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Find the sum and the product of the roots (real and complex) of x^3 + 3x^2 + 7x − 11 = 0

 Mar 19, 2024

Best Answer 

 #5
avatar+130458 
+1

x^3 + 3x^2  + 7x  - 11   =  0

By Vieta

Sum of the roots  = -3 / 1  =   -3

Product of the roots = -(-11) /  1 =  11

 

cool cool cool

 Mar 20, 2024
 #1
avatar+60 
+2

x3+3x2+7x11=0

simplifies to:

(x1)(x2+4x+11)=0

and 

x2+4x+11=0

which gives an imaginary number

but for the complex part, 

you get the roots of x as:

x=4±4241112

 

whic becomes

x=2±7

 

so the roots of 

x3+3x2+7x11=0

are

x=1

and

x=2±7

 Mar 19, 2024
 #2
avatar+11 
0

Was there a specific method you used to factor x - 1 out? I understand (x-1)(x^2 + 4x + 11) = x^3 + 3x^2 + 7x − 11 but just looking at  x^3 + 3x^2 + 7x − 11, I do not know how to get the x-1

theadfas  Mar 19, 2024
edited by theadfas  Mar 19, 2024
 #3
avatar+60 
+2

try it

 

smiley

hasAquestion  Mar 19, 2024
 #4
avatar+410 
+1

In response to your question:

A good way to do this is by the rational root theorem.

For any polynomial, anxn+an1xn1a1x1+a0, the rational roots can all be expressed as:±factor of a0factor of an.

Therefore in this polynomial, the only possible rational roots are ±1,±11. (factors of a0 are 1, 11 and the only factor of ais 1)

Plug these values into the function, to see which ones produce zero. ±1 is a really, really common ones, so always try those first.

Remainder theorem is always helpful. Remember if for a function f(x)f(a)=0, then xa is a factor of f(x).

 

Here is the workthrough of the first step:

Use rational root theorem, plug in 1 to f(x)=x3+3x2+7x11.

13+312+7111=0.

Therefore we know because f(1)=0, then x-1 is a factor of f(x), so we can factor it out.

 Mar 20, 2024
edited by hairyberry  Mar 20, 2024
 #5
avatar+130458 
+1
Best Answer

x^3 + 3x^2  + 7x  - 11   =  0

By Vieta

Sum of the roots  = -3 / 1  =   -3

Product of the roots = -(-11) /  1 =  11

 

cool cool cool

CPhill Mar 20, 2024

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