Question

7) find a polynomial with Real coefficients of degree 4, and with zeros:3, with multiplicity two; i with multiplicity one; containing the point (0, - 9).

Answer 1

The given polynomial is said to have a degree of 4. It has zeros 3 (multiplicity two) and i (multiplicity one) and contains the point (0,-9).

It is required to find the equation of the polynomial.

Recall the Complex Conjugate Theorem: A polynomial function or a polynomial equation with real coefficients that has a+bi as a complex zero, with b not zero, also has a-bi as a zero.

Since i is a zero of the polynomial, its conjugate -i is also a zero.

Hence, the zeros of the polynomial are 3,i,-i.

Recall from the factor theorem that if k is a zero of a polynomial with multiplicity m, then the expression,

`(x-k)^m`

is a factor of the polynomial.

It follows that the polynomial has factors:

`(x-3)^2,(x-i),(x+i)`

Hence, the polynomial can be written as a product of its factors and a real constant, a:

`y=a(x-3)^2(x-i)(x+i)`

Simplify the product on the right as follows:

`\begin{gathered} \text{ Use binomial expansion and difference of two squares:} \\ y=a(x^2-6x+9)(x^2-i^2) \\ \Rightarrow y=a(x^2-6x+9)(x^2-(-1)) \\ \Rightarrow y=a(x^2-6x+9)(x^2+1) \end{gathered}`

Expand the expression further:

`\begin{gathered} \Rightarrow y=a(x^4+x^2-6x^3-6x+9x^2+9) \\ \Rightarrow y=a(x^4-6x^3+x^2+9x^2-6x+9) \\ \Rightarrow y=a(x^4-6x^3+10x^2-6x+9) \end{gathered}`

Since it is given that the polynomial contains the point (0,-9), substitute (x,y)=(0,-9) into the equation to find the value of a:

`\begin{gathered} -9=a(0^4-6(0^3)+10(0^2)-6(0)+9) \\ \Rightarrow-9=a(9) \\ \Rightarrow9a=-9 \\ \Rightarrow(9a)/(9)=-(9)/(9) \\ \Rightarrow a=-1 \end{gathered}`

Substitute a=-1 back into the equation.

Hence, the required polynomial is:

`\begin{gathered} y=-1(x^4-6x^3+10x^2-6x+9) \\ \Rightarrow y=-x^4+6x^3-10x^2+6x-9 \end{gathered}`