document.write( "Question 854420: If I have the equation f(x)=3x^3-43x^2+161x-49 are the real zeros 7 and 1/3. If so how do you use the real zeros to factor the equation?\r
\n" ); document.write( "\n" ); document.write( "And if I have the equation f(x)=2x^3-4x^2-34x+68 are the real zeros positive and negative sqrt(17) and 2? And how do you use the real zeros to factor f?
\n" ); document.write( " HELP!!!!! \n" ); document.write( "
Algebra.Com's Answer #514668 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
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document.write( "Hi,
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document.write( "f(x)=3x^3-43x^2+161x-49    |Yes x = 7 is a root
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document.write( "Using synthetic Division to find the other two
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document.write( "7	3	-43	161	-49
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document.write( "		21	-154	49
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document.write( "	3	-22	7	 0
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document.write( "    3x^2 - 22 + 7
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document.write( "   ( 3x - 1)(x - 7) = 0 , x = 7 , x = 1/3
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document.write( "f(x)=3x^3-43x^2+161x-49 = (x-7)(x-7)(x - 1/3)\r
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document.write( "f(x)=2x^3-4x^2-34x+68  x = 2 is a root
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document.write( "Using synthetic Division to find the other two
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document.write( "2	2	-4	-34	68
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document.write( "		4	0	-68
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document.write( "	2	0	-34	0
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document.write( "        2x^2 = 34 = 0
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document.write( "         x = ± √17
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document.write( "f(x)=2x^3-4x^2-34x+68 = (x-2)(x + √17)( x -√17)  \n" );
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