Finding a third degree polynomial... stumped

[lashia]

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.

 

[kaliprasad]

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

 

[lashia]

Thank you so much.. what a relief!

 

[lashia]

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)

 

[Euge]

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 $A = 1/2$.

 

[MarkFL]

It doesn't seem tp pass through (2,-10)

Let's go back to:

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

Now, we set:

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

And so we have:

\(\displaystyle f(x)=\frac{1}{2}(x-6)\left(x^2+1\right)\) :D