Question 1205248
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I'll use the <a href="https://www.algebra.com/algebra/homework/playground/lessons/box-method.lesson">box method</a>


-x^2+2x-5 has 3 terms
3x+2 has 2 terms
We'll have a table that has 3 rows and 2 columns.
Place the terms as headers along the top and left side like so
<table border=1 cellpadding><tbody><tr><td></td><td>3x&nbsp;&nbsp;&nbsp;</td><td>2 &nbsp;&nbsp;&nbsp;</td></tr><tr><td>-x^2</td><td></td><td></td></tr><tr><td>2x</td><td></td><td></td></tr><tr><td>-5</td><td></td><td></td></tr></tbody></table>


To fill out this table, multiply the headers. 
Example: -x^2 times 3x = -3x^3 in the upper left corner.
<table border=1 cellpadding><tbody><tr><td></td><td>3x</td><td>2</td></tr><tr><td>-x^2</td><td>-3x^3</td><td>-2x^2</td></tr><tr><td>2x</td><td>6x^2</td><td>4x</td></tr><tr><td>-5</td><td>-15x</td><td>-10</td></tr></tbody></table>
Add up the results. Combine like terms.
-3x^3 + (-2x^2) + 6x^2 + 4x + (-15x) + (-10)
-3x^3 -2x^2 + 6x^2 + 4x  - 15x - 10
-3x^3 +(-2x^2 + 6x^2) + (4x  - 15x) - 10
<font color=red>-3x^3 + 4x^2 - 11x - 10</font>


Therefore,
(-x^2+2x-5)(3x+2) = <font color=red>-3x^3 + 4x^2 - 11x - 10</font>


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Let's look at another approach.
This time I'll use the distributive property.


Let y = 3x+2


(-x^2+2x-5)(3x+2) 
= (-x^2+2x-5)y
= y(-x^2+2x-5)
= -x^2*y + 2xy - 5y
= -x^2(3x+2) + 2x(3x+2) - 5(3x+2)
= -3x^3 - 2x^2 + 6x^2 + 4x - 15x - 10
= <font color=red>-3x^3 + 4x^2 - 11x - 10</font>


Note on the 2nd to last line, all of the terms mentioned are found inside the 6 inner boxes of the previous method.


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Or we could have these steps.


Let w = -x^2 + 2x - 5


(-x^2+2x-5)(3x+2)
= w(3x + 2)
= 3xw + 2w
= 3x(-x^2 + 2x - 5) + 2(-x^2 + 2x - 5)
= -3x^3 + 6x^2 - 15x - 2x^2 + 4x - 10
= <font color=red>-3x^3 + 4x^2 - 11x - 10</font>
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