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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+128794 
+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+45 
+1

\(x^3 + 3x^2 + 7x − 11 = 0\)

simplifies to:

\((x-1)(x^2+4x+11)=0\)

and 

\(x^2+4x+11=0\)

which gives an imaginary number

but for the complex part, 

you get the roots of x as:

\(x=\frac{-4±\sqrt{4^2-4*1*11}}{2}\)

 

whic becomes

\(x=2±\sqrt{-7}\)

 

so the roots of 

\(x^3 + 3x^2 + 7x − 11 = 0\)

are

\(x = 1\)

and

\(x=2±\sqrt{-7}\)

 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+45 
+1

try it

 

smiley

hasAquestion  Mar 19, 2024
 #4
avatar+394 
+1

In response to your question:

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

For any polynomial, \(a_nx^n+a_{n-1}x^{n-1}\dots a_1x^1+a_0\), the rational roots can all be expressed as:\(\pm\frac{\text{factor of }a_0}{\text{factor of }a_n}\).

Therefore in this polynomial, the only possible rational roots are \(\pm 1, \pm11\). (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. \(\pm1\) 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 \(x-a\) is a factor of \(f(x)\).

 

Here is the workthrough of the first step:

Use rational root theorem, plug in 1 to \(f(x)={x}^{3}+3x^{2}+7x-11\).

\({1}^{3}+3*{1}^{2}+7*1-11=0\).

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+128794 
+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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