document.write( "Question 1045631: How to factor8x^4-14x^2+3 \n" ); document.write( "
Algebra.Com's Answer #661110 by ikleyn(52858)\"\" \"About 
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document.write( "1.  Introduce new variable  u = \"x%5E2\".\r\n" );
document.write( "    Then the polynomial takes the form  \"8u%5E2-14u%2B3\".\r\n" );
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document.write( "2.  Find the roots of the polynomial  \"8u%5E2-14u%2B3\".\r\n" );
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document.write( "    For it, solve the quadratic equation  \"8u%5E2-14u%2B3\" = \"0\".\r\n" );
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document.write( "    Use the quadratic formula. You will get \r\n" );
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document.write( "    \"u%5B1%2C2%5D\" = \"%2814+%2B-+sqrt%2814%5E2+-+4%2A8%2A3%29%29%2F%282%2A8%29\" = \"%2814+%2B-+10%29%2F16\".\r\n" );
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document.write( "    So,  \"u%5B1%5D\" = \"3%2F2\"  and  \"u%5B2%5D\" = \"1%2F4\".\r\n" );
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document.write( "3.  Hence, the quadratic polynomial  \"8u%5E2-14u%2B3\"  is\r\n" );
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document.write( "    \"8u%5E2-14u%2B3\" = \"8%2A%28u-3%2F2%29%2A%28u-1%2F4%29\" = (2u-3)*(4u-1).\r\n" );
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document.write( "    Correspondingly,  the  original polynomial factorization is \r\n" );
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document.write( "    \"8x%5E4-14x%5E2%2B3\" = \"%282x%5E2-3%29%2A%284x%5E2-1%29\".\r\n" );
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document.write( "4.  Next, you can factor further each of the two factors above to get\r\n" );
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document.write( "    \"8x%5E4-14x%5E2%2B3\" = .\r\n" );
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document.write( "It is what you were asking for.\r\n" );
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