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 Find the zeros algebraically

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Find the zeros algebraically:

 

1.) p(x) = 6x^3 + 31x^2 + 4x - 5, p(-5) = 0

 

Answer: https://web2.0calc.com/questions/find-the-zeros-algebraically-last-one

Credit to: Rom

 Feb 19, 2019
edited by GAMEMASTERX40  Feb 19, 2019
edited by GAMEMASTERX40  Feb 19, 2019
edited by GAMEMASTERX40  Feb 19, 2019

Best Answer 

 #3
avatar+6251 
+2

\(p(x) = 6x^3 + 31x^2 + 4x-5\\ p(-5)=0 \Rightarrow (x+5) \text{ is a factor}\)

 

\(\dfrac{p(x)}{x+5} = 6 x^2+x-1\\ \text{we can factor this by eye}\\ 6 x^2+x-1 = (3x-1 )(2x +1)\\ p(x) = (3x-1)(2x+1)(x+5)\\ \text{with zeros at }x = \dfrac 1 3,~-\dfrac 1 2,~-5\)

.
 Feb 19, 2019

3+0 Answers

 #1
avatar+6251 
+2

that's the problem I answered for you over in the thread you changed.

 Feb 19, 2019
 #2
avatar+73 
-5

Yeah, its that I felt bad for changing the problem after all the work you did so I posted again so it will be different. Thanks by the way!

 

-- 7H3_5H4D0W

GAMEMASTERX40  Feb 19, 2019
edited by GAMEMASTERX40  Feb 19, 2019
 #3
avatar+6251 
+2
Best Answer

\(p(x) = 6x^3 + 31x^2 + 4x-5\\ p(-5)=0 \Rightarrow (x+5) \text{ is a factor}\)

 

\(\dfrac{p(x)}{x+5} = 6 x^2+x-1\\ \text{we can factor this by eye}\\ 6 x^2+x-1 = (3x-1 )(2x +1)\\ p(x) = (3x-1)(2x+1)(x+5)\\ \text{with zeros at }x = \dfrac 1 3,~-\dfrac 1 2,~-5\)

Rom Feb 19, 2019

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