$.

If $x in D$ then $(x + 1) > 0; (x -frac 13) > 0; (x-3)> 0$ and $(x+1)(x-frac 13)(x-3) > 0$ and $-6(x+1)(x-frac 13)(x-3) < 0$.

So $-6(x+1)(x-frac 13)(x-3) = -6x^3 + 14x^2 + 14x -6=A(x) - B(x) < 0$ if and only if $x in B$ or $xin D$. Or in othere words $A(x) - B(x) < 0 iff x in A cup B$.

2)

$D(x) = A(x) - B(x) = -6x^3 +6x^2 + 8x^2 + 8x + 6x - 6$

$= -6x^3 + 14x^2 + 14x - 6$

$= -6x^3 - 6x^2 + 20x^2 + 20x - 6x -6$

$=-6x^2(x+1) + 20x(x + 1) - 6(x+1)$

$=-2(3x^2 - 10x + 3)(x+1)$

$=-2(3x -1)(x -3)(x+1)$

$= -6(x+1)(x-frac 13)(x-3)$.

$D(x) < 0 $ when an odd number of $-6,x+1, x-frac 13, x-3$ are negative and the rest are positive. $-6$ is always negative we need an even number of $x+1, x-frac 13, x-3$ to be negative. In other words, none or two are negative.

As $x-3 < x-frac 13 < x+1$ to have none of them negative means $x -3 ge 0$ or $x ge 3$.

To have two of them negative is to have $x-3$ and $x - frac 13$ negative but $x + 1 > 0$. So $x-frac 13 < 0< x + 1$ or in other words $-1 < x < frac 13$.

3) Use the $D(x)$ of part 2)

Now $D(x) = 0$ if $x = -1, frac 13, $ or $3$ so those roots divide the real numbers into four regions and... continue as in part 1)

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answered Jul 14 at 16:33

fleablood

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Hint: the difference of your polynomials $$A(x),B(x)$$ can be written as $$-2, left( 3,x-1 right) left( x-3 right) left( x+1 right) $$

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answered Jul 14 at 15:08

Dr. Sonnhard Graubner

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

HINT:

Note that $$A(x)-B(x)=-6x^3+14x^2+14x-6=-2(3x^3-7x^2-7x+3)$$

The polynomial in brackets is palindromic so $x=-1$ is clearly a root. Factoring gives $$3x^3-7x^2-7x+3=(x+1)(x-3)(3x-1)$$ so $$A(x)-B(x)=-6(x+1)(x-3)left(x-frac13right)$$

This is negative only when $$(x+1)(x-3)left(x-frac13right)$$ is positive.

share|cite|improve this answer

answered Jul 14 at 15:16

TheSimpliFire

9,69261952

add a comment | 

up vote
1
down vote

HINT:

Note that $$A(x)-B(x)=-6x^3+14x^2+14x-6=-2(3x^3-7x^2-7x+3)$$

The polynomial in brackets is palindromic so $x=-1$ is clearly a root. Factoring gives $$3x^3-7x^2-7x+3=(x+1)(x-3)(3x-1)$$ so $$A(x)-B(x)=-6(x+1)(x-3)left(x-frac13right)$$

This is negative only when $$(x+1)(x-3)left(x-frac13right)$$ is positive.

share|cite|improve this answer

answered Jul 14 at 15:16

TheSimpliFire

9,69261952

add a comment | 

up vote
1
down vote

up vote
1
down vote

HINT:

Note that $$A(x)-B(x)=-6x^3+14x^2+14x-6=-2(3x^3-7x^2-7x+3)$$

The polynomial in brackets is palindromic so $x=-1$ is clearly a root. Factoring gives $$3x^3-7x^2-7x+3=(x+1)(x-3)(3x-1)$$ so $$A(x)-B(x)=-6(x+1)(x-3)left(x-frac13right)$$

This is negative only when $$(x+1)(x-3)left(x-frac13right)$$ is positive.

share|cite|improve this answer

answered Jul 14 at 15:16

TheSimpliFire

9,69261952

HINT:

Note that $$A(x)-B(x)=-6x^3+14x^2+14x-6=-2(3x^3-7x^2-7x+3)$$

The polynomial in brackets is palindromic so $x=-1$ is clearly a root. Factoring gives $$3x^3-7x^2-7x+3=(x+1)(x-3)(3x-1)$$ so $$A(x)-B(x)=-6(x+1)(x-3)left(x-frac13right)$$

This is negative only when $$(x+1)(x-3)left(x-frac13right)$$ is positive.

share|cite|improve this answer

answered Jul 14 at 15:16

TheSimpliFire

9,69261952

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 15:16

TheSimpliFire

9,69261952

answered Jul 14 at 15:16

TheSimpliFire

9,69261952

answered Jul 14 at 15:16

TheSimpliFire

9,69261952

9,69261952

add a comment | 

add a comment | 

up vote
1
down vote

1) $x:in :]-1,:frac13:[cup ]:3,:+infty[$

Means that if we divide the real numbers into four regions:

$A = (-infty, -1]; B= (-1,frac 13);C=[frac 13, 3]; D = (3, infty)$

If $x in A$ then $(x + 1) le 0; (x - frac 13) < 0; (x - 3) < 0$ and $(x+1)(x-frac 13)(x-3) le 0$ and $-6(x+1)(x-frac 13)(x-3) ge 0$.

If $x in B$ then $(x + 1) > 0; (x -frac 13) < 0; (x-3)< 0$ and $(x+1)(x-frac 13)(x-3) > 0$ and $-6(x+1)(x-frac 13)(x-3) < 0$.

If $x in C$ then $(x + 1) > 0; (x - frac 13) > 0; (x - 3)le 0$ and $(x+1)(x-frac 13)(x-3) le 0$ and $-6(x+1)(x-frac 13)(x-3) ge 0$.

If $x in D$ then $(x + 1) > 0; (x -frac 13) > 0; (x-3)> 0$ and $(x+1)(x-frac 13)(x-3) > 0$ and $-6(x+1)(x-frac 13)(x-3) < 0$.

So $-6(x+1)(x-frac 13)(x-3) = -6x^3 + 14x^2 + 14x -6=A(x) - B(x) < 0$ if and only if $x in B$ or $xin D$. Or in othere words $A(x) - B(x) < 0 iff x in A cup B$.

2)

$D(x) = A(x) - B(x) = -6x^3 +6x^2 + 8x^2 + 8x + 6x - 6$

$= -6x^3 + 14x^2 + 14x - 6$

$= -6x^3 - 6x^2 + 20x^2 + 20x - 6x -6$

$=-6x^2(x+1) + 20x(x + 1) - 6(x+1)$

$=-2(3x^2 - 10x + 3)(x+1)$

$=-2(3x -1)(x -3)(x+1)$

$= -6(x+1)(x-frac 13)(x-3)$.

$D(x) < 0 $ when an odd number of $-6,x+1, x-frac 13, x-3$ are negative and the rest are positive. $-6$ is always negative we need an even number of $x+1, x-frac 13, x-3$ to be negative. In other words, none or two are negative.

As $x-3 < x-frac 13 < x+1$ to have none of them negative means $x -3 ge 0$ or $x ge 3$.

To have two of them negative is to have $x-3$ and $x - frac 13$ negative but $x + 1 > 0$. So $x-frac 13 < 0< x + 1$ or in other words $-1 < x < frac 13$.

3) Use the $D(x)$ of part 2)

Now $D(x) = 0$ if $x = -1, frac 13, $ or $3$ so those roots divide the real numbers into four regions and... continue as in part 1)

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answered Jul 14 at 16:33

fleablood

60.5k22575

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up vote
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1) $x:in :]-1,:frac13:[cup ]:3,:+infty[$

Means that if we divide the real numbers into four regions:

$A = (-infty, -1]; B= (-1,frac 13);C=[frac 13, 3]; D = (3, infty)$

If $x in A$ then $(x + 1) le 0; (x - frac 13) < 0; (x - 3) < 0$ and $(x+1)(x-frac 13)(x-3) le 0$ and $-6(x+1)(x-frac 13)(x-3) ge 0$.

If $x in B$ then $(x + 1) > 0; (x -frac 13) < 0; (x-3)< 0$ and $(x+1)(x-frac 13)(x-3) > 0$ and $-6(x+1)(x-frac 13)(x-3) < 0$.

If $x in C$ then $(x + 1) > 0; (x - frac 13) > 0; (x - 3)le 0$ and $(x+1)(x-frac 13)(x-3) le 0$ and $-6(x+1)(x-frac 13)(x-3) ge 0$.

If $x in D$ then $(x + 1) > 0; (x -frac 13) > 0; (x-3)> 0$ and $(x+1)(x-frac 13)(x-3) > 0$ and $-6(x+1)(x-frac 13)(x-3) < 0$.

So $-6(x+1)(x-frac 13)(x-3) = -6x^3 + 14x^2 + 14x -6=A(x) - B(x) < 0$ if and only if $x in B$ or $xin D$. Or in othere words $A(x) - B(x) < 0 iff x in A cup B$.

2)

$D(x) = A(x) - B(x) = -6x^3 +6x^2 + 8x^2 + 8x + 6x - 6$

$= -6x^3 + 14x^2 + 14x - 6$

$= -6x^3 - 6x^2 + 20x^2 + 20x - 6x -6$

$=-6x^2(x+1) + 20x(x + 1) - 6(x+1)$

$=-2(3x^2 - 10x + 3)(x+1)$

$=-2(3x -1)(x -3)(x+1)$

$= -6(x+1)(x-frac 13)(x-3)$.

$D(x) < 0 $ when an odd number of $-6,x+1, x-frac 13, x-3$ are negative and the rest are positive. $-6$ is always negative we need an even number of $x+1, x-frac 13, x-3$ to be negative. In other words, none or two are negative.

As $x-3 < x-frac 13 < x+1$ to have none of them negative means $x -3 ge 0$ or $x ge 3$.

To have two of them negative is to have $x-3$ and $x - frac 13$ negative but $x + 1 > 0$. So $x-frac 13 < 0< x + 1$ or in other words $-1 < x < frac 13$.

3) Use the $D(x)$ of part 2)

Now $D(x) = 0$ if $x = -1, frac 13, $ or $3$ so those roots divide the real numbers into four regions and... continue as in part 1)

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answered Jul 14 at 16:33

fleablood

60.5k22575

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up vote
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up vote
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1) $x:in :]-1,:frac13:[cup ]:3,:+infty[$

Means that if we divide the real numbers into four regions:

$A = (-infty, -1]; B= (-1,frac 13);C=[frac 13, 3]; D = (3, infty)$

If $x in A$ then $(x + 1) le 0; (x - frac 13) < 0; (x - 3) < 0$ and $(x+1)(x-frac 13)(x-3) le 0$ and $-6(x+1)(x-frac 13)(x-3) ge 0$.

If $x in B$ then $(x + 1) > 0; (x -frac 13) < 0; (x-3)< 0$ and $(x+1)(x-frac 13)(x-3) > 0$ and $-6(x+1)(x-frac 13)(x-3) < 0$.

If $x in C$ then $(x + 1) > 0; (x - frac 13) > 0; (x - 3)le 0$ and $(x+1)(x-frac 13)(x-3) le 0$ and $-6(x+1)(x-frac 13)(x-3) ge 0$.

If $x in D$ then $(x + 1) > 0; (x -frac 13) > 0; (x-3)> 0$ and $(x+1)(x-frac 13)(x-3) > 0$ and $-6(x+1)(x-frac 13)(x-3) < 0$.

So $-6(x+1)(x-frac 13)(x-3) = -6x^3 + 14x^2 + 14x -6=A(x) - B(x) < 0$ if and only if $x in B$ or $xin D$. Or in othere words $A(x) - B(x) < 0 iff x in A cup B$.

2)

$D(x) = A(x) - B(x) = -6x^3 +6x^2 + 8x^2 + 8x + 6x - 6$

$= -6x^3 + 14x^2 + 14x - 6$

$= -6x^3 - 6x^2 + 20x^2 + 20x - 6x -6$

$=-6x^2(x+1) + 20x(x + 1) - 6(x+1)$

$=-2(3x^2 - 10x + 3)(x+1)$

$=-2(3x -1)(x -3)(x+1)$

$= -6(x+1)(x-frac 13)(x-3)$.

$D(x) < 0 $ when an odd number of $-6,x+1, x-frac 13, x-3$ are negative and the rest are positive. $-6$ is always negative we need an even number of $x+1, x-frac 13, x-3$ to be negative. In other words, none or two are negative.

As $x-3 < x-frac 13 < x+1$ to have none of them negative means $x -3 ge 0$ or $x ge 3$.

To have two of them negative is to have $x-3$ and $x - frac 13$ negative but $x + 1 > 0$. So $x-frac 13 < 0< x + 1$ or in other words $-1 < x < frac 13$.

3) Use the $D(x)$ of part 2)

Now $D(x) = 0$ if $x = -1, frac 13, $ or $3$ so those roots divide the real numbers into four regions and... continue as in part 1)

share|cite|improve this answer

answered Jul 14 at 16:33

fleablood

60.5k22575

1) $x:in :]-1,:frac13:[cup ]:3,:+infty[$

Means that if we divide the real numbers into four regions:

$A = (-infty, -1]; B= (-1,frac 13);C=[frac 13, 3]; D = (3, infty)$

If $x in A$ then $(x + 1) le 0; (x - frac 13) < 0; (x - 3) < 0$ and $(x+1)(x-frac 13)(x-3) le 0$ and $-6(x+1)(x-frac 13)(x-3) ge 0$.

If $x in B$ then $(x + 1) > 0; (x -frac 13) < 0; (x-3)< 0$ and $(x+1)(x-frac 13)(x-3) > 0$ and $-6(x+1)(x-frac 13)(x-3) < 0$.

If $x in C$ then $(x + 1) > 0; (x - frac 13) > 0; (x - 3)le 0$ and $(x+1)(x-frac 13)(x-3) le 0$ and $-6(x+1)(x-frac 13)(x-3) ge 0$.

If $x in D$ then $(x + 1) > 0; (x -frac 13) > 0; (x-3)> 0$ and $(x+1)(x-frac 13)(x-3) > 0$ and $-6(x+1)(x-frac 13)(x-3) < 0$.

So $-6(x+1)(x-frac 13)(x-3) = -6x^3 + 14x^2 + 14x -6=A(x) - B(x) < 0$ if and only if $x in B$ or $xin D$. Or in othere words $A(x) - B(x) < 0 iff x in A cup B$.

2)

$D(x) = A(x) - B(x) = -6x^3 +6x^2 + 8x^2 + 8x + 6x - 6$

$= -6x^3 + 14x^2 + 14x - 6$

$= -6x^3 - 6x^2 + 20x^2 + 20x - 6x -6$

$=-6x^2(x+1) + 20x(x + 1) - 6(x+1)$

$=-2(3x^2 - 10x + 3)(x+1)$

$=-2(3x -1)(x -3)(x+1)$

$= -6(x+1)(x-frac 13)(x-3)$.

$D(x) < 0 $ when an odd number of $-6,x+1, x-frac 13, x-3$ are negative and the rest are positive. $-6$ is always negative we need an even number of $x+1, x-frac 13, x-3$ to be negative. In other words, none or two are negative.

As $x-3 < x-frac 13 < x+1$ to have none of them negative means $x -3 ge 0$ or $x ge 3$.

To have two of them negative is to have $x-3$ and $x - frac 13$ negative but $x + 1 > 0$. So $x-frac 13 < 0< x + 1$ or in other words $-1 < x < frac 13$.

3) Use the $D(x)$ of part 2)

Now $D(x) = 0$ if $x = -1, frac 13, $ or $3$ so those roots divide the real numbers into four regions and... continue as in part 1)

share|cite|improve this answer

answered Jul 14 at 16:33

fleablood

60.5k22575

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answered Jul 14 at 16:33

fleablood

60.5k22575

answered Jul 14 at 16:33

fleablood

60.5k22575

answered Jul 14 at 16:33

fleablood

60.5k22575

60.5k22575

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Hint: the difference of your polynomials $$A(x),B(x)$$ can be written as $$-2, left( 3,x-1 right) left( x-3 right) left( x+1 right) $$

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answered Jul 14 at 15:08

Dr. Sonnhard Graubner

67k32660

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Hint: the difference of your polynomials $$A(x),B(x)$$ can be written as $$-2, left( 3,x-1 right) left( x-3 right) left( x+1 right) $$

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answered Jul 14 at 15:08

Dr. Sonnhard Graubner

67k32660

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up vote
0
down vote

up vote
0
down vote

Hint: the difference of your polynomials $$A(x),B(x)$$ can be written as $$-2, left( 3,x-1 right) left( x-3 right) left( x+1 right) $$

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answered Jul 14 at 15:08

Dr. Sonnhard Graubner

67k32660

Hint: the difference of your polynomials $$A(x),B(x)$$ can be written as $$-2, left( 3,x-1 right) left( x-3 right) left( x+1 right) $$

share|cite|improve this answer

answered Jul 14 at 15:08

Dr. Sonnhard Graubner

67k32660

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 15:08

Dr. Sonnhard Graubner

67k32660

answered Jul 14 at 15:08

Dr. Sonnhard Graubner

67k32660

answered Jul 14 at 15:08

Dr. Sonnhard Graubner

67k32660

67k32660

add a comment | 

add a comment |