Polynomials

Solution

A polynomial is an expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. Polynomials are classified based on their degree. A polynomial of degree 1 is called a linear polynomial, e.g., 2x + 3. A polynomial of degree 2 is called a quadratic polynomial, e.g., x^2 - 3x + 2. A polynomial of degree 3 is called a cubic polynomial, e.g., 2x^3 - x^2 + x - 5. Polynomials are used in various fields like physics, engineering, and economics to model real-world situations.

Solution

To find the zeroes of the polynomial x^2 - 5x + 6, we first factorize it: x^2 - 5x + 6 = (x - 2)(x - 3). Setting each factor equal to zero gives x = 2 and x = 3 as the zeroes. The sum of the zeroes is 2 + 3 = 5, which is equal to the negative of the coefficient of x divided by the coefficient of x^2, i.e., -(-5)/1 = 5. The product of the zeroes is 2 * 3 = 6, which is equal to the constant term divided by the coefficient of x^2, i.e., 6/1 = 6. This verifies the relationship between the zeroes and the coefficients.

Solution

The zeroes of a quadratic polynomial are the x-coordinates of the points where the graph of the polynomial intersects the x-axis. For example, the polynomial x^2 - 3x - 4 has zeroes at x = -1 and x = 4, which means its graph intersects the x-axis at (-1, 0) and (4, 0). The graph of a quadratic polynomial is a parabola, and the number of zeroes corresponds to the number of times the parabola intersects the x-axis. If the parabola intersects the x-axis at two points, the polynomial has two distinct real zeroes. If it touches the x-axis at one point, the polynomial has one real zero (a repeated root). If it does not intersect the x-axis, the polynomial has no real zeroes.

Solution

Given the sum of the zeroes (α + β) = 4 and the product of the zeroes (αβ) = 1, the quadratic polynomial can be written as x^2 - (sum of zeroes)x + (product of zeroes) = x^2 - 4x + 1. Therefore, the required quadratic polynomial is x^2 - 4x + 1. This polynomial will have zeroes that satisfy the given sum and product conditions.

Solution

The division algorithm for polynomials states that given two polynomials p(x) and g(x), where g(x) ≠ 0, there exist unique polynomials q(x) and r(x) such that p(x) = g(x) * q(x) + r(x), where the degree of r(x) is less than the degree of g(x). For example, let p(x) = x^3 - 3x^2 + 3x - 1 and g(x) = x - 1. Dividing p(x) by g(x) gives q(x) = x^2 - 2x + 1 and r(x) = 0, since (x - 1)(x^2 - 2x + 1) = x^3 - 3x^2 + 3x - 1. Here, the remainder r(x) is 0, indicating that g(x) is a factor of p(x).

Solution

Given that 1 and -2 are zeroes of the polynomial x^3 - 4x^2 - 7x + 10, we can factor out (x - 1) and (x + 2) from the polynomial. First, perform polynomial division or use synthetic division to divide the polynomial by (x - 1)(x + 2) = x^2 + x - 2. This gives the quotient as x - 5. Therefore, the polynomial can be written as (x - 1)(x + 2)(x - 5). Setting each factor equal to zero gives the zeroes x = 1, x = -2, and x = 5. Thus, all the zeroes of the polynomial are 1, -2, and 5.

Solution

The polynomial x^2 + 1 has no real zeroes because the equation x^2 + 1 = 0 implies x^2 = -1. In the real number system, the square of any real number is always non-negative, so there is no real number x that satisfies x^2 = -1. Therefore, the polynomial x^2 + 1 does not intersect the x-axis and has no real zeroes. Its zeroes are complex numbers, specifically x = i and x = -i, where i is the imaginary unit.

Solution

Given the zeroes of the polynomial x^3 - 3x^2 + x + 1 are a - b, a, and a + b, the sum of the zeroes is (a - b) + a + (a + b) = 3a. According to the relationship between the zeroes and coefficients, the sum of the zeroes is equal to the negative of the coefficient of x^2 divided by the coefficient of x^3, which is -(-3)/1 = 3. Therefore, 3a = 3, so a = 1. The product of the zeroes is (a - b) * a * (a + b) = a(a^2 - b^2). According to the coefficient relationship, the product of the zeroes is equal to the negative of the constant term divided by the coefficient of x^3, which is -1/1 = -1. Substituting a = 1 gives 1(1 - b^2) = -1, so 1 - b^2 = -1, leading to b^2 = 2 and b = ±√2. Thus, a = 1 and b = ±√2.

Solution

To verify that 2, -1, and -1/2 are the zeroes of the polynomial 2x^3 - x^2 - 5x - 2, we substitute each value into the polynomial. For x = 2: 2(2)^3 - (2)^2 - 5(2) - 2 = 16 - 4 - 10 - 2 = 0. For x = -1: 2(-1)^3 - (-1)^2 - 5(-1) - 2 = -2 - 1 + 5 - 2 = 0. For x = -1/2: 2(-1/2)^3 - (-1/2)^2 - 5(-1/2) - 2 = -1/4 - 1/4 + 5/2 - 2 = (-1/4 - 1/4) + (5/2 - 2) = -1/2 + 1/2 = 0. Since all three values satisfy the polynomial equation, they are indeed the zeroes of the polynomial.

Solution

First, find the zeroes of the polynomial x^2 - 4x + 3. Factorizing gives (x - 1)(x - 3) = 0, so the zeroes are x = 1 and x = 3. The reciprocals of these zeroes are 1/1 = 1 and 1/3. The sum of the reciprocals is 1 + 1/3 = 4/3, and the product is 1 * 1/3 = 1/3. The quadratic polynomial with these sum and product of zeroes is x^2 - (sum of zeroes)x + (product of zeroes) = x^2 - (4/3)x + 1/3. To eliminate fractions, multiply by 3 to get 3x^2 - 4x + 1. Therefore, the required quadratic polynomial is 3x^2 - 4x + 1.

Solution

The remainder theorem states that if a polynomial p(x) is divided by (x - a), the remainder is p(a). This theorem is significant because it provides a quick way to evaluate the remainder without performing the entire division process. For example, consider the polynomial p(x) = x^3 - 2x^2 + 3x - 4. To find the remainder when p(x) is divided by (x - 2), we simply evaluate p(2) = (2)^3 - 2(2)^2 + 3(2) - 4 = 8 - 8 + 6 - 4 = 2. Therefore, the remainder is 2. The remainder theorem is also useful in factor theorem, where if p(a) = 0, then (x - a) is a factor of p(x).

Solution

A linear polynomial is of degree 1 and its general form is ax + b, where a ≠ 0. Example: 2x + 3. Its graph is a straight line. A quadratic polynomial is of degree 2 and its general form is ax² + bx + c, where a ≠ 0. Example: x² - 3x - 4. Its graph is a parabola. The linear polynomial intersects the x-axis at one point, while the quadratic polynomial can intersect at two points, one point, or not at all, depending on the discriminant.

Solution

The polynomial x² - 5x + 6 can be factored as (x - 2)(x - 3). Thus, the zeroes are x = 2 and x = 3. Sum of zeroes = 2 + 3 = 5 = -(-5)/1 = -b/a. Product of zeroes = 2 * 3 = 6 = 6/1 = c/a. This verifies the relationships.

Solution

Given that 2 is a zero, substitute x = 2 into the polynomial: 2² + 3*2 + k = 0 → 4 + 6 + k = 0 → k = -10. The polynomial becomes x² + 3x - 10. The other zero can be found by factoring: (x + 5)(x - 2) = 0 → x = -5 or x = 2. Thus, the other zero is -5.

Solution

A linear polynomial, being of degree 1, can have exactly one zero. Example: 3x + 2 has zero at x = -2/3. A cubic polynomial, being of degree 3, can have up to three zeroes. Example: x³ - 6x² + 11x - 6 has zeroes at x = 1, x = 2, and x = 3. The number of zeroes is at most equal to the degree of the polynomial.

Solution

A quadratic polynomial with sum of zeroes (α + β) = -3 and product (αβ) = 2 is given by x² - (α + β)x + αβ = x² - (-3)x + 2 = x² + 3x + 2.

Solution

The polynomial x⁴ + 1 can be written as (x²)² + 1. Since x² is always non-negative for real x, x⁴ is also non-negative. Thus, x⁴ + 1 ≥ 1 for all real x, meaning it never equals zero. Therefore, the polynomial has no real zeroes.

Solution

Given that 4 is a zero, we can factor the polynomial as (x - 4)(2x² + 3x - 2). Factoring further: (x - 4)(2x - 1)(x + 2). Thus, the zeroes are x = 4, x = 1/2, and x = -2.

Solution

The zeroes of a polynomial p(x) are the x-coordinates of the points where the graph of y = p(x) intersects the x-axis. For example, the linear polynomial y = 2x + 3 intersects the x-axis at (-3/2, 0), so -3/2 is its zero. The quadratic polynomial y = x² - 3x - 4 intersects the x-axis at (-1, 0) and (4, 0), so -1 and 4 are its zeroes.

Solution

Given α + β = 6 and αβ = 8. Then, α² + β² = (α + β)² - 2αβ = 6² - 2*8 = 36 - 16 = 20.

Solution

The polynomial x³ - 4x can be factored as x(x² - 4) = x(x - 2)(x + 2). Setting each factor to zero gives x = 0, x = 2, and x = -2. Thus, the polynomial has three real zeroes at x = -2, x = 0, and x = 2.

Solution

To find the zeroes, factorize the polynomial to get (2x - 2)(x - 3). The zeroes are x = 1 and x = 3. Sum of zeroes = 1 + 3 = 4 = -(-8)/2. Product of zeroes = 1 * 3 = 3 = 6/2, verifying the relationships.

Solution

First, find α + β = 5 and αβ = 6. Then, α² + β² = (α + β)² - 2αβ = 25 - 12 = 13.

Solution

Factorize the polynomial as (x - 1)³. Thus, x = 1 is a root with multiplicity 3, proving it's a triple root.

Solution

Let the zeroes be α and β. The new zeroes are 1/α and 1/β. The polynomial is x² - (1/α + 1/β)x + (1/αβ) = x² - (α + β)/αβ x + 1/αβ = x² - (3/5)x - 2/5.

Solution

Let the zeroes be α and 1/α. Then, product of zeroes = α * (1/α) = 1 = 4k/(k² + 4). Solving gives k² - 4k + 4 = 0, so k = 2.

Solution

Let y = x². The equation becomes y² + 4y + 5 = 0. The discriminant D = 16 - 20 = -4 < 0, so no real roots for y, hence no real zeroes for x.

Solution

Let the zeroes be mα and nα. Then, sum of zeroes = (m + n)α = -b/a and product = mnα² = c/a. Eliminating α gives the condition b²mn = ac(m + n)².

Solution

Using the relationships, α + β + γ = 3, αβ + βγ + γα = 1, αβγ = -1. Then, 1/α + 1/β + 1/γ = (αβ + βγ + γα)/αβγ = 1/-1 = -1.

Solution

Let the zeroes be α, β, γ. The new zeroes are α², β², γ². Use the relationships to find the sum, sum of products, and product of the new zeroes to construct the polynomial.

Solution

Assume such a polynomial exists. Then, p(x) - x - 1 has roots at x = 1 and x = 2, so p(x) = (x - 1)(x - 2)q(x) + x + 1. Evaluating at x = 1 gives p(1) = 2, but p(1) must be 2, leading to a contradiction if q(x) is non-zero.