document.write( "Question 825413: Find all the real zeros of the function.\r
\n" ); document.write( "\n" ); document.write( "h(x)= $\"x%5E4-4x%5E3-9x%5E2%2B16x%2B20\"$ \n" ); document.write( "

Algebra.Com's Answer #497355 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
To find the zeros of h(x) we need to factor it. And of the factoring methods, I can only see that factoring by trial and error of the possible rational roots/zeros could work (to start).

\n" ); document.write( "The possible rational roots/zeros of a polynomial are all the ratios, positive and negative, which can be formed using a factor of the constant term (at the end) in the numerator and a factor of the leading coefficient (at the beginning) in the denominator. With h(x)'s constant term of 20 and leading coefficient of 1 we get the following list of possible rational roots/zeros for h(x):
\n" ); document.write( "+1/1, +2/1, +4/1, +5/1, +10/1 and +20/1
\n" ); document.write( "which reduce to:
\n" ); document.write( "+1, +2, +4, +5, +10 and +20

\n" ); document.write( "Now we can try these to see if any of them are actual root/zero. We can try 1 mentally since powers of 1 are extremely easy. We shoujld find that h(1) is not zero. So 1 is not a root/zero. Now we'll try 2. For this we will use synthetic division:
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\r\n" );
document.write( "2  |   1   -4   -9   16   20\r\n" );
document.write( "----        2   -4  -26  -20\r\n" );
document.write( "      -----------------------\r\n" );
document.write( "       1   -2  -13  -10    0\r\n" );
document.write( "

And we have a winner! The zero in the lower right corner is the remainder. And if the remainder is zero then the division divided evenly. And what we divided by was (x-2) so 2 is a root/zero of h(x)! Not only that but the rest of the bottom row is the quotient/other factor. The \"1 -2 -13 -10\" translate into $\"x%5E3-2x%5E2-13x-10\"$ . So now:
\n" ); document.write( " $\"h%28x%29+=+x%5E4-4x%5E3-9x%5E2%2B16x%2B20+=+%28x-2%29%28x%5E3-2x%5E2-13x-10%29\"$

\n" ); document.write( "As we look for more roots/zeros, we only need to factor the second factor above. (This means that +20 are no longer possible roots/zeros.) Let's try 2 again. (Roots/zeros can be roots/zeros multiple times!)
\n" ); document.write( "

\r\n" );
document.write( "2  |   1   -2   -13   -10\r\n" );
document.write( "----        2     0   -26\r\n" );
document.write( "      ---------------------\r\n" );
document.write( "       1    0   -13   -36\r\n" );
document.write( "No. If we try 4, it doesn't work either. But 5...\r\n" );
document.write( "5  |   1   -2   -13   -10\r\n" );
document.write( "----        5    15    10\r\n" );
document.write( "      ---------------------\r\n" );
document.write( "       1    3     2     0\r\n" );
document.write( "

And we have a second root/zero. Now:
\n" ); document.write( "

\n" ); document.write( "Now we factor $\"x%5E2%2B3x%2B2\"$ . This is a quadratic (so we no longer have to use the trial and error method). It factors easily. So:
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\n" ); document.write( "This makes the roots/zeros of h(x): 2, 5, -1 and -2.

\n" ); document.write( "P.S. Please post your problems in an appropriate category. \"test\" should be used only if there is no relevant, established category. This problem should be the the \"Polynomials\" category.

\n" ); document.write( "Problems posted in relevant categories get quicker responses! \n" ); document.write( "

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