/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 A polynomial P is given. (a) Fin... [FREE SOLUTION] | 91Ó°ÊÓ

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A polynomial P is given. (a) Find all zeros of P, real and complex. (b) Factor P completely. \(P(x)=x^{4}-x^{2}-2\)

Short Answer

Expert verified
Zeros: \( \sqrt{2}, -\sqrt{2}, i, -i \). Factor: \((x^2 - 2)(x^2 + 1)\).

Step by step solution

01

Understand the Polynomial

The given polynomial is \(P(x) = x^4 - x^2 - 2\). This is a quartic polynomial because the highest power of \(x\) is 4.
02

Substitute and Simplify

To solve \(P(x)\), notice it can be rewritten in the form of a quadratic. Substitute \(y = x^2\). The polynomial becomes \(P(y) = y^2 - y - 2\).
03

Solve the Quadratic Equation

The quadratic equation \(y^2 - y - 2 = 0\) can be solved using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -1\), \(c = -2\).
04

Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac = (-1)^2 - 4 \times 1 \times (-2) = 1 + 8 = 9\). Since the discriminant is positive, there are two real roots.
05

Find the Roots of the Quadratic

Using the quadratic formula: \(y = \frac{-(-1) \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2}\). The solutions are \(y = 2\) and \(y = -1\).
06

Solve for x

Recall that \(y = x^2\). Thus: For \(y = 2\), \(x^2 = 2\) giving roots \(x = \sqrt{2}\) and \(x = -\sqrt{2}\). For \(y = -1\), \(x^2 = -1\) yields complex roots \(x = i\) and \(x = -i\), where \(i\) is the imaginary unit.
07

List All Zeros

The zeros of \(P(x)\) are \(\sqrt{2}, -\sqrt{2}, i,\) and \(-i\).
08

Factor the Polynomial Completely

Given zeros, the factorization of \(P(x)\) is \((x - \sqrt{2})(x + \sqrt{2})(x - i)(x + i)\). This simplifies to \((x^2 - 2)(x^2 + 1)\).

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Key Concepts

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

Quartic Polynomials
A quartic polynomial is a polynomial with a degree of four, meaning it is expressed in the form of \(ax^4 + bx^3 + cx^2 + dx + e\). In this particular exercise, the polynomial \(P(x) = x^4 - x^2 - 2\) is already in standard form. Understanding its components is key to solving it. The highest power, which is 4, indicates that this is a quartic polynomial. Such equations can potentially have up to four zeros, which could be a mix of real and complex numbers depending on the coefficients and the discriminant of any transformed quadratic expressions. They are an integral part of algebra due to their complexity and the variety of methods needed for factorization.
Complex Roots
Complex roots occur when a polynomial equation has solutions that involve the imaginary unit \(i\), defined as \(i^2 = -1\). In the context of quadratic equations derived from quartic polynomials, complex roots often emerge through solutions of the square root of a negative number. In the exercise, after substitution, we find the quadratic equation \(y = x^2 = -1\), resulting in \(x = i\) and \(x = -i\). These roots reflect the fact that the quartic polynomial does not only have real roots but includes these complex numbers, demonstrating a richer polynomial algebraic structure. Complex roots always appear in conjugate pairs, such as \(i\) and \(-i\), ensuring the solution sets remain real when multiplied back into polynomial form.
Quadratic Substitution
Quadratic substitution is a strategic method used to simplify solving higher degree polynomials. By substituting a variable to replace part of a quartic polynomial equation, the problem is transformed into a quadratic format, which is easier to handle. For example, we use \(y = x^2\) to convert the quartic \(P(x) = x^4 - x^2 - 2\) into the quadratic \(y^2 - y - 2 = 0\). This technique allows us to apply familiar solutions like the quadratic formula to find the roots of the equation, which can then be interpreted for the original variable \(x\). This substitution not only simplifies the calculations but also reveals any hidden complex roots in the solution.
Zeros of a Polynomial
Zeros of a polynomial, also known as roots or solutions, are the values of \(x\) for which the polynomial equals zero. They are vital in determining the polynomial's factorization and the shape of its graph. For \(P(x) = x^4 - x^2 - 2\), finding the zeros involves solving both real and complex equations resulting from the quadratic method. After solving the quadratic versions of the polynomial, zeros like \(\sqrt{2}, -\sqrt{2}, i, \text{and}\ -i\) emerge. These zeros provide essential insight into the factorization of the polynomial, which reveals the polynomial's complete expression as a product of its linear factors \((x - \sqrt{2})(x + \sqrt{2})(x - i)(x + i)\). Understanding zeros is crucial as they directly relate to the solutions and the real-world applications of polynomial equations.

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Most popular questions from this chapter

Find all rational zeros of the polynomial, and write the polynomial in factored form. $$ P(x)=x^{5}+3 x^{4}-9 x^{3}-31 x^{2}+36 $$

Volume of a Rocket A rocket consists of a right circular cylinder of height 20 \(\mathrm{m}\) surmounted by a cone whose height and diameter are equal and whose radius is the same as that of the cylindrical section. What should this radius be (rounded to two decimal places) if the total volume is to be 500\(\pi / 3 \mathrm{m}^{3}\) ?

Use Descartes’ Rule of Signs to determine how many positive and how many negative real zeros the polynomial can have. Then determine the possible total number of real zeros. $$ P(x)=x^{4}+x^{3}+x^{2}+x+12 $$

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How Many Real Zeros Can a Polynomial Have? Give examples of polynomials that have the following properties, or explain why it is impossible to find such a polynomial. (a) A polynomial of degree 3 that has no real zeros (b) A polynomial of degree 4 that has no real zeros (c) A polynomial of degree 3 that has three real zeros, only one of which is rational (d) \(\mathrm{A}\) polynomial of degree 4 that has four real zeros, none of which is rational What must be true about the degree of a polynomial with integer coefficients if it has no real zeros?

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