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<font size="4"><span style="font-family: times new roman,times,serif;"><span style="font-style: italic;">La factorización de la expresión</span> <span class="nolink">«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»#eje«/mi»«/math»</span> </span><span style="font-family: times new roman,times,serif; font-style: italic;">es:</span></font><br /><br /><br />
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<font size="4" style="font-style: italic;"><span style="font-family: times new roman,times,serif;">Solución:<br /><br style="font-family: times new roman,times,serif;" /></span><span style="font-family: times new roman,times,serif;"> Debemos convertir al trinomio</span></font><span style="font-style: italic;"> </span><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»#eje«/mi»«/math»</span><span style="font-style: italic;"> </span><font size="4" style="font-style: italic;"><span style="font-family: times new roman,times,serif;">en el producto de dos factores, pues este caso de factorización corresponde al trinomio de la forma</span></font><span style="font-style: italic;"> </span><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mi»b«/mi»«mi»x«/mi»«mo»+«/mo»«mi»c«/mi»«/math»</span><span style="font-style: italic;">.</span><br style="font-style: italic;" /><font size="4" style="font-style: italic;"><span style="font-family: times new roman,times,serif;">cuya descomposición es:</span></font><span style="font-style: italic;"> </span><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mi»b«/mi»«mi»x«/mi»«mo»+«/mo»«mi»c«/mi»«/math»</span><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mo»=«/mo»«mo»(«/mo»«mi»x«/mi»«mo»+«/mo»«mo»§nbsp;«/mo»«mo»?«/mo»«mo»§nbsp;«/mo»«mo»)«/mo»«mo»§nbsp;«/mo»«mo»(«/mo»«mi»x«/mi»«mo»+«/mo»«mo»§nbsp;«/mo»«mo»?«/mo»«mo»)«/mo»«/math»</span><br style="font-style: italic;" /><font size="4" style="font-family: times new roman,times,serif; font-style: italic;">Para </font><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»#eje«/mi»«/math»</span><span style="font-style: italic;"> =</span><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mo»(«/mo»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mo»?«/mo»«mo»§nbsp;«/mo»«mo»)«/mo»«mo»§nbsp;«/mo»«mo»(«/mo»«mi»x«/mi»«mo»+«/mo»«mo»§nbsp;«/mo»«mo»?«/mo»«mo»)«/mo»«/math»</span><font size="4" style="font-family: times new roman,times,serif; font-style: italic;"> debemos encontrar dos números que cumplan las siguientes condiciones a la vez:<br /><br /> Primero: cuya suma (o diferencia) sea igual a </font><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»#sum«/mi»«/math»</span><span style="font-style: italic;"> </span><br style="font-style: italic;" /> <font size="4" style="font-style: italic;"><span style="font-family: times new roman,times,serif;">Segundo: cuyo producto de igual a</span></font><span style="font-style: italic;"> </span><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»#pro«/mi»«/math»</span><br style="font-style: italic;" /><font size="4" style="font-style: italic;"><span style="font-family: times new roman,times,serif;">Los números</span></font><span style="font-style: italic;"> </span><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»#k1«/mi»«mo»§nbsp;«/mo»«mi»y«/mi»«mo»§nbsp;«/mo»«mi»#k2«/mi»«/math»</span><span style="font-style: italic;"> </span><font size="4" style="font-style: italic;"><span style="font-family: times new roman,times,serif;">cumplen ambas condiciones, puesto que</span></font><br style="font-style: italic;" /><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»#k1«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mi»#z1«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mi»#sum«/mi»«/math»</span><span style="font-style: italic;"> </span><font size="4" style="font-style: italic;"><span style="font-family: times new roman,times,serif;">y a la vez </span></font><span style="font-style: italic;"> </span><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»#k1«/mi»«mo»§nbsp;«/mo»«mo»·«/mo»«mo»§nbsp;«/mo»«mi»#z1«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mi»#pro«/mi»«/math»</span><br /><font size="4" style="font-style: italic;"><span style="font-family: times new roman,times,serif;">Finalmente, la descomposición en factores queda: </span></font><br /><br style="font-style: italic;" /> <div style="text-align: center;"><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»#eje«/mi»«mo»=«/mo»«mi»#opa«/mi»«/math»</span><br /></div><br /><br />
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Choice 1
Answer
#opa
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<font size="4" style="font-style: italic;"><span style="font-family: times new roman,times,serif;">¡Muy bien</span></font><font size="4" style="font-style: italic;"><span style="font-family: times new roman,times,serif;">!</span></font><font size="4" style="font-style: italic;"><span style="font-family: times new roman,times,serif;"> Sigue así. </span></font>
Choice 2
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#altb
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<font size="4"><span style="font-family: times new roman,times,serif;"><span style="font-style: italic;">Al parecer no estás factorizando en forma adecuada.</span><br style="font-style: italic;" /><span style="font-style: italic;">Desarrolla el cuadrado de binomio obtenido y comprueba que concuerde con la pregunta planteada.</span><br /></span></font>
Choice 3
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#altc
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<font size="4" style="font-style: italic;"><span style="font-family: times new roman,times,serif;">Al parecer no reconoces el tipo de factorización que se solicita. Revisa la solución.</span></font>
Choice 4
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#altd
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<font size="4" style="font-style: italic;"><span style="font-family: times new roman,times,serif;">Recuerda que debes descomponer en factores la expresión dada.</span></font>
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Choice 6
Answer
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Choice 7
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Feedback
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Friday, 23 August 2013, 3:10 PM
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