document.write( "Question 114403This question is from textbook Fund of Alg and Trig
\n" ); document.write( ": Show that the eqs has no rational root.\r
\n" ); document.write( "\n" ); document.write( "2x^5+3x^3+7=0\r
\n" ); document.write( "\n" ); document.write( "I know that I am supposed to use synthetic division.
\n" ); document.write( "I know the constant is 7 thus 1,7\r
\n" ); document.write( "\n" ); document.write( "I know the numbers I should divide into are as follows: 2 0 3 0 7\r
\n" ); document.write( "\n" ); document.write( "I tried to use both 1, and 7 but It doesn't appear to work. Can anyone help. Thanks. \n" ); document.write( "

Algebra.Com's Answer #83265 by solver91311(24713) $\"\"$ $\"About$

You can put this solution on YOUR website!
I take your comment that \"it doesn't appear to work\" as meaning that you get a remainder when dividing by either 7 or 1. Since you are trying to show that the equation has no rational root, this is to be expected. The fact that the only possible factors of the constant term, 7, are 7 and 1 as you stated, coupled with the fact that the synthetic division you performed resulted in a remainder when you divided by either of those factors is proof that the equation does not, in fact, have rational roots.\r
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\n" ); document.write( "\n" ); document.write( "Just for fun, here is a graph of the function, $\"f%28x%29=2x%5E5%2B3x%5E3%2B7\"$ . Not only does it not have any rational roots, it only has one real number root; the other four roots are two conjugate pairs of complex numbers.\r
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\n" ); document.write( "\n" ); document.write( " $\"graph%28600%2C600%2C-4%2C4%2C-2%2C12%2C2x%5E5%2B3x%5E3%2B7%29\"$ \n" ); document.write( "

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