document.write( "Question 717927: find the real zeros of the function. f(x)=x^8+8x^7-28x^6-56x^5+70x^4+56x^3-28x^2-8x+1 \n" ); document.write( "

Algebra.Com's Answer #440597 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
Either there is a mistake in
\n" ); document.write( " $\"f%28x%29=x%5E8%2B8x%5E7-28x%5E6-56x%5E5%2B70x%5E4%2B56x%5E3-28x%5E2-8x%2B1\"$
\n" ); document.write( "or I don't believe it is possible to find the exact real roots for this polynomial.

\n" ); document.write( "If what you posted is correct, then there are no rational roots. The only possible rational roots are 1 and -1 and neither of them work out to be roots. So the only real roots of this function would be irrational. The best one could do, I believe, is to use a graphing calculator:

Graph $\"y=x%5E8%2B8x%5E7-28x%5E6-56x%5E5%2B70x%5E4%2B56x%5E3-28x%5E2-8x%2B1\"$
Then use the trace function to find decimal approximations for the x-coordinates of all the places where the graph intersects the x-axis. (If the graph never intersects the x-axis then there are no real roots.) I'll show you a graph at then end.

\n" ); document.write( "I suspect that the function is supposed to be:
\n" ); document.write( " $\"f%28x%29=x%5E8-8x%5E7%2B28x%5E6-56x%5E5%2B70x%5E4-56x%5E3%2B28x%5E2-8x%2B1\"$
\n" ); document.write( "Again, the only possible rational roots are 1 and -1. But this time 1 is actually a root. Here's the synthetic division:
\n" ); document.write( "

\r\n" );
document.write( "1  |   1   -8   28   -56   70   -56   28   -8   1\r\n" );
document.write( "----        1   -7    21  -35    35  -21    7  -1\r\n" );
document.write( "      --------------------------------------------\r\n" );
document.write( "       1   -7   21   -35   35   -21    7   -1   0  <== a root!\r\n" );
document.write( "Trying 1 again:\r\n" );
document.write( "1  |   1   -7   21   -35   35   -21    7   -1 \r\n" );
document.write( "----        1   -6    15  -20    15   -6    1    \r\n" );
document.write( "      ----------------------------------------\r\n" );
document.write( "       1   -6   15   -20   15    -6    1    0  <== a root!\r\n" );
document.write( "Trying 1 again:\r\n" );
document.write( "1  |   1   -6   15   -20   15    -6    1\r\n" );
document.write( "----        1   -5    10  -10     5    1    \r\n" );
document.write( "      -----------------------------------\r\n" );
document.write( "       1   -5   10   -10    5    -1    0  <== a root!\r\n" );
document.write( "Trying 1 again:\r\n" );
document.write( "1  |   1   -5   10   -10    5    -1 \r\n" );
document.write( "----        1   -4     6   -4     1        \r\n" );
document.write( "      ------------------------------\r\n" );
document.write( "       1   -4    6    -4    1     0  <== a root!\r\n" );
document.write( "

I think by now you can see where this is going. 1 is a root 8 times. In other words it is a root of multiplicity 8. And f(x) in factored form is:
\n" ); document.write( " $\"f%28x%29+=+%28x-1%29%5E8\"$

\n" ); document.write( "FWIW, The following is the graph of the function you posted. It shows 5 points where the graph intersects the x-axis. But there are more such points outside of what we see on this graph. There will be 1, 2 or 3 additional points where the graph intersects the x-axis. We know there is at least 1 because an 8th degree polynomial which has a positive leading coefficient will go toward infinity for large positive and negative x's. So that steep \"dive\" we see at about x = 1.2 has to eventually come back up and cross the x-axis somewhere out to the right.
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