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Help web 2.0 calc you're my only hope!

 

Would it be possible to answer this question with all steps and how and why you did it? Big thanks!

 

Find the possible rational zeros,number of positive and negative real zeros and the exact value for every zero for each polynomial. Also factor the polynomial.

 

Thank you all so much in advance!!!

dom6547  Feb 18, 2018
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5+0 Answers

 #1
avatar+86649 
+2

For which polynomials ??/

 

 

cool cool cool

CPhill  Feb 18, 2018
 #2
avatar+90 
0

Oh shoot how did I forget the most important part!!! 

 

They are 

 

 \(x^4+7x^3+9x^2-17x-20\)

 

\(x^3-2x^2+3x-6\)

 

\(x^6-x^4-x^2+1\)

 

real sorry about that!

dom6547  Feb 18, 2018
 #3
avatar+86649 
+2

LOL!!!.....yep, that helps  !!!

 

x^4 + 7x^3  + 9x^2  - 17x - 20

 

The possible rational zeroes are  ±  ( 1, 2, 4, 5, 10 and 20)

 

We have  one sign change ..so we have 1  positive root 

 

To  find the number of possible negative roots look at f(-x)  =

 

(-x)^4 + 7(-x)^3 + 9(-x)^2 - 17(-x) -20  =

x^4 - 7x^3 + 9x^2 + 17x - 20

We have hree sign changes so we either have 3 or 1  negative roots

 

Let's write the above in a slightly different manner

 

x^4 + 7x^3 + 6x^2 + 3x^2 - 17x - 20        factor this

 

x^2(x^2 + 7x + 6)  +  (3x - 20)) (x + 1)  =

 

x^2 ( x + 1) (x + 6) + (x + 1) (3x - 20)   factor out the ( x + 1)

 

(x + 1)  [ x^2 (x + 6) + (3x - 20)  ]

 

(x + 1) [ x^3 + 6x^2 + 3x - 20 ]  the second polynomial factors as  (x + 4) (x^2 + 2x - 5)

 

So   we have

 

(x + 1) (x + 4)(x^2 + 2x - 5)

 

Setting the first two  factors to 0  and solving for x produces   x = - 1   and x =  4

 

The other two roots  come from using the quadratic formula to solve the 3rd polynomial...these are

 

x  = √6 - 1    (p0sitive)   and  x  =    -√ 6  -  1   (negative)

 

So....we have 1 positive  root and 3 negative roots

 

 

 

cool cool cool

CPhill  Feb 18, 2018
 #4
avatar
+1

Solve for x:

x^3 - 2 x^2 + 3 x - 6 = 0

 

The left hand side factors into a product with two terms:

(x - 2) (x^2 + 3) = 0

 

Split into two equations:

x - 2 = 0 or x^2 + 3 = 0

 

Add 2 to both sides:

x = 2 or x^2 + 3 = 0

 

Subtract 3 from both sides:

x = 2 or x^2 = -3

 

Take the square root of both sides:

x = 2      or      x = i sqrt(3)      or      x = -i sqrt(3)

 

 

Solve for x:

x^6 - x^4 - x^2 + 1 = 0

 

The left hand side factors into a product with three terms:

(x - 1)^2 (x + 1)^2 (x^2 + 1) = 0

 

Split into three equations:

(x - 1)^2 = 0 or (x + 1)^2 = 0 or x^2 + 1 = 0

 

Take the square root of both sides:

x - 1 = 0 or (x + 1)^2 = 0 or x^2 + 1 = 0

 

Add 1 to both sides:

x = 1 or (x + 1)^2 = 0 or x^2 + 1 = 0

 

Take the square root of both sides:

x = 1 or x + 1 = 0 or x^2 + 1 = 0

 

Subtract 1 from both sides:

x = 1 or x = -1 or x^2 + 1 = 0

 

Subtract 1 from both sides:

x = 1 or x = -1 or x^2 = -1

 

Take the square root of both sides:

x = 1      or      x = -1      or      x = i      or      x = -i

Guest Feb 19, 2018
 #5
avatar+90 
0

Thanks to you both so much!

dom6547  Feb 19, 2018

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