Solving Polynomial Roots: Sum of Cubes and Fourth Powers - Further Maths Help

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In summary: Also, you haven't answered the questions I asked you in my previous response. Please do that before responding further.In summary, the problem is to show that the sum of the cubes of the roots of the equation x^3 + (lambda)x + 1 = 0 is -3 and that there is no real value of (lambda) for which the sum of the fourth powers of the roots is negative. The method for solving this problem is not clear, but it may involve using a formula for finding the sum of the squares of the roots of a cubic equation and showing that the sum of the three roots is 0. Further clarification is needed on the exact wording of the problem and the equations involved.
  • #1
rdajunior95
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Show that the sum of the cubes of the root of the equation

x3 + (lambda)x + 1 = 0

is -3

Show also that there is no real value of (lambda) for which the sum of the fourth powers of the roots is negative.

This question is in one of the past papers of further maths and I don't know how to start solving this problem!

So please help me :)

P.S. Sorry if this is the wrong section, I am new here. ;)
 
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  • #2
If m is a root of the equation, what do you know is true about m3?
 
  • #3
rdajunior95 said:
Show that the sum of the cubes of the root of the equation

x3 + (lambda)x + 1 = 0

is -3
Your problem statement threw me off for a minute. I'm sure you mean "the sum of the cubes of the roots of the equation.."

Let r1, r2, and r3 be the three roots of this equation.

1) What does it mean for r1 to be a root of that equation?
2) Same for r2.
3) Same for r3.
4) What do you get if you add together r13 + r23 + r33?
rdajunior95 said:
Show also that there is no real value of (lambda) for which the sum of the fourth powers of the roots is negative.

This question is in one of the past papers of further maths and I don't know how to start solving this problem!

So please help me :)

P.S. Sorry if this is the wrong section, I am new here. ;)
 
  • #4
Mark44 said:
Your problem statement threw me off for a minute. I'm sure you mean "the sum of the cubes of the roots of the equation.."

Let r1, r2, and r3 be the three roots of this equation.

1) What does it mean for r1 to be a root of that equation?
2) Same for r2.
3) Same for r3.
4) What do you get if you add together r13 + r23 + r33?

Yes we have to show that r13 + r23 + r33 is -3
 
  • #6
when you equate x=r1, the equation would be equal to 0
 
  • #7
rdajunior95 said:
when you equate x=r1, the equation would be equal to 0
That makes no sense. An equation can't be equal to something. An equation is true or it isn't true, one or the other.

What does it mean to say that r1 is a root of the equation?
 
  • #8
Yeah like when x=r1, r2 or r3 than the equation is true!

Sorry about the confusion!
 
  • #9
Which equations are you talking about? (These are what I'm trying to get you to show me, without much success so far.)
 
  • #10
(x^3) + (lambda)x + 1 = 0 this is the equation which is satisfied when x=r1, r2 or r3We have to show that (r1)^3+(r2)^3+(r3)^3 = -3
 
  • #11
What equation do you get when x = r1?
What equation do you get when x = r2?
What equation do you get when x = r3?

Instead of just telling me about it, please write the equations.
 
  • #12
I think you have to use some kind of formula like when a cubic equation has roots (alpha), (beta) and (gamma) than to find the sum of α2 + β2 + γ2 you basically use

α2 + β2 + γ2 = (α + β + γ)2 - 2(αβ - βγ - αγ)
 
  • #13
I hope this helps! :)
 
  • #14
Helps? This was your question and you haven't done what was suggested yet.

What does it mean to say that [itex]x= r_1[/itex] is a root of the equation [itex]x^3+ \lambda x+ 1= 0[/itex]? And, as Mark44 said, "Instead of just telling me about it, please write the equations."
 
  • #15
rdajunior95 said:
I think you have to use some kind of formula like when a cubic equation has roots (alpha), (beta) and (gamma) than to find the sum of α2 + β2 + γ2 you basically use

α2 + β2 + γ2 = (α + β + γ)2 - 2(αβ - βγ - αγ)
I don't think that this will get you anywhere, even after fixing the sign error. The identity is α2 + β2 + γ2 = (α + β + γ)2 - 2(αβ + βγ + αγ).

If you can show that the sum of the three roots is 0, it's simple to show that the sum of the cubes of the roots is -3. Are you sure that what you have posted is the exact wording of the problem?
 

FAQ: Solving Polynomial Roots: Sum of Cubes and Fourth Powers - Further Maths Help

What are polynomials?

Polynomials are algebraic expressions that consist of variables, coefficients, and exponents. They are made up of one or more terms that are combined using addition, subtraction, and multiplication.

What is the degree of a polynomial?

The degree of a polynomial is the highest exponent of the variable in the expression. For example, the degree of the polynomial 3x^2 + 2x + 1 is 2.

How do you add or subtract polynomials?

To add or subtract polynomials, you must combine like terms. This means that you add or subtract the coefficients of terms with the same variable and exponent. If there are no like terms, the polynomial cannot be simplified further.

How do you multiply polynomials?

To multiply polynomials, you can use the FOIL method, which stands for First, Outer, Inner, Last. This method involves multiplying the first terms, then the outer terms, then the inner terms, and finally the last terms. Then, you combine like terms to simplify the expression.

What is long division of polynomials?

Long division of polynomials is a method used to divide a polynomial by another polynomial. It involves dividing the first term of the dividend by the first term of the divisor, then multiplying the result by the divisor and subtracting it from the dividend. This process is repeated until the remainder has a lower degree than the divisor, and the final answer is the quotient.

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