Question 72343
<pre><font size = 4><b>
2x² + 7x + 3
<font color = "blue">2</font>x² + <font color = "green">7</font>x <font color = "purple">+</font> <font color = "red">3</font>

1.  Multiply the blue <font color = "blue">2</font> by the red <font color = "red">3</font>, getting 6.
2.  Think of two positive integers which have product 6 and
    which have SUM (since the purple sign is +) equal to the
    green <font color = "green">7</font>.  Such positive integers are 1 and 6 because 
    1×6 = 6 and 1+6 = <font color = "green">7</font>
3.  Rewrite the green <font color = "green">7</font> using 1 and 6 

2x² + (1 + 6)x + 3

4. Remove the parentheses using the distributive principle.

2x² + 1x + 6x + 3

5. Factor x out of the first two terms and factor 3 out of
   the last two terms:

x(2x + 1) + 3(2x + 1)

6. Factor the (2x + 1) out of both terms

(2x + 1)(x + 3)

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Another example:
6x² + x — 2
<font color = "blue">6</font>x² + <font color = "green">1</font>x <font color = "purple">—</font> <font color = "red">2</font>

1.  Multiply the blue <font color = "blue">6</font> by the red <font color = "red">2</font>, getting 12.
2.  Think of two positive integers which have product 12 and
    which have DIFFERENCE (since the purple sign is —) equal to the
    green <font color = "green">1</font>.  Such positive integers are 4 and 3 because 
    4×3 = 12 and 4—3 = <font color = "green">1</font>
3.  Rewrite the green <font color = "green">1</font> using 4 and 3 

6x² + (4 — 3)x — 2

4. Remove the parentheses using the distributive principle.

6x² + 4x — 3x — 2

5. Factor 2x out of the first two terms and factor -1 out of
   the last two terms:

2x(3x + 2) — 1(3x + 2)

6. Factor the (3x + 2) out of both terms

(3x + 2)(2x — 1)

Edwin</pre>