Polynomials in one variable

The governing concept is finding the zeroes of a quadratic polynomial by solving p(x)=0 with the quadratic formula. Set x²−8x+10=0, where a=1, b=−8, and c=10. Then x=[−b±√(b²−4ac)]/(2a)=[8±√(64−40)]/2=[8±√24]/2. Since √24=√(4×6)=2√6, the roots become [8±2√6]/2=4±√6. Thus the zeroes are 4+√6 and 4−√6, so option A is correct. Option B uses an incorrect discriminant and fails to divide correctly. Option C has the wrong half of the constant term, while option D replaces √6 with √10 without justification. Substitution into the polynomial also verifies both roots.

For a quadratic polynomial ax² + bx + c with zeroes α and β, the relation α + β = −b/a holds, even when a coefficient is irrational. In p(x) = 2x² − 4√2x + 4, we have a = 2 and b = −4√2. Therefore α + β = −(−4√2)/2 = 4√2/2 = 2√2. The constant term c = 4 determines the product αβ = c/a, not the sum. Hence option A is correct. Option B results from forgetting to divide by the leading coefficient, option C comes from an incorrect division, and option D confuses the constant term with the sum. The standard coefficient-zero relation remains valid without any change.

If α and β are the zeroes of a monic quadratic, its standard form is x² − (α + β)x + αβ. The leading coefficient is 1 because the polynomial is monic. Here α + β = 2√5 and αβ = 4. Substituting these values gives x² − 2√5x + 4, which is option A. Option B has the wrong sign before the coefficient of x, so it would represent a sum of −2√5. Option C places the given sum and product in the wrong positions, while option D has incorrect signs and a wrong constant term. Irrational coefficients are allowed in a polynomial, so 2√5 causes no difficulty.

The required quantity is the sum of the two given zeroes, so we add the radicals after simplifying them. First, √8 = √(4 × 2) = √4 × √2 = 2√2. Thus the sum is √2 + √8 = √2 + 2√2 = 3√2. Therefore option A is correct. Option C incorrectly treats √8 as 3√2, while option B combines the radicands as though addition of radicals were multiplication. Option D also does not represent the sum. The governing concept is that like surds can be added by adding their coefficients, but the radicands must first be reduced to the same square-free form.

Use the perfect-square identity (x − a)² = x² − 2ax + a². Taking a = √2 gives (x − √2)² = x² − 2√2x + (√2)² = x² − 2√2x + 2, exactly the given polynomial. Therefore option A is correct. Option B expands to x² + 2√2x + 2 and has the wrong middle-term sign. Option C expands to x² − 4x + 4, which has different coefficients. Option D uses the difference-of-squares identity and expands to x² − 2, so it also does not match. The factorization further shows that √2 is a repeated irrational zero.

For a monic quadratic with zeroes α and β, the required polynomial is x² − (α + β)x + αβ. Let α = 1 + √2 and β = 1 − √2. Their sum is α + β = 2 because the irrational terms cancel. Their product is αβ = (1 + √2)(1 − √2) = 1 − 2 = −1, using the difference-of-squares identity. Hence the polynomial is x² − 2x − 1, so option A is correct and its coefficients are rational. Option B has the wrong sign for the linear coefficient. Option C has neither the correct sum-product pair nor the correct constant term. Option D has constant term 3, whereas the product of the zeroes must be −1.

The governing concept is the difference-of-squares identity u²−v²=(u−v)(u+v). Let u=x⁴ and v=9. Then x⁸−81=(x⁴)²−9²=(x⁴−9)(x⁴+9). Substituting this factorisation into the fraction gives [(x⁴−9)(x⁴+9)]/(x⁴−9). The stated condition x⁴≠9 guarantees that the denominator is not zero, so cancellation of the common factor is valid. The simplified expression is therefore x⁴+9, making option B correct. Option A is the factor that cancels, option C uses an incorrect power of x, and option D corresponds neither to the difference-of-squares factorisation nor to the original quotient.

The expression (3x^2-5x+7) has only (x) and non-negative integer powers. In exams, the variable power must not be negative or fractional.

The degree of a non-zero polynomial in one variable is the greatest exponent of the variable whose coefficient is non-zero. In p(x) = 4x^3 − 2x + 9, the term 4x^3 has exponent 3, the term −2x has exponent 1, and the constant 9 has exponent 0. Since the coefficient of x^3 is 4, which is non-zero, the greatest relevant exponent is 3. Therefore the degree is 3, so option A is correct. Option B may be chosen by imagining a missing x^2 term, but absent terms do not determine degree. Option C is only the degree of the linear term, and 9 is a constant value, not the degree of the polynomial.

The constant term is the term that does not contain the variable \(x\). In the given polynomial, \(7x^2\) and \(5x\) contain \(x\), whereas \(-11\) does not; therefore, the constant term is \(-11\). In an exam, identify the term independent of the variable.

The relevant concept is the coefficient of a specified power of a variable. In the polynomial 5x^4 − 3x^2 + x − 6, the term containing x^2 is −3x^2. The numerical factor multiplying x^2, including its negative sign, is therefore −3. Hence option B is correct. The number 5 is the coefficient of x^4, not x^2. The term x can be written as 1x, so 1 is the coefficient of x^1, not x^2. The number −6 is the constant term and has no variable factor. The requested power must be matched exactly, and the sign must also be retained when identifying a coefficient.

A quadratic polynomial is a polynomial whose highest power of the variable is 2. In \(x^2-4x+1\), the highest power of \(x\) is 2, so it is quadratic. Option A is linear, option C is cubic, and option D is a constant polynomial. Exam tip: identify the degree by looking at the term with the highest exponent.

The term (x^{-1}) has a negative power of the variable, so it is not a polynomial. In exams, powers should be like (0,1,2,\ldots).

To find \(p(1)\), substitute \(x=1\): \(p(1)=2(1)^2-3(1)+1=2-3+1=0\). Therefore, the correct answer is 0. Choosing 1 usually results from incorrectly combining the constant and negative terms. Exam tip: substitute the given value at every occurrence of \(x\) and evaluate powers before performing addition or subtraction.

The governing concept is that separate polynomial terms are identified by addition or subtraction signs, while each term includes its sign when necessary. In q(x) = x^3 + 2x^2 − x + 4, the four terms are x^3, 2x^2, −x, and 4. Therefore the polynomial contains 4 terms, making option C correct. The minus sign before x does not create an additional term; it belongs to the coefficient of the term −x. Likewise, the constant 4 is itself one term. Counting only the variable terms would give 3, which is why option B is incomplete. The expression has no hidden fifth term, because there are only three separating signs and four resulting parts.

A linear polynomial has degree 1. In \(3x-8\), the highest power of the variable \(x\) is 1, so it is a linear polynomial. The degrees of \(x^2+1\) and \(x^3+x\) are 2 and 3, respectively, while \(6\) is a constant polynomial of degree 0. Exam tip: identify the highest power of the variable to determine the degree.

The term with the highest power is \\(6x^5\\). The numerical coefficient of this term is called the leading coefficient, so the correct answer is \\(6\\). Remember that \\(5\\) is the degree of the polynomial, not its leading coefficient.

A monomial is a polynomial containing exactly one term. The expression \(3x^2\) has only one term, so it is a monomial. The other options contain two or three terms. In an exam, identify a monomial by counting its terms, separated by plus or minus signs.

A binomial is a polynomial with exactly two unlike terms. In 5x - 7, the terms are 5x and -7, so it is a binomial. Option A has three terms, option C has one term, and option D has four terms. Exam tip: count the terms separated by plus or minus signs, ignoring the sign as part of the term.

A trinomial is a polynomial containing three unlike terms. In \(x^2+x+1\), the three terms are \(x^2\), \(x\), and \(1\), so it is a trinomial. Option B is a binomial, option C is a monomial, and option D is a four-term polynomial. Exam tip: count the terms separated by plus or minus signs.