document.write( "Question 694376: x^4-2x^3+3x^2-2x+1\r
\n" ); document.write( "\n" ); document.write( "How to solve without using trial and error method ?? \n" ); document.write( "

Algebra.Com's Answer #427953 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"P%28x%29=x%5E4-2x%5E3%2B3x%5E2-2x%2B1\"$
\n" ); document.write( "Maybe you just wanted to factor the polynomial $\"P%28x%29\"$ .
\n" ); document.write( "Or maybe you wanted to find its zeros, to solve $\"P%28x%29=0\"$
\n" ); document.write( "I'll try both.
\n" ); document.write( "
\n" ); document.write( "FINDING ZEROS:
\n" ); document.write( "We can use a trick that I (dimly) remember from high school.
\n" ); document.write( "Obviously $\"P%280%29=1\"$ , so $\"0\"$ is not a zero of the polynomial.
\n" ); document.write( "We can divide by $\"x\"$ (or by $\"x%5E2\"$ ) without fear.
\n" ); document.write( "When looking for zeros,
\n" ); document.write( " $\"x%5E4-2x%5E3%2B3x%5E2-2x%2B1=0\"$ <--> $\"%28x%5E4-2x%5E3%2B3x%5E2-2x%2B1%29%2Fx%5E2=0\"$ <--> $\"x%5E2-2x%2B3-2%281%2Fx%29%2B1%2Fx%5E2=0\"$
\n" ); document.write( "We can do a change of variable.
\n" ); document.write( "If we define $\"y=x%2B1%2Fx\"$ , then $\"y%5E2=x%5E2%2B1%2Fx%5E2%2B2\"$ <--> $\"x%5E2%2B1%2Fx%5E2=y%5E2-2\"$
\n" ); document.write( "We can re-write $\"x%5E2-2x%2B3-2%281%2Fx%29%2B1%2Fx%5E2=0\"$ in terms of $\"y\"$ as
\n" ); document.write( " $\"y%5E2-2%2B2y%2B3=0\"$ <--> $\"y%5E2%2B2y%2B1=0\"$ <--> $\"%28y%2B1%29%5E2=0\"$ <--> $\"highlight%28y=-1%29\"$
\n" ); document.write( "Going back to $\"x\"$ , remembering we defined $\"y=x%2B1%2Fx\"$ ,
\n" ); document.write( " $\"x%2B1%2Fx=-1\"$ , and multiplying by $\"x\"$ we get
\n" ); document.write( " $\"x%5E2%2B1=-x\"$ <--> $\"x%5E2%2Bx%2B1=0\"$
\n" ); document.write( "Applying the quadratic formula we solve to get
\n" ); document.write( "

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\n" ); document.write( "\n" ); document.write( "FACTORING (no tricks):
\n" ); document.write( "We realize that it is so very symmetrical that if some non-zero $\"a\"$ is a zero of $\"P%28x%29\"$
\n" ); document.write( " $\"P%28a%29=a%5E4-2a%5E3%2B3a%5E2-2a%2B1=0\"$ --> $\"P%28a%29%2Fa%5E4=a%5E4%2Fa%5E4-2a%5E3%2Fa%5E4%2B3a%5E2%2Fa%5E4-2a%2Fa%5E4%2B1%2Fa%5E4=0\"$ --> $\"1-2%2Fa%2B3%2Fa%5E2-2%2Fa%5E3%2B1%2Fa%5E4=0\"$ --> $\"1-2%281%2Fa%29%2B3%281%2Fa%29%5E2-2%281%2Fa%29%5E3%2B%281%2Fa%29%5E4=P%281%2Fa%29=0\"$
\n" ); document.write( "So the 4 complex zeros of the polynomial will come in pairs $\"a\"$ with $\"1%2Fa\"$ and $\"b\"$ with $\"1%2Fb\"$ ,
\n" ); document.write( "and the full factoring of the polynomial would be
\n" ); document.write( " $\"P%28x%29=%28x-a%29%28x-1%2Fa%29%28x-b%29%28x-1%2Fb%29\"$
\n" ); document.write( "Multiplying pairs we get
\n" ); document.write( " $\"P%28x%29=%28x%5E2-%28a%2B1%2Fa%29x%2B1%29%28x%5E2-%28b%2B1%2Fb%29x%2B1%29\"$
\n" ); document.write( "Maybe we could find $\"m=a%2B1%2Fa\"$ and $\"n=b%2B1%2Fb\"$ and have a partial factoring with real coefficients.
\n" ); document.write( "It would be
\n" ); document.write( "

\n" ); document.write( "If it must be

for all $\"x\"$ ,
\n" ); document.write( "then $\"m%2Bn=2\"$ and $\"mn%2B2=3\"$ <--> $\"mn=1\"$
\n" ); document.write( "The solution to $\"system+%28m%2Bn=2%2Cmn=1%29\"$ is $\"m=n=1\"$ ,
\n" ); document.write( "so substituting into $\"P%28x%29=%28x%5E2-mx%2B1%29%28x%5E2-nx%2B1%29\"$
\n" ); document.write( "the factoring with real coefficients is
\n" ); document.write( " $\"highlight%28P%28x%29=%28x%5E2-x%2B1%29%28x%5E2-x%2B1%29%29\"$ or $\"highlight%28P%28x%29=%28x%5E2-x%2B1%29%5E2%29\"$
\n" ); document.write( "The factoring with real coefficients is done.
\n" ); document.write( "To factor further we need to go beyond real numbers.
\n" ); document.write( "Since $\"x%5E2-x%2B1%2B0\"$ does not have real solutions,
\n" ); document.write( "any linear $\"%28x-a%29\"$ factor would have an imaginary $\"a\"$ coefficient. \n" ); document.write( "

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