Algebra.Com's Answer #137400 by stanbon(75887)![\"\"](\"/images/stars/stars-13.gif\")  
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Find all rational zeros of the polynomial (using synthetic division)
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document.write( "22. p(x)=  \r
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document.write( "What I did: multiples of 1= +-1
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document.write( " multiples of 4= +-1 +-4 +-2
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document.write( "possible zeros: +-1 +-4 +-2 (after dividing constant over leading coeficient)
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document.write( "I used synthetic division to test which are zeros. +1 worked, +4 did not, +2 using the quotient of +1 (1 -1 -4 4) did not work gave me remainder of 8, but when I used the original coeficients (1 -2 -3 8 -4)it gave me a zero. I was told by my professor that either way it should work.
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document.write( "Since the coefficients add up to zero, x = 1 is a zero.\r
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document.write( "1)....1....-2....-3....8....-4
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document.write( ".......1....-1....-4...4..|..0\r
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document.write( "Since the quotient coefficients add up to zero. x = 1 is AGAIN a zero.\r
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document.write( "1)....1....-1....-4....4
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document.write( ".......1.....0....-4...|..0\r
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document.write( "Now you have a quadratic which says x^2-4 = 0
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document.write( "Factor to get (x-2)(x+2)=0
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document.write( "x = 2 or x = -2
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document.write( "The 4 zeros are 1,1,2,-2
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document.write( "Cheers,
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document.write( "Stan H. \n" );
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