document.write( "Question 139381: y^3 =125 solve the equation for y. seeing if i am doing this right. my answer was y=5 \n" ); document.write( "
Algebra.Com's Answer #101647 by solver91311(24713)\"\" \"About 
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Your answer is correct -- as far as it goes. If your assignment is to find just the real number roots, then you are done. But if you were asked to completely solve the equation, including finding any complex number roots, you are a long way from done.\r
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\n" ); document.write( "\n" ); document.write( "The Fundamental Theorem of Algebra tells us a couple of important things in this context. First, any polynomial equation of degree n has exactly n roots. You have a 3rd degree equation, so there must be 3 roots. Furthermore, if \"a\" is a root of the polynomial equation, then \"x-a\" is a factor of the polynomial. That is to say that, given a polynomial function \"P%28x%29\", if \"P%28a%29=0\", then there exists a polynomial function \"Q%28x%29\" such that \"P%28x%29=%28x-a%29Q%28x%29\".\r
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\n" ); document.write( "\n" ); document.write( "Knowing that 5 is one of the roots of your equation, and we know this because \"P%28y%29=y%5E3-125\" and \"P%285%29=0\", makes the task of finding the other two roots much simpler. Solving the general cubic is a horror that I wouldn't wish on anyone, even if I particularly disliked them.\r
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\n" ); document.write( "\n" ); document.write( "We can use the fact that \"y-5\" is a factor of \"P%28y%29=y%5E3-125\" and the process of polynomial long division to find the other factor of \"P%28y%29\" which will be a quadratic polynomial. But first we need to put \"P%28y%29\" into standard form by putting in placeholders for the zero coefficient terms, thus:
\n" ); document.write( "\"P%28y%29=y%5E3%2B0y%5E2%2B0y-125\"\r
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\n" ); document.write( "\n" ); document.write( "\"y\" goes into \"y%5E3\" \"y%5E2\" times. \"y%5E2\" is the first term of the quotient.\r
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\n" ); document.write( "\n" ); document.write( "\"y%5E2\" times \"y-5\" is \"y%5E3-5y%5E2\". Subtract this from the first two terms of \"P%28y%29\" to get \"5y%5E2\" (remember: change the sign and add).\r
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\n" ); document.write( "\n" ); document.write( "Bring down the \"0y\".\r
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\n" ); document.write( "\n" ); document.write( "\"y\" goes into \"5y%5E2\" \"5y\" times. \"5y\" is the second term of the quotient.\r
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\n" ); document.write( "\n" ); document.write( "\"5y\" times \"y-5\" is \"5y%5E2-25y\". Subtract this from \"5y%5E2%2B0y\" to get \"25y\".\r
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\n" ); document.write( "\n" ); document.write( "Bring down the \"-125\".\r
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\n" ); document.write( "\n" ); document.write( "\"y\" goes into \"25y\" \"25\" times. \"25\" is the third term of the quotient.\r
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\n" ); document.write( "\n" ); document.write( "\"25\" times \"y-5\" is \"25y-125\". Subtract this from \"25y-125\" to get \"0\" -- meaning no remainder.\r
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\n" ); document.write( "\n" ); document.write( "The quotient polynomial is then \"Q%28y%29=y%5E2%2B5y%2B25\", and the zeros of this polynomial are the other two zeros of your original polynomial.\r
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\n" ); document.write( "\n" ); document.write( "Using the quadratic formula:
\n" ); document.write( "\"y+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"\r
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\n" ); document.write( "\n" ); document.write( "\"y+=+%28-5+%2B-+sqrt%28+%285%29%5E2-4%2A%281%29%2A%2825%29+%29%29%2F%282%2A%281%29%29+\"\r
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\n" ); document.write( "\n" ); document.write( "\"y+=+%28-5+%2B-+sqrt%28+25-100+%29%29%2F2+\"\r
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\n" ); document.write( "\n" ); document.write( "\"y+=+%28-5+%2B-+sqrt%28+-75+%29%29%2F2+\"\r
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\n" ); document.write( "\n" ); document.write( "\"y+=+%28-5+%2B+5i%2Asqrt%283%29%29%2F2+\" or \"y+=+%28-5+-+5i%2Asqrt%283%29%29%2F2+\"\r
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\n" ); document.write( "\n" ); document.write( "Where \"i\" is the imaginary number defined by \"i%5E2=-1\"\r
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\n" ); document.write( "\n" ); document.write( "In summary, your three roots are:\r
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\n" ); document.write( "\n" ); document.write( "\"y%5B1%5D=5\", \"y%5B2%5D+=+%28-5+%2B+5i%2Asqrt%283%29%29%2F2+\", \"y%5B3%5D+=+%28-5+-+5i%2Asqrt%283%29%29%2F2+\"
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