document.write( "Question 711966: I have to write a polynomial Function using these roots
\n" ); document.write( "-1,-1,1
\n" ); document.write( "How do I do this?? \n" ); document.write( "
Algebra.Com's Answer #437775 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
In general, a polynomial that has r as a root will have a factor of (x-r). So the polynomial we are looking for will have factors of:
\n" ); document.write( "\"P%28x%29+=+%28x-%28-1%29%29%28x-%28-1%29%29%28x-1%29\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"P%28x%29+=+%28x%2B1%29%28x%2B1%29%28x-1%29\"

\n" ); document.write( "Now all we have to is multiply this out. We can take advantage of the \"%28a%2Bb%29%5E2+=+a%5E2%2B2ab%2Bb%5E2\" pattern to multiply the first two factors. Or we can take advantage of the \"%28a%2Bb%29%28a-b%29+=+a%5E2-b%5E2\" pattern to multiply the last two factors. (Remember, with multiplication the order does not matter!) Since the second pattern has a simpler result I'm going to use it to multiply the last two factors:
\n" ); document.write( "\"P%28x%29+=+%28x%2B1%29%28%28x%29%5E2-%281%29%5E2%29\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"P%28x%29+=+%28x%2B1%29%28x%5E2-1%29\"
\n" ); document.write( "Now we multiply the remaining factors. They do not fit any pattern so we must resort to FOIL:
\n" ); document.write( "\"P%28x%29+=+x%2Ax%5E2-x%2A1%2B1%2Ax%5E2-1%2A1\"
\n" ); document.write( "which simplifies...
\n" ); document.write( "\"P%28x%29+=+x%5E3-x%2Bx%5E2-1\"
\n" ); document.write( "Rearranging the terms so they are in standard form:
\n" ); document.write( "\"P%28x%29+=+x%5E3%2Bx%5E2-x-1\"

\n" ); document.write( "P.S. When I described the factored form of a polynomial, I left out a minor detail. The general factored form of a polynomial is:
\n" ); document.write( "P(x) = a * (x-r) ... (with as many (x-r) factors as there are roots).
\n" ); document.write( "The \"a\" can be any non-zero number and it is the part I left out. The answer we got while ignoring the \"a\" has, in effect, an \"a\" of 1. Fell free to re-do the solution and pick a different \"a\". \n" ); document.write( "
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