MHB Finding a third degree polynomial... stumped

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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.
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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.
 
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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
 
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