document.write( "Question 778860: Using the factor theorem , factorise the polynomial 2y^3+y^2-2y-1 \n" ); document.write( "

Algebra.Com's Answer #474926 by Edwin McCravy(20055) $\"\"$ $\"About$

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document.write( "2y³+y²-2y-1\r\n" );
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document.write( "Using the factor theorem:\r\n" );
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document.write( "If r is a root then (y-r) is a factor\r\n" );
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document.write( "The candidates for the roots are ± the factors each of whose\r\n" );
document.write( "numerator is a factor of 1, the constant term, and whose \r\n" );
document.write( "denominator is a factor of 2, the coefficient of the term\r\n" );
document.write( "with the largest power of the variable, 2y³.\r\n" );
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document.write( "The candidates for roots are ±1, \r\n" );
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document.write( "We try 1 using synthetic division:\r\n" );
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document.write( "1|2  1  -2 -1\r\n" );
document.write( " |   2   3  1\r\n" );
document.write( "  2  3   1  0\r\n" );
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document.write( "The remainder is 0, so 1 is a root and we have now\r\n" );
document.write( "factorised 2y³+y²-2y-1 as\r\n" );
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document.write( "(y-1)(2y²+3y+1)\r\n" );
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document.write( "We now factorise the trinomial 2x²+3x+1 as (2x+1)(x+1)\r\n" );
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document.write( "(y-1)(2y+1)(y+1) \r\n" );
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document.write( "Without using the factor theorem:\r\n" );
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document.write( "2y³+y²-2y-1\r\n" );
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document.write( "Factorise out y² from the first two terms\r\n" );
document.write( "Factorise out -1 from the last two terms\r\n" );
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document.write( "y²(2y+1)-1(2y+1)\r\n" );
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document.write( "Factorise out (2y+1)\r\n" );
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document.write( "(2y+1)(y²-1)\r\n" );
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document.write( "Factorise y²-1 as the difference of squares (y-1)(y+1)\r\n" );
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document.write( "(2y+1)(y-1)(y+1)\r\n" );
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document.write( "Edwin

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