document.write( "Question 117516:
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Algebra.Com's Answer #85518 by jim_thompson5910(35256)\"\" \"About 
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\"a%2Bb%2Bc%2Bd=0\" Start with the given equation\r
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\n" ); document.write( "\n" ); document.write( "\"d=-a-b-c\" Solve for d\r
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\n" ); document.write( "\n" ); document.write( "\"ax%5E3%2Bbx%5E2%2Bcx-a-b-c=0\" Plug in \"d=-a-b-c\" into \"ax%5E3%2Bbx%5E2%2Bcx%2Bd=0\" \r
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\n" ); document.write( "\n" ); document.write( "\"%28ax%5E3-a%29%2B%28bx%5E2-b%29%2B%28cx-c%29=0\" Group like terms\r
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\n" ); document.write( "\n" ); document.write( "\"a%28x%5E3-1%29%2Bb%28x%5E2-1%29%2Bc%28x-1%29=0\" Factor out the GCF from each individual group\r
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\n" ); document.write( "\n" ); document.write( "\"a%28x-1%29%28x%5E2%2Bx%2B1%29%2Bb%28x%5E2-1%29%2Bc%28x-1%29=0\" Factor \"x%5E3-1\" by using the difference of cubes formula to get \"%28x-1%29%28x%5E2%2Bx%2B1%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"a%28x-1%29%28x%5E2%2Bx%2B1%29%2Bb%28x-1%29%28x%2B1%29%2Bc%28x-1%29=0\" Factor \"x%5E2-1\" by using the difference of squares formula to get \"%28x-1%29%28x%2B1%29\"\r
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\n" ); document.write( "\n" ); document.write( "Notice how we have the common term \"x-1\". We can factor this term out.\r
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\n" ); document.write( "\n" ); document.write( "\"%28x-1%29%28a%28x%5E2%2Bx%2B1%29%2Bb%28x%2B1%29%2Bc%29=0\" Factor out the GCF \"x-1\"\r
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\n" ); document.write( "\n" ); document.write( "Now if we let \"H=x-1\" and \"K=a%28x%5E2%2Bx%2B1%29%2Bb%28x%2B1%29%2Bc\" we'll get\r
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\n" ); document.write( "\n" ); document.write( "\"H%2AK=0\"\r
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\n" ); document.write( "\n" ); document.write( "So by the zero product property we get\r
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\n" ); document.write( "\n" ); document.write( "\"H=0\" or \"K=0\"\r
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\n" ); document.write( "\n" ); document.write( "but since \"H=x-1\", this means\r
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\n" ); document.write( "\n" ); document.write( "\"x-1=0\"\r
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\n" ); document.write( "\n" ); document.write( "Now solve for x\r
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\n" ); document.write( "\n" ); document.write( "\"x=1\"\r
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\n" ); document.write( "\n" ); document.write( "So \"x=1\" is a zero of \"ax%5E3%2Bbx%5E2%2Bcx%2Bd=0\" if the coefficients satisfy the equation \"a%2Bb%2Bc%2Bd=0\"\r
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\n" ); document.write( "\n" ); document.write( "Check:\r
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\n" ); document.write( "\n" ); document.write( "Unfortunately, there is no formal check for this type of problem. But if we make sure that the coefficients satisfy the equation \"a%2Bb%2Bc%2Bd=0\", then we can graph the equation and see that one root is \"x=1\". For instance, let \"a=b=1\" and \"c=d=-1\" so we'll get\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E3%2Bx%5E2-x-1\" (notice how \"1%2B1-1-1=2-2=0\" which satisfies the equation \"a%2Bb%2Bc%2Bd=0\")\r
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\n" ); document.write( "\n" ); document.write( "Now graph the polynomial to get\r
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\n" ); document.write( "\n" ); document.write( "\"+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E3%2Bx%5E2-x-1%29+\"\r
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\n" ); document.write( "\n" ); document.write( "and we can see that one root is \"x=1\"\r
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\n" ); document.write( "\n" ); document.write( "If we try different values of a,b,c, and d that will satisfy the equation \"a%2Bb%2Bc%2Bd=0\", we'll see that \"x=1\" is a root every time. So in a way, our answer has been verified.
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