document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=4743>Question 4743</A>: <I><SPAN STYLE=\"word-break: break-all;\"> solve each equation for x using factoring. please show work.\r<BR>\n" );
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document.write( "x2-4x+1=13\r<BR>\n" );
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document.write( "x2+2x+48=15x\r<BR>\n" );
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document.write( "5x2+7=52\r<BR>\n" );
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document.write( "3x2=4-11x\r<BR>\n" );
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document.write( "2x3+4x=6x2\r<BR>\n" );
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document.write( "2x2-4=7x </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #2989 by </B><A HREF=/tutors/aboutme.mpl?userid=guapa><B>guapa(62)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=guapa><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=2989\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1)x²-4x+1=13 In order to factor the right side has to be 0. So, we substract 13 from both sides.<BR>\n" );
document.write( "x²-4x+1-13=13-13<BR>\n" );
document.write( "x²-4x-12=0 Now we need 2 numbers whose sum is -4 and whose factor is -12. To find them we express 12 in its prime factors. 2*2*3 By grouping these factors in various ways we find that 6(2*3), 2(2)and 2*6=12 Since we need -12 one of these factors has to be negative. -6*2=-12 and -6+2=-4 Now we can express the middle terms as -6x+2x. Therefore, we can complete the factoring as follows:<BR>\n" );
document.write( "x²-6x+2x-12=0 Factor by grouping(x from first two terms and 2 from last two terms)<BR>\n" );
document.write( "x(x-6)+2(x-6)=0 Factor common binominal factor of (x-6) from both terms.<BR>\n" );
document.write( "(x-6)(x+2)=0  For all real numbers: ab=0 if and only if a=0 or b=0<BR>\n" );
document.write( "Let's use this technique:<BR>\n" );
document.write( "a=0, (x-6)=0 , x=6  ->(6-6)=0<BR>\n" );
document.write( "b=0, (x+2)=0 , x=-2 ->(-2+2)=0<BR>\n" );
document.write( "The solution set is (-2,6)<BR>\n" );
document.write( "Check:<BR>\n" );
document.write( "6²-4(6)+1=13<BR>\n" );
document.write( "36-24+1=13<BR>\n" );
document.write( "13=13<BR>\n" );
document.write( "(-2)²-4(-2)+1=13<BR>\n" );
document.write( "4-(-8)+1=13<BR>\n" );
document.write( "4+8+1=13<BR>\n" );
document.write( "13=13<BR>\n" );
document.write( "Use this process for all following problems\r<BR>\n" );
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document.write( "2)x²+2x+48=15x<BR>\n" );
document.write( "x²-13x+48=0<BR>\n" );
document.write( "48 expressed in its prime factors: 2*2*2*2*3. The solution to this problem is not factorable since there are no two numbers whose sum is -13 and whose product is 48.\r<BR>\n" );
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document.write( "3)5x²+7=52<BR>\n" );
document.write( "5x²+7-52=52-52<BR>\n" );
document.write( "5x²-45=0 Divide both sides by 5<BR>\n" );
document.write( "(5x²-45)/5=0/5<BR>\n" );
document.write( "x²-9=0 The difference of two squares {a²-b²= (a-b)(a+b)}<BR>\n" );
document.write( "(x-3)(x+3)=0<BR>\n" );
document.write( "x-3=0, x=3<BR>\n" );
document.write( "x+3=0, x=-3 Solution set is (-3,3)\r<BR>\n" );
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document.write( "4)3x²=4-11x<BR>\n" );
document.write( "3x²+11x-4=0  (2 numbers whose sum is 11 and whose product is -12(3*-4)<BR>\n" );
document.write( "3x²+12x-x-4=0<BR>\n" );
document.write( "3x(x+4)-1(x+4)=0<BR>\n" );
document.write( "(x+4)(3x-1)=0<BR>\n" );
document.write( "x+4=0, x=-4<BR>\n" );
document.write( "3x-1=0, x= 1/3 Solution set is (-4,1/3)\r<BR>\n" );
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document.write( "5)2x³+4x=6x²<BR>\n" );
document.write( "2x³-6x²+4x=0  Divide both sides by 2<BR>\n" );
document.write( "x³-3x²+2x=0   Factor out x<BR>\n" );
document.write( "x(x²-3x+2)=0<BR>\n" );
document.write( "x(x²-2x-x+2)=0<BR>\n" );
document.write( "x(x-2)(x-1)=0<BR>\n" );
document.write( "x-2=0, x=2<BR>\n" );
document.write( "x-1=0, x=1 Solution set is (0,1,2)\r<BR>\n" );
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document.write( "6)2x²-4=7x<BR>\n" );
document.write( "2x²-7x-4=0<BR>\n" );
document.write( "2x²-8x+x-4=0<BR>\n" );
document.write( "2x(x-4)+1(x-4)=0<BR>\n" );
document.write( "(x-4)(2x+1)=0<BR>\n" );
document.write( "x-4=0, x=4<BR>\n" );
document.write( "2x+1=0, x= -(1/2)  Solution set is (-1/2,4)<BR>\n" );
document.write( "I hope this helps\r<BR>\n" );
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document.write( " </SPAN>\n" );
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