document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=747293>Question 747293</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let f(x) = 4x - 7 and g(x) = 2x - 3. Find.<BR>\n" );
document.write( "(f*g)(4) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #454849 by </B><A HREF=/tutors/aboutme.mpl?userid=RedemptiveMath><B>RedemptiveMath(80)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=RedemptiveMath><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=454849\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let f(x) = 4x - 7 and g(x) = 2x - 3. Find.<BR>\n" );
document.write( "(f*g)(4)\r<BR>\n" );
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document.write( "All we need to do is plug what f and g are into the multiplication problem. In terms of x (when x is in the domain of f and g), (f*g) is\r<BR>\n" );
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document.write( "(4x-7)(2x-3).\r<BR>\n" );
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document.write( "Multiplying this out, we have \r<BR>\n" );
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document.write( "(4x-7)(2x-3) = 8x^2 - 26x + 21.\r<BR>\n" );
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document.write( "Since there is another factor, 4, being multiplied or multiplying in the problem, we need to multiply what we have just found by that number:\r<BR>\n" );
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document.write( "(8x^2 - 26x + 21)(4) = 32x^2 - 104x + 84.\r<BR>\n" );
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document.write( "Combining all of these parts, we can see the entire picture in the next series of steps:\r<BR>\n" );
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document.write( "(f*g)(4) = (4x-7)(2x-3)(4) = (8x^2-26x+21)(4) = 32x^2 - 104x + 84.\r<BR>\n" );
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document.write( "We could also multiply 4 and (4x-7) or 4 and (2x-3) first and then proceeded to multiply the rest of the problem. No matter which two sets of parentheses you start with, the final answer should be the same. </SPAN>\n" );
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