Polynomials in one variable

Here (\alpha+\beta=\frac{10}{3}) and (\alpha\beta=\frac{7}{3}). Hence (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\frac{100}{9}-\frac{14}{3}=\frac{58}{9}).

The governing concept is the zero test: a number r is a zero of p(x) exactly when p(r) = 0. Test the options by substitution. For r = -2, p(-2) = (-2)^3 - 2(-2)^2 - 13(-2) - 10 = -8 - 8 + 26 - 10 = 0. Therefore -2 is a zero and option D is correct. For comparison, p(5) = 125 - 50 - 65 - 10 = 0 as well, so 5 is also a zero. This means the question has two correct options, A and D, contrary to the supplied key. Indeed, the polynomial factors as (x - 5)(x + 2)(x + 1), confirming both 5 and -2 are zeros.

The product of zeroes is (2\cdot(-3)=-6), and it equals (\frac{c}{a}). Pay special attention to the sign in products.

Let the two zeroes be \(t\) and \(t+2\). Their sum is \(6\), so \(t+(t+2)=6\), giving \(t=2\) and the other zero as \(4\). The product of the zeroes equals \(s\); hence, \(s=2\times4=8\). Exam tip: For \(x^2+bx+c\), the sum of the zeroes is \(-b\) and their product is \(c\).

(\alpha+\beta=\frac{5}{2}) and (\alpha\beta=-\frac{3}{2}). (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=-\frac{37}{6}).

The governing idea is the factor theorem and polynomial division. Since (x − 1) is a factor, divide p(x) by (x − 1), or use synthetic division with 1. The coefficients 1, −4, −7, 10 give a quotient with coefficients 1, −3, −10 and remainder 0. Therefore p(x) = (x − 1)(x² − 3x − 10). Multiplication confirms this: x³ − 3x² − 10x − x² + 3x + 10 = x³ − 4x² − 7x + 10. Thus option A is correct. Option B has the wrong sign for the middle term, option C does not reproduce the cubic, and option D is only part of the original coefficient pattern, not a quadratic quotient.

For the polynomial to have degree 3, the coefficient of \(x^5\) must be zero so that the fifth-degree term disappears. Thus, \(k-3=0\), giving \(k=3\). The polynomial then becomes \(2x^3-x+9\), whose degree is indeed 3. Exam tip: set the coefficient of every term with a degree higher than the required degree to zero; for \(k=2\), the degree would remain 5.

(\alpha+\beta=9) and (\alpha\beta=20). ((\alpha+2)(\beta+2)=\alpha\beta+2(\alpha+\beta)+4=20+18+4=42).

The sum of reciprocals is (\frac{\alpha+\beta}{\alpha\beta}). Here (\alpha+\beta=4) and (\alpha\beta=1), so the answer is (4).

Let the zeroes be \(\alpha=2+\sqrt{3}\) and \(\beta=2-\sqrt{3}\). Their sum is \(\alpha+\beta=4\), and their product is \(\alpha\beta=2^2-(\sqrt{3})^2=1\). A monic quadratic with zeroes \(\alpha\) and \(\beta\) is \(x^2-(\alpha+\beta)x+\alpha\beta\), so it is \(x^2-4x+1\). Option B has the wrong sign for the sum. Exam tip: for a monic quadratic, the coefficient of \(x\) is the negative of the sum of the zeroes, while the constant term is their product.

We can factor the polynomial as \(x^3-6x^2+12x-8=(x-2)^3\), using the identity \((a-b)^3=a^3-3a^2b+3ab^2-b^3\). Therefore, \(p(x)=0\) only when \(x=2\), and the factor \((x-2)\) occurs three times; hence the zero is 2 with multiplicity 3. Option B reverses the zero and its multiplicity. In an exam, first try to express a cubic as a perfect cube to identify repeated zeros quickly.

Here, \(a=3\), \(b=-12\), and \(c=15\). The discriminant is \(D=b^2-4ac=(-12)^2-4(3)(15)=-36<0\), so the quadratic polynomial has no real zeroes. Equivalently, \(p(x)=3[(x-2)^2+1]\), which is positive for every real \(x\). Thus option B is incorrect because equal real zeroes require \(D=0\). Exam tip: for a quadratic polynomial, \(D<0\) means that it has no real zeroes.

For a quadratic polynomial ax² + bx + c with zeroes α and β, Vieta’s relation gives αβ = c/a. In this polynomial, a = 1 and c = 16, so αβ = 16, independently of the value of b. If two nonzero numbers are reciprocals of each other, they have the form t and 1/t, and their product is exactly 1. Since the required product would have to be both 16 and 1, the condition is impossible for every value of b. The coefficient b controls the sum α + β = −b, but it cannot alter the fixed product. Thus option A is correct. Options B, C, and D merely assign particular values to b and do not remove the contradiction in the product of the roots.

The zeroes of (p(x)) are (5) and (6). For (p(x-2)=0), (x-2=5) or (x-2=6), so the zeroes are (7,8).

Factoring the polynomial gives \(p(x)=(x-7)(x+4)\). Therefore, its zeroes are \(7\) and \(-4\). Their distance is \(|7-(-4)|=11\), so option A is correct. Exam tip: For two real zeroes, calculate the absolute value of their difference to obtain the distance.

For a quadratic polynomial \(ax^2+bx+c\), the product of its zeroes is \(\frac{c}{a}\). Here, the product is \(\frac{18}{2}=9\). If the other zero is \(\alpha\), then \(3\alpha=9\), giving \(\alpha=3\). Thus, the correct answer is 3. Exam tip: Find the product of the zeroes by dividing the constant term by the coefficient of \(x^2\).

(4x^2-4x+1=(2x-1)^2), so the zero (\frac{1}{2}) occurs twice. Perfect square form gives equal zeroes directly.

The polynomial factors as \(p(x)=(x^2-1)(x^2-4)\). Thus, \(x^2=1\) gives \(x=\pm1\), and \(x^2=4\) gives \(x=\pm2\). Therefore, the real zeroes are \((-2,-1,1,2)\). Option B is incorrect because \(x^2=4\) has roots \(\pm2\), not \(\pm4\). Exam tip: For a polynomial involving only even powers of \(x\), factor it by treating \(x^2\) as a single variable.

The sum (1+3-4=0), so (a=0). The sum of pairwise products is (3-4-12=-13), so (b=-13) and (a+b=-13).

Factoring gives \(p(x)=x(x^2-9)=x(x-3)(x+3)\). Thus, the zeroes are \(x=0,3,-3\), which are three distinct real numbers. Therefore, the correct answer is 3. Exam tip: when counting distinct zeroes, count a repeated zero only once.