document.write( "Question 746886: Hello, I would really appreciate if someone could show me how to factor the following equation using the ac method. 8x^2-22x+5. I have been trying but can not figure out how to do it. Thank you so much for your help. \n" ); document.write( "
Algebra.Com's Answer #454534 by Edwin McCravy(20056)![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( "To factor: 8x²-22x+5\r\n" ); document.write( "\r\n" ); document.write( "Multiply the 8 by the 5 ignoring signs. Get 40\r\n" ); document.write( "\r\n" ); document.write( "Write down all the ways to have two positive integers\r\n" ); document.write( "which have product 40, starting with 40*1\r\n" ); document.write( "\r\n" ); document.write( "40*1\r\n" ); document.write( "20*2\r\n" ); document.write( "10*4\r\n" ); document.write( " 8*5\r\n" ); document.write( "\r\n" ); document.write( "Since the last sign in 8x²-22x+5 is +, ADD them,\r\n" ); document.write( "and place the SUM out beside that:\r\n" ); document.write( "\r\n" ); document.write( "40*1 40+1=41\r\n" ); document.write( "20*2 20+2=22\r\n" ); document.write( "10*4 10+4=14\r\n" ); document.write( " 8*5 8+5=13\r\n" ); document.write( "\r\n" ); document.write( "Now, again ignoring signs, we find in that list of\r\n" ); document.write( "sums the coefficient of the middle term in 8x²-22x+5\r\n" ); document.write( "\r\n" ); document.write( "So we replace the number 22 by 20+2\r\n" ); document.write( "\r\n" ); document.write( "8x²-22x+5\r\n" ); document.write( "8x²-(20+2)x+5\r\n" ); document.write( "\r\n" ); document.write( "Then we distribute to remove the parentheses:\r\n" ); document.write( "\r\n" ); document.write( "8x²-20x-2x+5\r\n" ); document.write( "\r\n" ); document.write( "Factor the first two terms 8x²-20x by taking out the\r\n" ); document.write( "greatest common factor, getting 4x(2x-5)\r\n" ); document.write( "\r\n" ); document.write( "Factor the last two terms -2x+5 by taking out the\r\n" ); document.write( "greatest common factor, -1, getting -1(2x-5)\r\n" ); document.write( "\r\n" ); document.write( "So we have\r\n" ); document.write( "\r\n" ); document.write( "4x(2x-5)-1(2x-5)\r\n" ); document.write( "\r\n" ); document.write( "Notice that there is a common factor, (2x-5)\r\n" ); document.write( "\r\n" ); document.write( "4x(2x-5)-1(2x-5)\r\n" ); document.write( "\r\n" ); document.write( "which we can factor out leaving the 4x and the -1 to put \r\n" ); document.write( "in parentheses:\r\n" ); document.write( "\r\n" ); document.write( "(2x-5)(4x-1)\r\n" ); document.write( "\r\n" ); document.write( "To see a couple of other examples just like that using the AC\r\n" ); document.write( " method, go here:\r\n" ); document.write( "\r\n" ); document.write( "http://www.algebra.com/my/change_this_name32371.lesson?content_action=show_dev\r\n" ); document.write( "\r\n" ); document.write( "Edwin\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |