document.write( "Question 985639: Factor the polynomial P(x). Then solve the equation P(x)=0.
\n" ); document.write( "1.P(x)=x^3+4x^2+x-6
\n" ); document.write( "2.P(x)=x^3-6x^2-x+6
\n" ); document.write( "3.P(x)=x^3-x^2-x+1
\n" ); document.write( "4.P(x)=2x^3-3x^2-3x+2 \n" ); document.write( "
Algebra.Com's Answer #606472 by ikleyn(52781)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "Factor the polynomial  P(x).  Then solve the equation  P(x)=0.\r
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\n" ); document.write( "\n" ); document.write( "1.  P(x) = \"x%5E3%2B4x%5E2%2Bx-6\"
\n" ); document.write( "2.  P(x) = \"x%5E3-6x%5E2-x%2B6\"
\n" ); document.write( "3.  P(x) = \"x%5E3-x%5E2-x%2B1\"
\n" ); document.write( "4.  P(x) = \"2x%5E3-3x%5E2-3x%2B2\"
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\n" ); document.write( "\n" ); document.write( "1.  P(x) = \"x%5E3+%2B+4x%5E2+%2B+x+-+6\".\r
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\n" ); document.write( "\n" ); document.write( " The integer  1  is the root.  Indeed,  P(1) = \"1%5E3+%2B+4%2A1%5E2+%2B+1+-+6\" = \"1+%2B+4+%2B+1+-+6\" = \"0\".\r
\n" ); document.write( "\n" ); document.write( "It means that the binomial  (x-1)  divides the polynomial  P(x):  P(x) = (x-1)*Q(x),  where  Q(x)  is a quadratic polynomial.  (See the  Remainder Theorem  in the lesson \r
\n" ); document.write( "\n" ); document.write( "Divisibility of polynomial f(x) by binomial x-a  in this site). \r
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\n" ); document.write( "\n" ); document.write( "If you make the long division,  you will get  Q(x) = \"x%5E2+%2B+5x+%2B+6\". \r
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\n" ); document.write( "\n" ); document.write( "The quadratic polynomial  Q(x)  has the roots  -2  and  -3:  Q(-2) = 0  and  Q(-3) = 0.\r
\n" ); document.write( "\n" ); document.write( "You can use the quadratic formula to find the roots  (see the lesson  Introduction into Quadratic Equations  in this site)  or \r
\n" ); document.write( "\n" ); document.write( "the  Vieta's Theorem  (see the lesson  Solving quadratic equations without quadratic formula).  You also can check it directly. \r
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\n" ); document.write( "\n" ); document.write( "This means that the binomials  (x+2)  and  (x+3)  divide the polynomial  Q(x),  so  Q(x) = (x+2)*(x+3). \r
\n" ); document.write( "\n" ); document.write( "You also can check this factorization immediately. \r
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\n" ); document.write( "\n" ); document.write( "It implies that the polynomial  P(x)  has the factorization  P(x) = (x-1)*(x+2)*(x+3). \r
\n" ); document.write( "\n" ); document.write( "Hence,  its roots are  1,  -2 and  -3.\r
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\n" ); document.write( "\n" ); document.write( "2.  P(x) = \"x%5E3+-+6x%5E2+-+x+%2B+6\". \r
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\n" ); document.write( "\n" ); document.write( "Re-group:  P(x) = \"%28x%5E3+-+6x%5E2%29\" - \"%28x+-+6%29\" = \"x%5E2%2A%28x-6%29+-+%28x-6%29\" = \"%28x-6%29%2A%28x%5E2-1%29\" = (x-6)*(x-1)*(x+1).\r
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\n" ); document.write( "\n" ); document.write( "The roots of the polynomial  P(x)  are  6,  1  and  -1. \r
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\n" ); document.write( "\n" ); document.write( "3.  P(x) = \"x%5E3+-+x%5E2+-+x+%2B+1\". \r
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\n" ); document.write( "\n" ); document.write( "Re-group:  P(x) = \"%28x%5E3+-+x%5E2%29\" - \"%28x+-+1%29\" = \"x%5E2%2A%28x-1%29+-+%28x-1%29\" = \"%28x-1%29%2A%28x%5E2-1%29\" = (x-1)*(x-1)*(x+1).\r
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\n" ); document.write( "\n" ); document.write( "The roots of the polynomial  P(x)  are  1  (multiplicity 2),  and  -1. \r
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\n" ); document.write( "\n" ); document.write( "4. P(x) = \"2x%5E3+-+3x%5E2+-+3x+%2B+2\". \r
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\n" ); document.write( "\n" ); document.write( "One root is  2:  P(2) = 0  (check it yourself). \r
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\n" ); document.write( "\n" ); document.write( "Hence,  P(x)  is divided by  (x-2):  P(x) = (x-2)*Q(x),  where  Q(x)  is a quadratic polynomial.  (By the same reason as in the  n.1  above). \r
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\n" ); document.write( "\n" ); document.write( "Long division gives  Q(x) = \"2x%5E2+%2B+x+-+1\". \r
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\n" ); document.write( "\n" ); document.write( "Hence,  P(x) = \"%28x-2%29%2A%282x%5E2+%2B+x+-+1%29\". \r
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\n" ); document.write( "\n" ); document.write( "The quadratic polynomial  \"2x%5E2+%2B+x+-+1\"  has the roots  -1  and  \"1%2F2\".  You can find them using the same methods as in the  n.1  above.\r
\n" ); document.write( "\n" ); document.write( "So,  Q(x) = \"2x%5E2+%2B+x+-+1\" = \"2%2A%28x%2B1%29%2A%28x-1%2F2%29\". \r
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\n" ); document.write( "\n" ); document.write( "Thus  P(x) = \"2%2A%28x-2%29%2A%28x%2B1%29%2A%28x-1%2F2%29\"  is the final factorization of the polynomial  P(x)  over the real domain. \r
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\n" ); document.write( "\n" ); document.write( "The polynomial  P(x)  has the roots  2,  -1  and  \"1%2F2\".\r
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