document.write( "Question 308790: If one of the zeros of y=x^3+bx+1 is 1, determine the value of b. then solve x^3+bx+1=0 express your answer as exect values. \n" ); document.write( "

Algebra.Com's Answer #220792 by Fombitz(32388) $\"\"$ $\"About$

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$\"0=1%5E3%2Bb%281%29%2B1\"$
\n" ); document.write( " $\"b%2B2=0\"$
\n" ); document.write( " $\"b=-2\"$
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\n" ); document.write( "Then solve for
\n" ); document.write( " $\"%28x%5E3-2x%2B1%29%2F%28x-1%29\"$
\n" ); document.write( "The first multiplier would be $\"x%5E2\"$ and yield $\"x%5E2%28x-1%29=x%5E3-x%5E2\"$ .
\n" ); document.write( "Subtracting from the polynomial,
\n" ); document.write( " $\"%28x%5E3-2x%2B1%29-%28x%5E3-x%5E2%29=-x%5E2-2x%2B1\"$
\n" ); document.write( "The second multiplier would be $\"-x\"$ and yield $\"-x%28x-1%29=-x%5E2%2Bx\"$
\n" ); document.write( " $\"%28-x%5E2-2x%2B1%29-%28-x%5E2%2Bx%29=-x%2B1\"$
\n" ); document.write( "The final multiplier would be $\"-1\"$ and yield $\"-1%28x-1%29=-x%2B1\"$
\n" ); document.write( " $\"%28-x%2B1%29-%28-x%2B1%29=0\"$
\n" ); document.write( "So then,
\n" ); document.write( " $\"%28x%5E3-2x%2B1%29%2F%28x-1%29=x%5E2-x-1\"$
\n" ); document.write( "Use the quadratic formula to solve for these roots.
\n" ); document.write( " $\"x+=+%28-%28-1%29+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-1%29+%29%29%2F%282%2A1%29+\"$
\n" ); document.write( " $\"x+=+%281+%2B-+sqrt%28+5+%29%29%2F%282%29+\"$
\n" ); document.write( "The three roots of $\"x%5E3-2x%2B1\"$ are then
\n" ); document.write( "( $\"1\"$ , $\"%281%2Bsqrt%285%29%29%2F2\"$ , $\"%281-sqrt%285%29%29%2F2\"$ )
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