document.write( "Question 193068: THE FACTOR THEOREM\r
\n" ); document.write( "\n" ); document.write( "Factor P(x) = 6x^3 + 31x^2 + 4x -5 given that x+5 is one factor.\r
\n" ); document.write( "\n" ); document.write( "Factor R(x) = x^4 -2x3 + x^2 -4, given that x+1 and x-2 are factors. \n" ); document.write( "

Algebra.Com's Answer #144914 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
# 1\r
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\n" ); document.write( "\n" ); document.write( "To factor $\"6x%5E3+%2B+31x%5E2+%2B+4x+-+5\"$ , we can use synthetic division\r
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\n" ); document.write( "\n" ); document.write( "First, let's find our test zero:\r
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\n" ); document.write( "\n" ); document.write( " $\"x%2B5=0\"$ Set the given factor $\"x%2B5\"$ equal to zero\r
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\n" ); document.write( "\n" ); document.write( " $\"x=-5\"$ Solve for x.\r
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\n" ); document.write( "\n" ); document.write( "so our test zero is -5\r
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\n" ); document.write( "\n" ); document.write( "Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of $\"6x%5E3+%2B+31x%5E2+%2B+4x+-+5\"$ to the right of the test zero.\n" ); document.write( "\n" ); document.write( "

-5	\|	6	31	4	-5
	\|

\r
\n" ); document.write( "\n" ); document.write( "Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 6)\r
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( "\n" ); document.write( "

-5	\|	6	31	4	-5
	\|
		6

\r
\n" ); document.write( "\n" ); document.write( " Multiply -5 by 6 and place the product (which is -30) right underneath the second coefficient (which is 31)\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

-5	\|	6	31	4	-5
	\|		-30
		6

\r
\n" ); document.write( "\n" ); document.write( " Add -30 and 31 to get 1. Place the sum right underneath -30.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

-5	\|	6	31	4	-5
	\|		-30
		6	1

\r
\n" ); document.write( "\n" ); document.write( " Multiply -5 by 1 and place the product (which is -5) right underneath the third coefficient (which is 4)\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

-5	\|	6	31	4	-5
	\|		-30	-5
		6	1

\r
\n" ); document.write( "\n" ); document.write( " Add -5 and 4 to get -1. Place the sum right underneath -5.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

-5	\|	6	31	4	-5
	\|		-30	-5
		6	1	-1

\r
\n" ); document.write( "\n" ); document.write( " Multiply -5 by -1 and place the product (which is 5) right underneath the fourth coefficient (which is -5)\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

-5	\|	6	31	4	-5
	\|		-30	-5	5
		6	1	-1

\r
\n" ); document.write( "\n" ); document.write( " Add 5 and -5 to get 0. Place the sum right underneath 5.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

-5	\|	6	31	4	-5
	\|		-30	-5	5
		6	1	-1	0

\r
\n" ); document.write( "\n" ); document.write( "Since the last column adds to zero, we have a remainder of zero. This means $\"x%2B5\"$ is a factor of $\"6x%5E3+%2B+31x%5E2+%2B+4x+-+5\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now lets look at the bottom row of coefficients:\r
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\n" ); document.write( "\n" ); document.write( "The first 3 coefficients (6,1,-1) form the quotient\r
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\n" ); document.write( "\n" ); document.write( " $\"6x%5E2+%2B+x+-+1\"$ \r
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\n" ); document.write( "\n" ); document.write( "So $\"6x%5E3+%2B+31x%5E2+%2B+4x+-+5\"$ factors to $\"%28x%2B5%29%286x%5E2+%2B+x+-+1%29\"$ \r
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\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "In other words, $\"6x%5E3+%2B+31x%5E2+%2B+4x+-+5=%28x%2B5%29%286x%5E2+%2B+x+-+1%29\"$ \r
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\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "I'll let you continue the factorization....\r
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\r
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\n" ); document.write( "\n" ); document.write( "# 2\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "First lets find our test zero:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"x%2B1=0\"$ Set the denominator $\"x%2B1\"$ equal to zero\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"x=-1\"$ Solve for x.\r
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\n" ); document.write( "\n" ); document.write( "so our test zero is -1\r
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\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of $\"x%5E4+-2x3+%2B+x%5E2+-4\"$ to the right of the test zero.(note: remember if a polynomial goes from $\"1x%5E2\"$ to $\"-4x%5E0\"$ there is a zero coefficient for $\"x%5E1\"$ . This is simply because $\"x%5E4+-+2x%5E3+%2B+x%5E2+-+4\"$ really looks like $\"1x%5E4%2B-2x%5E3%2B1x%5E2%2B0x%5E1%2B-4x%5E0\"$ \n" ); document.write( "\n" ); document.write( "

-1	\|	1	-2	1	0	-4
	\|

\r
\n" ); document.write( "\n" ); document.write( "Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)\r
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( "\n" ); document.write( "

-1	\|	1	-2	1	0	-4
	\|
		1

\r
\n" ); document.write( "\n" ); document.write( " Multiply -1 by 1 and place the product (which is -1) right underneath the second coefficient (which is -2)\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

-1	\|	1	-2	1	0	-4
	\|		-1
		1

\r
\n" ); document.write( "\n" ); document.write( " Add -1 and -2 to get -3. Place the sum right underneath -1.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

-1	\|	1	-2	1	0	-4
	\|		-1
		1	-3

\r
\n" ); document.write( "\n" ); document.write( " Multiply -1 by -3 and place the product (which is 3) right underneath the third coefficient (which is 1)\r
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-1	\|	1	-2	1	0	-4
	\|		-1	3
		1	-3

\r
\n" ); document.write( "\n" ); document.write( " Add 3 and 1 to get 4. Place the sum right underneath 3.\r
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-1	\|	1	-2	1	0	-4
	\|		-1	3
		1	-3	4

\r
\n" ); document.write( "\n" ); document.write( " Multiply -1 by 4 and place the product (which is -4) right underneath the fourth coefficient (which is 0)\r
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-1	\|	1	-2	1	0	-4
	\|		-1	3	-4
		1	-3	4

\r
\n" ); document.write( "\n" ); document.write( " Add -4 and 0 to get -4. Place the sum right underneath -4.\r
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-1	\|	1	-2	1	0	-4
	\|		-1	3	-4
		1	-3	4	-4

\r
\n" ); document.write( "\n" ); document.write( " Multiply -1 by -4 and place the product (which is 4) right underneath the fifth coefficient (which is -4)\r
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-1	\|	1	-2	1	0	-4
	\|		-1	3	-4	4
		1	-3	4	-4

\r
\n" ); document.write( "\n" ); document.write( " Add 4 and -4 to get 0. Place the sum right underneath 4.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

-1	\|	1	-2	1	0	-4
	\|		-1	3	-4	4
		1	-3	4	-4	0

\r
\n" ); document.write( "\n" ); document.write( "Since the last column adds to zero, we have a remainder of zero. This means $\"x%2B1\"$ is a factor of $\"x%5E4+-+2x%5E3+%2B+x%5E2+-+4\"$ \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Now lets look at the bottom row of coefficients:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "The first 4 coefficients (1,-3,4,-4) form the quotient\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"x%5E3+-+3x%5E2+%2B+4x+-+4\"$ \r
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\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So $\"x%5E4+-+2x%5E3+%2B+x%5E2+-+4\"$ factors to $\"%28x%2B1%29%28x%5E3+-+3x%5E2+%2B+4x+-+4%29\"$ \r
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\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "In other words, $\"x%5E4+-+2x%5E3+%2B+x%5E2+-+4=%28x%2B1%29%28x%5E3+-+3x%5E2+%2B+4x+-+4%29\"$ \r
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\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Now let's use the factor $\"x-2\"$ to factor $\"x%5E3+-+3x%5E2+%2B+4x+-+4\"$ \r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "First lets find our test zero:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"x-2=0\"$ Set the denominator $\"x-2\"$ equal to zero\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"x=2\"$ Solve for x.\r
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\n" ); document.write( "\n" ); document.write( "so our test zero is 2\r
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\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of $\"x%5E3+-+3x%5E2+%2B+4x+-+4\"$ to the right of the test zero.\n" ); document.write( "\n" ); document.write( "

2	\|	1	-3	4	-4
	\|

2	\|	1	-3	4	-4
	\|
		1

\r
\n" ); document.write( "\n" ); document.write( " Multiply 2 by 1 and place the product (which is 2) right underneath the second coefficient (which is -3)\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

2	\|	1	-3	4	-4
	\|		2
		1

\r
\n" ); document.write( "\n" ); document.write( " Add 2 and -3 to get -1. Place the sum right underneath 2.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

2	\|	1	-3	4	-4
	\|		2
		1	-1

\r
\n" ); document.write( "\n" ); document.write( " Multiply 2 by -1 and place the product (which is -2) right underneath the third coefficient (which is 4)\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

2	\|	1	-3	4	-4
	\|		2	-2
		1	-1

\r
\n" ); document.write( "\n" ); document.write( " Add -2 and 4 to get 2. Place the sum right underneath -2.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

2	\|	1	-3	4	-4
	\|		2	-2
		1	-1	2

\r
\n" ); document.write( "\n" ); document.write( " Multiply 2 by 2 and place the product (which is 4) right underneath the fourth coefficient (which is -4)\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

2	\|	1	-3	4	-4
	\|		2	-2	4
		1	-1	2

2	\|	1	-3	4	-4
	\|		2	-2	4
		1	-1	2	0

\r
\n" ); document.write( "\n" ); document.write( "Since the last column adds to zero, we have a remainder of zero. This means $\"x-2\"$ is a factor of $\"x%5E3+-+3x%5E2+%2B+4x+-+4\"$ \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Now lets look at the bottom row of coefficients:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "The first 3 coefficients (1,-1,2) form the quotient\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"x%5E2+-+x+%2B+2\"$ \r
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\n" ); document.write( "\n" ); document.write( "So $\"%28x%5E3+-+3x%5E2+%2B+4x+-+4%29%2F%28x-2%29=x%5E2+-+x+%2B+2\"$ \r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Basically $\"x%5E3+-+3x%5E2+%2B+4x+-+4\"$ factors to $\"%28x-2%29%28x%5E2+-+x+%2B+2%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "So $\"x%5E3+-+3x%5E2+%2B+4x+-+4=%28x-2%29%28x%5E2+-+x+%2B+2%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "This means that $\"x%5E4+-+2x%5E3+%2B+x%5E2+-+4=%28x%2B1%29%28x%5E3+-+3x%5E2+%2B+4x+-+4%29\"$ then becomes\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"x%5E4+-+2x%5E3+%2B+x%5E2+-+4=%28x%2B1%29%28x-2%29%28x%5E2+-+x+%2B+2%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "So all you have to do now is factor $\"x%5E2+-+x+%2B+2\"$ (I'll let you do that) \n" ); document.write( "

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