document.write( "Question 547646: F(x)= x^4-6x^3+7x^2+6x-8 .....find all the zeros of the polynomial function \n" ); document.write( "

Algebra.Com's Answer #357022 by KMST(5328) $\"\"$ $\"About$

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If the zeros are rational, they would be fractions whose numerator is a factor of 8 (the constant term) and whose denominator is a factor of 1 (the leading coeficient. The choices are 1, -1, 2, -2, 4, -4, 8, and -8.
\n" ); document.write( "We can see that
\n" ); document.write( " $\"F%281%29=+1%5E4-6%2A1%5E3%2B7%2A1%5E2%2B6%2A1-8=1-6%2B7%2B6-8=0\"$ and
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\n" ); document.write( "So, we can divide the polynomial by $\"x-1\"$ and $\"x=1\"$ , or by their product
\n" ); document.write( " $\"%28x-1%29%28x%2B1%29=x%5E2-1\"$
\n" ); document.write( "to get a second degree polynomial, whose roots (real or not) we know we can find.
\n" ); document.write( "Alternately, we can try our luck with the other possible integer roots.
\n" ); document.write( "Dividing, by whatever method we choose, we get
\n" ); document.write( " $\"%28x%5E4-6x%5E3%2B7x%5E2%2B6x-8%29%2F%28%28x-1%29%28x%2B1%29%29=x%5E2-6x%2B8\"$
\n" ); document.write( "That second degree polynomial is factoring friendly, and we easily see that
\n" ); document.write( " $\"x%5E2-6x%2B8=%28x-2%29%28x-4%29\"$
\n" ); document.write( "So we've found all four roots: -1, 1, 2, and 4,
\n" ); document.write( "and they were all integers.
\n" ); document.write( "A meaner problem would have you ending up with a second degree polynomials with irrational roots, or with no real roots. \n" ); document.write( "

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