<PRE><B>Question 366571</B>
The second one, {{{2x^2 + 27x + 13}}}, can be solved by
careful inspection.
First I notice that {{{2*13 + 1 = 27}}}. I want to arrange
the factors so I get this result for the coefficient of {{{x}}}
{{{(2x + 1)*(x + 13)}}} accomplishes this.
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For the 4th one, {{{10x^2  - 7x  - 3}}} , I try {{{10x}}} and {{{x}}}
for the factors of  {{{10x^2}}}. Then I try {{{-1}}} and {{{3}}}
for the factors of {{{-3}}}. This gives me
{{{(10x + 3)*(x - 1)}}} as the answer with {{{-7}}} as the coefficient of {{{x}}}
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