/*! 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 120 Let $$ p(x)=x^{2}+\sqrt{2} x... [FREE SOLUTION] | 91Ó°ÊÓ

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Let $$ p(x)=x^{2}+\sqrt{2} x+1 $$ Find a polynomial \(q(x)\) such that the product \(p(x) q(x)\) has integer coefficients. $$ \triangle $$

Short Answer

Expert verified
Let \( q(x) = x^2 - \sqrt{2}x + 1 \). The product \( p(x)q(x) \) is \( x^4 - x^2 + 1 \), an integer-coefficient polynomial.

Step by step solution

01

Understand the Problem

We are given the polynomial \( p(x) = x^2 + \sqrt{2} x + 1 \). The goal is to find another polynomial \( q(x) \) such that the product \( p(x)q(x) \) has integer coefficients. The presence of \( \sqrt{2} \) in \( p(x) \) indicates that we need to construct \( q(x) \) in such a way to eliminate \( \sqrt{2} \) from the product.
02

Consider the Conjugate

To eliminate the \( \sqrt{2} \) term, consider using its conjugate. The conjugate of \( \sqrt{2} \) is \( -\sqrt{2} \). Construct the polynomial \( q(x) = x^2 - \sqrt{2} x + 1 \). Notice that this polynomial is similar to \( p(x) \) but with the middle term negated.
03

Multiply the Polynomials

Multiply \( p(x) = x^2 + \sqrt{2}x + 1 \) with \( q(x) = x^2 - \sqrt{2}x + 1 \). Calculate each term of the product:- \((x^2)(x^2) = x^4\)- \((x^2)(-\sqrt{2}x) = -\sqrt{2}x^3\)- \((x^2)(1) = x^2\)- \((\sqrt{2}x)(x^2) = \sqrt{2}x^3\)- \((\sqrt{2}x)(-\sqrt{2}x) = -2x^2\)- \((\sqrt{2}x)(1) = \sqrt{2}x\)- \((1)(x^2) = x^2\)- \((1)(-\sqrt{2}x) = -\sqrt{2}x\)- \((1)(1) = 1\)Add all these terms together.
04

Simplify the Product

Combine like terms:\[ x^4 + (-\sqrt{2}x^3 + \sqrt{2}x^3) + (x^2 - 2x^2 + x^2) + (\sqrt{2}x - \sqrt{2}x) + 1 \]Notice that the \( \sqrt{2}x^3 \) and \( \sqrt{2}x \) terms cancel out, which simplifies to:\[ x^4 + 0x^3 + 0x^2 + 1 \]This results in the polynomial \( x^4 - x^2 + 1 \), which has integer coefficients.

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Key Concepts

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

Integer Coefficients
When working with polynomials, the coefficients are the numbers in front of the variable terms. Sometimes these coefficients can be whole numbers, which are called integer coefficients. Having integer coefficients is often desirable because they are straightforward to compute with and understand. In the exercise given, the goal was to manipulate the polynomial so its resulting product has integer coefficients.
To achieve this, you may need to remove or neutralize irrational numbers or radicals, like the square root of two (\( \sqrt{2} \)). By choosing a partner polynomial wisely, any terms that involve irrational coefficients can cancel out when multiplied, leaving behind a polynomial where all coefficients are integers. This is exactly what was done by selecting the conjugate polynomial and multiplying it with the original polynomial.
Conjugate Root Theorem
The Conjugate Root Theorem is a useful concept when dealing with polynomials that contain radicals. Basically, if you have a polynomial with real coefficients and a non-real root such as a complex number or a radical, then the conjugate of that root will also be a root of the polynomial.
For example, if you have a term with \( \sqrt{2} \), its conjugate would be \( -\sqrt{2} \). When constructing the polynomial \( q(x) \) in the exercise, we applied this theorem by considering the conjugate of the \( \sqrt{2} \) present in \( p(x) \). Hence, \( q(x) \) was constructed to eliminate the non-integer coefficients when multiplied with \( p(x) \), resulting in a polynomial with all integer coefficients.
Polynomial Simplification
Simplifying polynomials involves reducing them to their most compact and manageable form. This includes combining like terms and eliminating terms that cancel each other out. By doing this, you make the polynomial easier to analyze and work with.
In the exercise, after multiplying \( p(x) \) and \( q(x) \), the expanded product had terms like \( -\sqrt{2}x^3 + \sqrt{2}x^3 \), which canceled each other out. Similarly, \( \sqrt{2}x - \sqrt{2}x \) canceled out. These cancellations are a key aspect of polynomial simplification and lead to a simpler polynomial that is easier to interpret, verify, or use in further calculations.
Radicals in Polynomials
A radical in a polynomial often arises when the polynomial includes an expression under a square root, cube root, or any root. These radicals can make calculations more complicated. However, there are strategies to work with radicals effectively, one of which involves the conjugate.
By forming a conjugate polynomial, radicals can often be eliminated through multiplication. This was illustrated in the exercise by pairing \( p(x) \) with its conjugate \( q(x) \). Radicals in the resulting product were negated, thereby simplifying the polynomial to a form with integer coefficients. It's a common technique to ensure a polynomial is easier to work with by removing the complexities introduced by radicals.

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Most popular questions from this chapter

(a)(i) For any positive real numbers \(a, b,\) prove that $$ \sqrt{a}+\sqrt{b}=\sqrt{a+b+\sqrt{4 a b}} $$ (ii) Simplify \(\sqrt{5+\sqrt{24}}\). (b) (i) Find a similar formula for \(\sqrt{a}-\sqrt{b}\). (ii) Simplify \(\sqrt{5-\sqrt{16}}\) and \(\sqrt{6-\sqrt{20}}\).

(Integer polynomials with a given root) We know that \(\alpha=1\) is a root of the polynomial equation \(x^{2}-1=0 ;\) that \(\alpha=\sqrt{2}\) is a root of \(x^{2}-2=0 ;\) and that \(\alpha=\sqrt{3}\) is a root of \(x^{2}-3=0\). (a) Find a quadratic polynomial with integer coefficients which has $$ \alpha=1+\sqrt{2} $$ as a root. (b) Find a quadratic polynomial with integer coefficients which has $$ \alpha=1+\sqrt{3} $$ as a root. (c) Find a polynomial with integer coefficients which has $$ \alpha=\sqrt{2}+\sqrt{3} $$ as a root. What are the other roots of this polynomial? (d) Find a polynomial with integer coefficients which has $$ \alpha=\sqrt{2}+\frac{1}{\sqrt{3}} $$ as a root. What are the other roots of this polynomial?

(a) I am thinking of two numbers, and am willing to tell you their sum \(s\) and their product \(p\). Express the following procedure algebraically and explain why it will always determine my two unknown numbers. Halve the sum \(s,\) and square the answer. Then subtract the product \(p\) and take the square root of the result, to get the answer. Add "the answer" to half the sum and you have one unknown number; subtract "the answer" from half the sum and you have the other unknown number. (b) I am thinking of the length of one side of a square. All I am willing to tell you are two numbers \(b\) and \(c,\) where when I add \(b\) times the side length to the area I get the answer \(c\). Express the following procedure algebraically and explain why it will always determine the side length of my square. Take one half of \(b\), square it and add the result to \(c\). Then take the square root. Finally subtract half of \(b\) from the result. (c) A regular pentagon \(A B C D E\) has sides of length 1 . (i) Prove that the diagonal \(A C\) is parallel to the side \(E D\). (ii) If \(A C\) and \(B D\) meet at \(X,\) explain why \(A X D E\) is a rhombus. (iii) Prove that triangles \(A D X\) and \(C B X\) are similar. (iv) If \(A C\) has length \(x,\) set up an equation and find the exact value of \(x\). Problem \(121(a),\) (b) link to Problem \(111(\) a) (and to Problem 129 below) in relating the roots and the coefficients of a quadratic. If we forget for the moment that the coefficients are usually known, while the roots are unknown, then we see that if \(\alpha\) and \(\beta\) are the roots of the quadratic $$ x^{2}+b x+c $$ then $$ (x-\alpha)(x-\beta)=x^{2}+b x+c $$ so $$ \alpha+\beta=-b \text { and } \alpha \beta=c $$ In other words, the two coefficients \(b, c\) are equal to the two simplest symmetric expressions in the two roots \(\alpha\) and \(\beta .\) Part (a) of the next problem is meant to suggest that all other symmetric expressions in \(\alpha\) and \(\beta\) (that is, any expression that is unchanged if we swap \(\alpha\) and \(\beta\) ) can then be written in terms of \(b\) and \(c\). The full result proving this fact is generally attributed to Isaac Newton \((1642-1727) .\) Part (b) suggests that, provided one is willing to allow case distinctions, something similar may be true of anti-symmetric expressions (where the effect of swapping \(\alpha\) and \(\beta\) is to multiply the expression by "-1").

Complex numbers \(a+b i\), where both \(a\) and \(b\) are integers, are called Gaussian integers. Try to formulate a version of the "division algorithm" for "division with remainder" (where the remainder is always "less than" the divisor in some sense) for pairs of Gaussian integers. Extend this to construct a version of the Euclidean algorithm to find the HCF of two given Gaussian integers. It is a profoundly erroneous truism ... that we should cultivate the habit of thinking what we are doing. The precise opposite is the case. Civilisation advances by extending the number of important operations which we can perform without thinking about them. Alfred North Whitehead \((1861-1947)\)

Numbers are assigned (secretly) to the vertices of a polygon. Each edge of the polygon is then labelled with the sum of the numbers at its two end vertices. (a) If the polygon is a triangle \(A B C,\) and the labels on the three sides are \(c\) (on \(A B), b\) (on \(A C)\), and \(a\) (on \(B C\) ), what were the numbers written at each of the three vertices? (b) If the polygon is a quadrilateral \(A B C D,\) and the labels on the four sides are \(w\) (on \(A B), x(\) on \(B C), y(\) on \(C D),\) and \(z(\) on \(D A),\) what numbers were written at each of the four vertices? (c) If the polygon is a pentagon \(A B C D E,\) and the labels on the five sides are \(d\) (on \(A B), e(\) on \(B C), a\) (on \(C D), b\) (on \(D E)\), and \(c\) (on \(E A\) ), what numbers were written at each of the five vertices?

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