Chapter 2.5, Problem 52E

a.

To determine

To write the factored form of the polynomial function of degree 4 whose zeros are $1,-4,\sqrt{3}i$ .

Also, $f(0)=-6$ .

Answer to Problem 52E

  $f(x)=\frac{1}{2}(x-1)(x+4)(x-\sqrt{3}i)(x+\sqrt{3}i)$

Explanation of Solution

Given:

Zeros of polynomial function: $1,-4,\sqrt{3}i$

Value: $f(0)=-6$

Concept used:

Linear Factorization Theorem:

If $f(x)$ is a polynomial of degree n, where n > 0, then f has precisely a linear factors

  $f(x)=a(x-c_1)(x-c_2)\mathrm{........}(x-c_n)$

Calculation:

As $\sqrt{3}i$ is a zero of the polynomial, the conjugate of $\sqrt{3}i$ is also a zero of the polynomial.

Conjugate of $\sqrt{3}i$ is $-\sqrt{3}i$ .

So, the zeros are $1,-4,\sqrt{3}i,-\sqrt{3}i$

Using above concept, $f(x)$ can be written as,

  $\begin{array}{l}\Rightarrow f(x)=a(x-(1))(x-(-4))(x-(\sqrt{3}i))(x-(-\sqrt{3}i))\\ \Rightarrow f(x)=a(x-1)(x+4)(x-\sqrt{3}i)(x+\sqrt{3}i)\end{array}$

According to question, $f(0)=-6,$

  $\begin{array}{l}\Rightarrow f(0)=-6\\ \Rightarrow a(0-1)(0+4)(0-\sqrt{3}i)(0+\sqrt{3}i)=-6\\ \Rightarrow a(-1)(4)(-3i^2)=-6\\ \Rightarrow a(-1)(4)(-3(-1))=-6\\ \Rightarrow a(-1)(4)(3)=-6\\ \Rightarrow a(-12)=-6\\ \Rightarrow a=\frac{1}{2}\end{array}$

So, polynomial function can be written as $f(x)=\frac{1}{2}(x-1)(x+4)(x-\sqrt{3}i)(x+\sqrt{3}i)$ .

Conclusion:

Therefore, the polynomial function in factored form is $f(x)=\frac{1}{2}(x-1)(x+4)(x-\sqrt{3}i)(x+\sqrt{3}i).$

b.

To determine

To write the polynomial form of the polynomial function of degree 4 whose zeros are $1,-4,\sqrt{3}i$ .

Answer to Problem 52E

  $f(x)=\frac{x^4}{2}+\frac{3x^3}{2}-\frac{x^2}{2}+\frac{9x}{2}-6$

Explanation of Solution

Calculation:

From above answer, the factored form of polynomial function is

  $f(x)=\frac{1}{2}(x-1)(x+4)(x-\sqrt{3}i)(x+\sqrt{3}i).$

Now, writing the above form as polynomial form,

  $\begin{array}{l}\Rightarrow f(x)=\frac{1}{2}(x-1)(x+4)(x-\sqrt{3}i)(x+\sqrt{3}i)\\ \Rightarrow f(x)=\frac{1}{2}(x^2+4x-x-4)(x^2+\sqrt{3}xi-\sqrt{3}xi-3i^2)\\ \Rightarrow f(x)=\frac{1}{2}(x^2+3x-4)(x^2-3(-1))[\because i^2=-1]\\ \Rightarrow f(x)=\frac{1}{2}(x^2+3x-4)(x^2+3)\\ \Rightarrow f(x)=\frac{1}{2}(x^4+3x^2+3x^3+9x-4x^2-12)\\ \Rightarrow f(x)=\frac{1}{2}(x^4+3x^3-x^2+9x-12)\\ \Rightarrow f(x)=\frac{x^4}{2}+\frac{3x^3}{2}-\frac{x^2}{2}+\frac{9x}{2}-6\end{array}$

Conclusion:

Therefore, the polynomial function in polynomial form is $f(x)=\frac{x^4}{2}+\frac{3x^3}{2}-\frac{x^2}{2}+\frac{9x}{2}-6.$