MHB Finding a third degree polynomial... stumped

  • Thread starter Thread starter lashia
  • Start date Start date
  • Tags Tags
    Degree Polynomial
AI Thread Summary
To find a third-degree polynomial with rational coefficients that has zeros at 6 and -i, and passes through the point (2, -10), the initial attempt was to use the polynomial form (x-6)(x^2+1). However, this resulted in the polynomial passing through (2, -20) instead. The correct approach involves introducing a constant A, leading to the polynomial f(x) = A(x-6)(x^2+1). By solving for A using the point (2, -10), it is determined that A should be 1/2, resulting in the final polynomial f(x) = (1/2)(x-6)(x^2+1). This solution successfully meets the criteria of the problem.
lashia
Messages
3
Reaction score
0
Problem:

Find a third degree polynomial with rational coefficients if two of its zeros
are 6 and – 𝑖 and it passes through the point (2, -10)So far, I have came up with this:
(x-6)(x^2+1) however, instead of passing through (2,-10), it passes through (2,-20)

Anyone know how to come up with a third degree polynomial with 6,-i as its zeros, and passes through (2,-10)? It has given me a headache!

Thank you in advance.
 
Mathematics news on Phys.org
lashia said:
Problem:

Find a third degree polynomial with rational coefficients if two of its zeros
are 6 and – 𝑖 and it passes through the point (2, -10)So far, I have came up with this:
(x-6)(x^2+1) however, instead of passing through (2,-10), it passes through (2,-20)

Anyone know how to come up with a third degree polynomial with 6,-i as its zeros, and passes through (2,-10)? It has given me a headache!

Thank you in advance.

you have done right except a small modification that you need to yo do

$f(x) = A(x-6)(x^2+1)$ where A is a constant
putting the values you get A = 2

so $f(x) = 2(x-6)(x^2+1)$
 
Last edited:
Thank you so much.. what a relief!
 
kaliprasad said:
you have done right except a small modification that you need to yo do

$f(x) = A(x-6)(x^2+1)$ where A is a constant
putting the values you get A = 2

so $f(x) = 2(x-6)(x^2+1)$

It doesn't seem tp pass through (2,-10) :confused:
 
Hi lashia,

kaliprasad's value of $A$ should instead be $A = 1/2$. Check now to see that the polynomial passes through $(2,-10)$ with $A = 1/2$.
 
lashia said:
It doesn't seem tp pass through (2,-10) :confused:

Let's go back to:

$$f(x)=A(x-6)\left(x^2+1\right)$$

Now, we set:

$$f(2)=A(2-6)\left(2^2+1\right)=-20A=-10\implies A=\frac{1}{2}$$

And so we have:

$$f(x)=\frac{1}{2}(x-6)\left(x^2+1\right)$$ :D
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top