Chapter 2.3, Problem 39E

To determine

To write the function $f(x)=x^3+3x^2-2x-14$ in the form of $f(x)=(x-k)q(x)+r(x)$ if the value of k is $\sqrt{2}$ . Also, demonstrate $f(k)=r$ using a graph.

Answer to Problem 39E

  $f(x)=(x-\sqrt{2})(x^2+(3+\sqrt{2})x+3\sqrt{2})-8$

Explanation of Solution

Given:

Function: $f(x)=x^3+3x^2-2x-14$

Value of k: $\sqrt{2}$

Calculation:

1. Expressing given function in the form of $f(x)=(x-k)q(x)+r(x)$

To express the given function in the above form, divide the given function with $x-k=x-\sqrt{2}$

  $\Rightarrow (x^3+3x^2-2x-14)÷(x-\sqrt{2})$

Dividend: $x^3+3x^2-2x-14$

Divisor: $x-\sqrt{2}$

Step 1: Writing coefficients of polynomial from given dividend on the right side and constant term of given divisor with opposite sign of the constant on the left side.

  $\begin{array}{ccccc}\sqrt{2}& 1& 3& -2& -14\\ & & & & \\ & & & & \end{array}$

Step 2: Now, adding terms in columns and multiplying the results by $\sqrt{2}$ .

  $\begin{array}{ccccc}\sqrt{2}& 1& 3& -2& -14\\ & & \sqrt{2}& 3\sqrt{2}+2& 6\\ & 1& 3+\sqrt{2}& 3\sqrt{2}& \boxed{-8}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

So, quotient is $x^2+(3+\sqrt{2})x+3\sqrt{2}$ and remainder is -8.

Hence, $f(x)=(x-\sqrt{2})(x^2+(3+\sqrt{2})x+3\sqrt{2})-8$

2. Demonstration of $f(\sqrt{2})=-8$ using graph.

Calculation for graph:

Consider $f(x)=x^3+3x^2-2x-14$

Values of x	Values of f (x)
0	-14
1	-12
-1	-10
2	2
-2	-6

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

Graph:

Interpretation:

   By observing graph, the value of $f(k)=f(\sqrt{2})$ is equal to -8.

Therefore, $f(\sqrt{2})=f(1.414)=-8.$

Conclusion:

Therefore, the given function can be written as $f(x)=(x-\sqrt{2})(x^2+(3+\sqrt{2})x+3\sqrt{2})-8$ .