Question

Find a polynomial with integer coefficients that satisfies the given conditions. Q has degree 3 and zeros −9 and 1 + i.Q(x) =

Answer 1

To find the polynomial with integer coefficients that satisfies the given conditions, we use the fact that the zeros of the polynomial correspond to its factors. The polynomial Q(x) is found to be x^2 + (8 - (1 + i))x - (9 + 9i).

Step-by-step explanation:

In this case, the polynomial Q(x) that satisfies the given conditions can be found using the fact that the zeros of the polynomial correspond to its factors. Since the zeros are -9 and 1 + i, we can write the polynomial as Q(x) = (x + 9)(x - (1 + i)).

Expanding this expression, we get Q(x) = (x + 9)(x - 1 - i). Using the distributive property, we can further simplify this to Q(x) = x^2 - x - ix + 9x - 9 - 9i. Combining like terms, we have Q(x) = x^2 + 8x - 9 - (1 + i)x - 9i.

Finally, rearranging the terms, we get the polynomial in the desired form: Q(x) = x^2 + (8 - (1 + i))x - (9 + 9i).

Answer 2

Notice that Q(x) has a complex root; therefore, in order to get a polynomial with integer coefficients (real coefficients), we need to consider another root to be the complex conjugate of the given complex root.

In general, the conjugate of a complex number is

`\begin{gathered} a+ib\rightarrow a-ib \\ \\ \\ \\ \end{gathered}`

Thus, in our case,

`1+i\rightarrow1-i`

Then, Q(x) is

`\Rightarrow Q(x)=(x+9)(x-1-i)(x-1+i)`

Expanding Q(x),

`\begin{gathered} Q(x)=(x+9)(x-(1+i))(x-(1-i)) \\ \Rightarrow Q(x)=(x+9)(x^2-((1+i)+(1-i))x+(1+i)(1-i) \\ \Rightarrow Q(x)=(x+9)(x^2-2x+2) \\ \end{gathered}`
`\Rightarrow Q(x)=x^3+7x^2-16x+18`

Thus, the answer is Q(x)=x^3+7x^2-16x+18

Calculations in more detail

`\begin{gathered} (x-i-1)(x-1+i) \\ Set \\ z=i+1\text{ for simplicity} \\ \Rightarrow z^*=1-i \end{gathered}`

Thus,

`\begin{gathered} \Rightarrow(x-i-1)(x-1+i)=(x-z)(x-z^*)=x^2-zx-z^*x+zz^* \\ =x^2-(z+z^*)x+zz^* \end{gathered}`

Then,

`\begin{gathered} zz^*=(1+i)(1-i)=1+i-i-i^2=1-i^2=1-(-1)=2 \\ and \\ z+z^*=(1+i)+(1-i)=1+i+1-i=2 \end{gathered}`

Therefore,

`\Rightarrow(x-i-1)(x-1+i)=x^2-(2)x+2=x^2-2x+2`

Then, we can express Q(x) as

`\Rightarrow Q(x)=(x+9)(x^2-2x+2)`

And we only need to multiply as shown above, but now all the numbers are integers; we do not need to deal with complex numbers anymore.