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How do I find the roots, solutions, zeros, or what have you, of 2x3+13x2+23x+12 without graphing it and observing the points where y=0?

Guest Jan 23, 2018
 #1
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2x^3+13x^2+23x+12  =  0

 

Let's see if we can find an alternative way to write this.

 

A little trial and error produces  a way to break up 13x^2   as  2x^2 + 11x^2

 

So we have

 

2x^3  + 2x^2  + 11x^2 + 23x + 12      factor by grouping

 

2x^2 ( x + 1)  +  (11x + 12) ( x + 1)        the common factor is  x + 1

 

So we have

 

(x + 1)  [  2x^2 + 11x + 12 ]  =  0        factor the second polynomial

 

( x + 1)  ( 2x  + 3) ( x + 4)  =  0

 

Setting each factor to o and solving for x  produces

 

x + 1  = 0             2x + 3  = 0           x  + 4  = 0

x = -1                    2x  =  -3              x  = -4

                               x  = -3/2

 

The  zeroes are in red

 

BTW  - the first factoring trick won't always work...but....it is something to try

 

 

 

cool cool cool

CPhill  Jan 24, 2018

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