document.write( "Question 986182: Which number is a root of f(x)=x^3+6x^2+9x that has multiplicity 1?\r
\n" ); document.write( "\n" ); document.write( " a. 3
\n" ); document.write( " b. 1
\n" ); document.write( " c. 0
\n" ); document.write( " d. –3\r
\n" ); document.write( "\n" ); document.write( "Do you just substitute one of the numbers for x?
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Algebra.Com's Answer #606983 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Replace f(x) with 0, then factor\r
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\n" ); document.write( "\n" ); document.write( "f(x)=x^3+6x^2+9x
\n" ); document.write( "0=x^3+6x^2+9x
\n" ); document.write( "x^3+6x^2+9x = 0
\n" ); document.write( "x(x^2+6x+9) = 0
\n" ); document.write( "x(x+3)(x+3) = 0\r
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\n" ); document.write( "\n" ); document.write( "Then use the zero product property to break up x(x+3)(x+3) = 0 into these three equations\r
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\n" ); document.write( "\n" ); document.write( "x=0 or x+3=0 or x+3=0\r
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\n" ); document.write( "\n" ); document.write( "The first equation x=0 means we have a root of 0. This root is of multiplicity 1\r
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\n" ); document.write( "\n" ); document.write( "The other two equations are exact duplicates of one another. They both lead to x=-3 being the other root of multiplicity 2.\r
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\n" ); document.write( "\n" ); document.write( "So in the end, the answer is c. 0\r
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\n" ); document.write( "\n" ); document.write( "The graph confirms this. Notice how we have a double root at x = -3. It doesn't cross over the x axis as it only touches the x axis here. Contrast that with the root at x = 0 where it crosses through the x axis one time.\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+0%2Cx%5E3%2B6x%5E2%2B9x%29+\"
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