document.write( "Question 93571: Can you solve by factoring the quadratic\r
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Algebra.Com's Answer #68150 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Can you solve by factoring the quadratic\r\n" );
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document.write( "5y³ + 34y² = 7y \r\n" );
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document.write( "and show every step?\r\n" );
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document.write( "     5y³ + 34y² = 7y\r\n" );
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document.write( "Get 0 on the right by adding -7y to both sides\r\n" );
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document.write( "5y³ + 34y² - 7y = 0\r\n" );
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document.write( "y is a common factor so factor out y\r\n" );
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document.write( "y(5y² + 34y - 7) = 0\r\n" );
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document.write( "Write this:\r\n" );
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document.write( "y( y    )( y    ) = 0\r\n" );
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document.write( "Now think of a way to write 5, the first\r\n" );
document.write( "coefficient as the product of two positive \r\n" );
document.write( "integers.  Since 5 is prime there is only \r\n" );
document.write( "one way, namely 1×5 or 5×1.  So put a 1 \r\n" );
document.write( "before the first y and a 5 before the second y.  \r\n" );
document.write( "(It would be OK to reverse these).\r\n" );
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document.write( "y(1y    )(5y    ) = 0\r\n" );
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document.write( "Now think of a way to write 7, the last\r\n" );
document.write( "term, as the product of two positive \r\n" );
document.write( "integers.  Since 7 is also prime there is \r\n" );
document.write( "only one way, namely 1×7 or 7×1.  So try \r\n" );
document.write( "putting a 1 at the end of the first\r\n" );
document.write( "parentheses and a 7 at the end of the \r\n" );
document.write( "second parentheses: \r\n" );
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document.write( "y(1y   1)(5y   7) = 0\r\n" );
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document.write( "Now check the OUTERS and INNERS as though\r\n" );
document.write( "you were going to multiply that out by FOIL,\r\n" );
document.write( "ignoring the signs.  The OUTER product would\r\n" );
document.write( "be 1y times 7 or 7y and the INNER product \r\n" );
document.write( "would be 1 times 5y or 5y.\r\n" );
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document.write( "But there is no way that 7y and 5y could \r\n" );
document.write( "combine to give the mddle term +34y no matter\r\n" );
document.write( "what signs we gave them.\r\n" );
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document.write( "So let's discard this\r\n" );
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document.write( "y(1y   1)(5y   7) = 0\r\n" );
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document.write( "and reverse the 1 and the 7.\r\n" );
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document.write( "y(1y   7)(5y   1) = 0 \r\n" );
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document.write( "Now again check the OUTERS and INNERS as though\r\n" );
document.write( "you were going to multiply that out by FOIL,\r\n" );
document.write( "ignoring the signs.  The OUTER product would\r\n" );
document.write( "be 1y times 1 or 1y and the INNER product \r\n" );
document.write( "would be 7 times 5y or 35y.\r\n" );
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document.write( "This time there IS A way that 1y and 35y could \r\n" );
document.write( "combine to give the middle term +34y.  That would\r\n" );
document.write( "be if we gave the 35y a positive sign and the 1y a\r\n" );
document.write( "negative sign, since +35y-1y does give the middle\r\n" );
document.write( "term +34y. So we put a + by the 7 to make the INNERS\r\n" );
document.write( "be +35y and we put a - by the 1 in the second \r\n" );
document.write( "parentheses to make the OUTERS be -1y\r\n" );
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document.write( "y(1y + 7)(5y - 1) = 0\r\n" );
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document.write( "Of course we can erase the 1 before the y in \r\n" );
document.write( "the first parentheses:\r\n" );
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document.write( "y(y + 7)(5y - 1) = 0\r\n" );
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document.write( "So the left side is now completely factored.\r\n" );
document.write( "The factors are y, y + 7, and 5y - 1\r\n" );
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document.write( "We set each factor = 0\r\n" );
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document.write( "Setting the factor y = 0 just gives y = 0\r\n" );
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document.write( "Setting the factor y + 7 = 0, gives y = -7\r\n" );
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document.write( "Setting 5y - 1 = 0 gives 5y = 1 or y = 1/5 \r\n" );
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document.write( "So there are three solutions: 0, -7, and 1/5\r\n" );
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document.write( "Edwin

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