Chapter 2, Problem 62RE

a.

To determine

To verify whether $(x-2),(x-5)$ is a factor to the function $f(x)=x^4-11x^3+41x^2-61x+30$ .

Answer to Problem 62RE

Factor

Explanation of Solution

Given:

Function: $f(x)=x^4-11x^3+41x^2-61x+30$

Factor: $(x-2),(x-5)$

Calculation:

Verifying whether $x-2$ is a factor to given function or not by using synthetic division method.

Step 1: Writing coefficients of polynomial function on the right side and value of constant of factor with opposite sign on the left side.

  $\begin{array}{cccccc}2& 1& -11& 41& -61& 30\\ & & & & & \\ & & & & & \end{array}$

Step 2: Now, adding terms in columns and multiplying the results by 2.

  $\begin{array}{cccccc}2& 1& -11& 41& -61& 30\\ & & 2& -18& 46& -30\\ & 1& -9& 23& -15& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, the remainder is zero, it means that $x-2$ is a factor to given polynomial function.

Verifying whether $x-5$ is a factor to given function or not by using synthetic division method.

Step 1: Writing coefficients of polynomial function on the right side and value of constant of factor with opposite sign on the left side.

  $\begin{array}{cccccc}5& 1& -11& 41& -61& 30\\ & & & & & \\ & & & & & \end{array}$

Step 2: Now, adding terms in columns and multiplying the results by 5.

  $\begin{array}{cccccc}5& 1& -11& 41& -61& 30\\ & & 5& -30& 55& -30\\ & 1& -6& 11& -6& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, the remainder is zero, it means that $x-5$ is a factor to given polynomial function.

Conclusion:

Therefore, $(x-2),(x-5)$ are factors to given polynomial function.

b.

To determine

Find the remaining factors of the function $f(x)=x^4-11x^3+41x^2-61x+30$ .

Answer to Problem 62RE

  $(x-3),(x-1)$

Explanation of Solution

Given:

Function: $f(x)=x^4-11x^3+41x^2-61x+30$

Calculation:

  $(x-2)(x-5)=x^2-2x-5x+10=x^2-7x+10$

Now, dividing above result with given function to find unfactored part.

Dividend: $f(x)=x^4-11x^3+41x^2-61x+30$

Divisor: $x^2-7x+10$

  $x^2-7x+10\begin{array}{c}\hfill x^2-4x+3\\ \hfill \overline{)\begin{array}{l}{x}^4-11x^3+41x^2-61x+30\\ \begin{array}{c}{x}^4-7x^3+10x^2\end{array}\\ -4x^3+31x^2-61x+30\\ \begin{array}{c}-4x^3+28x^2-40x\end{array}\\ 3x^2-21x+30\\ \begin{array}{c}3x^2-21x+30\end{array}\\ 0\end{array}}\end{array}$   $\begin{array}{l}\frac{x^4}{x^2}=x^2\\ \text{Multiply}{x}^2(x^2-7x+10)\text{and subtract}\text{.}\\ \frac{\begin{array}{l}\\ -4x^3\end{array}}{x^2}=-4x\\ \text{Multiply}-4x(x^2-7x+10)\text{and subtract}\text{.}\\ \\ \frac{3x^2}{x^2}=3\\ \text{Multiply}3(x^2-7x+10)\text{and subtract}\text{.}\end{array}$

So, quotient is $x^2-4x+3$ and remainder is 0.

Hence, $f(x)=x^4-11x^3+41x^2-61x+30\iff f(x)=(x-2)(x-5)(x^2-4x+3)$

  $\begin{array}{l}\Rightarrow f(x)=(x-2)(x-5)(x^2-4x+3)\\ \Rightarrow f(x)=(x-2)(x-5)(x^2-3x-x+3)\\ \Rightarrow f(x)=(x-2)(x-5)(x(x-3)-1(x-3))\\ \Rightarrow f(x)=(x-2)(x-5)(x-3)(x-1)\\ \Rightarrow f(x)=(x-1)(x-2)(x-3)(x-5)\end{array}$

Conclusion:

Therefore, $(x-3),(x-1)$ are the remaining factors to the given polynomial function.

c.

To determine

Write the complete factorization of function $f(x)=x^4-11x^3+41x^2-61x+30$ .

Answer to Problem 62RE

  $f(x)=(x-1)(x-2)(x-3)(x-5)$

Explanation of Solution

Given:

Function: $f(x)=x^4-11x^3+41x^2-61x+30$

Calculation:

From above answers, the complete factorization of given polynomial function is

  $f(x)=(x-1)(x-2)(x-3)(x-5)$ .

Conclusion:

Therefore, $f(x)=(x-1)(x-2)(x-3)(x-5)$ is complete factorization to given polynomial function.

d.

To determine

Find the zeros of the function $f(x)=x^4-11x^3+41x^2-61x+30$ .

Answer to Problem 62RE

  $1,2,3,5$

Explanation of Solution

Given:

Function: $f(x)=x^4-11x^3+41x^2-61x+30$

Calculation:

To find the zeros, put $f(x)=0.$

  $\begin{array}{l}\Rightarrow (x-1)(x-2)(x-3)(x-5)=0\\ \Rightarrow x=1,2,3,5\end{array}$

Calculation for graph:

Consider $f(x)=x^4-11x^3+41x^2-61x+30$

Values of x	Values of f ( x )
0	30
1	0
-1	144
2	0
-2	420

By taking different values of x , the graph can be plotted.

Graph:

Interpretation:

By observing graph, it is clear that the curve of the function meets x -axis at $x=1,2,3\text{and}5.$

Hence, the zeros of the function are $x=1,2,3\text{and}5.$

Conclusion:

Therefore, the zeros of given polynomial function are $x=1,2,3\text{and}5.$