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Find all values of \(k\) such that \(f(x)\) is divisible by the given linear polynomial. $$f(x)=k x^{3}+x^{2}+k^{2} x+3 k^{2}+11 ; \quad x+2$$

Short Answer

Expert verified
The values of \(k\) that make \(f(x)\) divisible by \(x+2\) are \(k = 3\) and \(k = 5\).

Step by step solution

01

Understand Divisibility Criteria

If the polynomial \(f(x)\) is divisible by \(x+2\), it means that \(x = -2\) is a root of \(f(x)\). Therefore, \(f(-2) = 0\).
02

Substitute -2 into f(x)

Substitute \(-2\) for \(x\) in the polynomial \(f(x) = k x^{3} + x^{2} + k^{2} x + 3 k^{2} + 11\). This gives us \(f(-2) = k(-2)^3 + (-2)^2 + k^2(-2) + 3k^2 + 11\).
03

Simplify and Set the Expression to Zero

Simplify the expression from Step 2:\[-8k + 4 - 2k^2 + 3k^2 + 11\]Combine like terms: \(-8k + k^2 + 15\).Set this equal to zero: \[k^2 - 8k + 15 = 0\].
04

Solve the Quadratic Equation

Solve the quadratic equation \(k^2 - 8k + 15 = 0\) using the formula:\(k = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where \(a=1\), \(b=-8\), and \(c=15\).Calculate the discriminant: \((-8)^2 - 4 \times 1 \times 15 = 64 - 60 = 4\).Find the roots: \(k = \frac{8 \pm \sqrt{4}}{2}\).
05

Calculate the Roots

Calculate the two possible values for \(k\):\(k = \frac{8 + 2}{2} = 5\), and \(k = \frac{8 - 2}{2} = 3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Roots of Polynomials
When we talk about the roots of a polynomial, we are referring to the values of the variable that make the polynomial equal to zero. Imagine you have a polynomial function, let's call it \(f(x)\). If \(f(a) = 0\), then \(a\) is considered a root of the polynomial. This principle is often used to determine factors of the polynomial as well.

For instance, if a polynomial \(f(x)\) is divisible by another polynomial like \(x+2\), it means that \(-2\) is a root of \(f(x)\). Essentially, when \(-2\) is substituted into \(f(x)\), the result should be zero.

Finding the roots helps in factorizing the polynomial and is a crucial step in identifying divisibility. Roots also play an essential role in solving complex equations because they provide the solutions to the polynomial equations. Solving for roots might involve quadratic equations and sometimes, higher-degree polynomials in more complex scenarios.
Quadratic Equation
Quadratic equations are fundamental in algebra and come in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. The solutions for \(x\) in this equation are the values that satisfy the equation, often referred to as the roots.

To solve a quadratic equation, one of the most common methods is using the quadratic formula:
  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
You can also factor the quadratic equation or complete the square, though the quadratic formula is usually the most straightforward approach.

In the example problem, the quadratic equation is derived as \(k^2 - 8k + 15 = 0\), and we solve it using the quadratic formula. Here, the discriminant \(b^2 - 4ac\) determines the nature of the roots, whether they are real or complex. A positive discriminant, like 4 in this case, indicates two distinct real roots.
Polynomial Division
Polynomial division is quite similar to numerical division that you might be familiar with. It involves dividing a polynomial by another polynomial, which might sometimes leave a remainder. However, when a polynomial is exactly divisible by a divisor, the remainder is zero.

In our given problem, the polynomial \(f(x)\) is divided by a linear polynomial \(x+2\). To determine divisibility, we substitute \(-2\) into \(f(x)\) and simplify. If the result yields zero, this means \(f(x)\) is divisible by \(x+2\), confirming that \(-2\) is a root of \(x+2\).

Polynomial division becomes particularly significant when simplifying equations or solving for unknowns, as it helps in breaking down complex polynomials into simpler factors and terms. Comprehension of polynomial division allows for deeper insights into the nature and behavior of polynomials.

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Most popular questions from this chapter

A right triangle has area \(30 \mathrm{ft}^{2}\) and a hypotenuse that is 1 foot longer than one of its sides. (a) If \(x\) denotes the length of this side, then show that \(2 x^{3}+x^{2}-3600=0\). (b) Show that there is a positive root of the equation in part (a) and that this root is less than 13. (c) Find the lengths of the sides of the triangle.

Exer. 31-34: Examine the expression for the given set of data points of the form \((x, y) .\) Find the constant of variation and a formula that describes how \(y\) varles with respect to \(x .\) $$\begin{aligned} &x y ;\\{(0.2,-26.5),(0.4,-13.25),(0.8,-6.625)\\\ &(1.6,-3.3125),(3.2,-1.65625)\\} \end{aligned}$$

Constructing a crate The frame for a shipping crate is to be constructed from 24 feet of \(2 \times 2\) lumber. Assuming the crate is to have square ends of length \(x\) feet, determine the value(s) of \(x\) that result(s) in a volume of \(4 \mathrm{ft}^{3}\).

A scientist has limited data on the temperature \(T\) (in \(^{\circ} \mathrm{C}\) ) during a 24 -hour period. If \(t\) denotes time in hours and \(t=0\) corresponds to midnight, find the fourthdegree polynomial that fits the information in the following table. $$\begin{array}{|l|lllll|} \hline t \text { (hours) } & 0 & 5 & 12 & 19 & 24 \\ \hline T\left(^{\circ} \mathrm{C}\right) & 0 & 0 & 10 & 0 & 0 \\ \hline \end{array}$$

Refer to Exercise 86 in Section 2.4 The distance \(D\) (in feet) required for a car to safely stop varies directly with its speed \(S\) (in mi/hr). Use the table to determine an approximate value for \(k\) in the variation formula \(D=k S^{23}\) $$\begin{array}{|ccccccc|} \hline \boldsymbol{s} & 20 & 30 & 40 & 50 & 60 & 70 \\ \hline \boldsymbol{D} & 33 & 86 & 167 & 278 & 414 & 593 \\ \hline \end{array}$$ (b) Check your approximation by graphing both the data and \(D\) on the same coordinate axes.

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