



 
Hi Pedro, You want to write $6 x^2  19 x + 3 = (ax + b)(cx + d)$ for some numbers $a, b, c \mbox{ and } d.$ If you imagine multiplying the right side to get the left side you have $a \times c = 6, \mbox{ and } b \times d = 3.$ From the second of these equations you must have $b$ and $d$ either 1 and 3 or 1 and 3. If a, b, c and d are all positive then expanding $(ax + b)(cx + d)$ will yield all positive terms but the original expression has a negative middle term, $19 x$ and hence I am going to try $b = 1$ and $d = 3.$ Thus I have \[6 x^2  19 x + 3 = (ax 1)(cx 3).\] But I know that $a \times c = 6.$ What are the possibilities for $a$ and $c?$ Try them and expand the right side to see which choice of $a$ and $c$ give a middle term of $19 x.$ Penny  


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