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What's the best way to find the zeros?

 Feb 18, 2019
 #1
avatar+5770 
+2

\(\text{in (a), for example, you are told that }p(-5)=0\\ \text{thus }-5 \text{ is a zero and }x+5 \text{ is a factor}\\ \text{so divide }p(x) \text{ by }(x+5) \text{ using polynomial division}\\ \dfrac{p(x)}{x+5} = 6 x^2+x-1\\ \text{this should be easy enough to factor by eye}\\ 6x^2 + x -1 = (3x -1)(2x+1) \\ \text{and it's easy enough to read off the zeros in this form as }\\ x = \dfrac 1 3,~x = -\dfrac 1 2\\ \text{follow this procedure for (b), which I leave to you}\)

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 Feb 18, 2019
 #2
avatar+102995 
+4

Second one

 

We know that      f(3) = 0.....so.....x = 3 is one zero

 

Using some synthetic division, we have

 

3 [  2       1           -16           -15 ]

                6            21            15

     _________________________

      2        7            5                0

 

The remaining polynomial  is  ( 2x^2 + 7x + 5)

To factor this......we ca break up 7x  as 2x + 5x

 

2x^2 + 2x + 5x + 5           factor by grouping

 

2x (  x + 1)  + 5 ( x + 1)    =

 

(x + 1) (2x + 5)

 

Setting both factors to 0  and solving for x produces the other two zeros  = 

 

x = 1   and    x = -5/2

 

 

cool cool cool

 Feb 18, 2019

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