Polynomials in one variable

Using the given zeroes, the polynomial can be written as \((x-1)(x-2)(x-3)\). On expansion, this becomes \(x^3-6x^2+11x-6\). Comparing coefficients gives \(p=-6\) and \(q=11\), so \(p+q=-6+11=5\). Exam tip: For a monic cubic polynomial, compare the coefficients of \(x^2\) and \(x\) directly after forming the product of the corresponding linear factors.

For zeroes \(\alpha=\frac{2}{3}\) and \(\beta=-\frac{5}{4}\), their sum is \(-\frac{7}{12}\) and their product is \(-\frac{5}{6}\). For a quadratic polynomial \(ax^2+bx+c\), the sum of zeroes is \(-\frac{b}{a}\) and their product is \(\frac{c}{a}\). Option A gives exactly these two values, so it is correct. In option B, the sign of the middle term is wrong, which changes the sum of the zeroes. Exam tip: calculate the sum and product of the given zeroes first, then compare them with the options.

For a quadratic polynomial \(x^2+ax+b\), the sum of the zeroes is \(-a\) and their product is \(b\). Here, \(4+(-7)=-3\), so \(-a=-3\) and hence \(a=3\). Also, \(b=4\times(-7)=-28\). Therefore, \(a-b=3-(-28)=31\), so option B is correct. Exam tip: In \(x^2+px+q\), the sum of the zeroes is \(-p\), while their product is \(q\).

The degree of a polynomial is determined by the highest power whose coefficient is nonzero. A term with coefficient zero is actually zero and does not contribute to the polynomial or its degree. In option A, the apparent fifth-degree term is \\(0x^5\\), so it disappears. The remaining highest nonzero power is \\(x^4\\), giving degree 4.

Option A can be viewed as \\(7x^4-3x+1\\), because \\(0x^5=0\\). Its highest nonzero exponent is 4. Option B has degree 3 after removing \\(0x^6\\); option C has degree 2; and option D, the nonzero constant 9, has degree 0. Therefore only option A has degree 4.

\(p(3)=3^2-5(3)+6=9-15+6=0\) and \(p(2)=2^2-5(2)+6=4-10+6=0\). Therefore, \(p(3)-p(2)=0-0=0\), so option A is correct. Option C, \(-1\), can result from a common subtraction or substitution error. Exam tip: when evaluating a polynomial, substitute the value for the variable carefully in every term before simplifying.

For a quadratic polynomial \(ax^2+bx+c\), the sum of its zeroes is \(-\frac{b}{a}\). Here, \(a=2\) and \(b=k\), so \(-\frac{k}{2}=3\), which gives \(k=-6\). Exam tip: use \(-b/a\) for the sum of zeroes; \(c/a\) gives their product, not their sum.

For a quadratic polynomial \(ax^2+bx+c\), the product of its zeroes is \(\frac{c}{a}\). Here, \(a=3\) and \(c=m\), so the product is \(\frac{m}{3}\). Therefore, \(\frac{m}{3}=\frac{4}{3}\), which gives \(m=4\). Exam tip: remember that the sum of zeroes is \(-\frac{b}{a}\), while their product is \(\frac{c}{a}\).

Factoring the polynomial gives \(x^3-6x^2+11x-6=(x-1)(x-2)(x-3)\). Setting each factor equal to zero gives \(x=1,2,3\), so option A is correct. The values in option C do not satisfy the required sum and product relationships for this cubic polynomial. Exam tip: For likely integer zeroes, substitute each candidate directly into the polynomial to verify it.

By the factor theorem, if \(p(a)=0\), then \(x-a\) is a factor of \(p(x)\). Here, \(p(2)=2^3-3(2)^2-4(2)+12=8-12-8+12=0\), so \(x-2\) is a factor. In fact, \(p(x)=(x-2)(x-3)(x+2)\); the other given options are not factors. Exam tip: to test a possible linear factor \(x-a\), calculate \(p(a)\) directly.

By the factor theorem, if \((x+1)\) is a factor, the polynomial must be zero at \(x=-1\). Thus, \(2(-1)^3+k(-1)^2-5(-1)+2=0\), giving \(-2+k+5+2=0\), so \(k+5=0\) and \(k=-5\). The value 5 is a close distractor caused by a sign error; remember that \(x+1=x-(-1)\), so substitute \(-1\), not 1.

By the Factor Theorem, if (x+1) is a factor of P(x), then P(-1)=0. Thus, P(-1)=2(-1)^3+k(-1)^2-5(-1)+2=-2+k+5+2=k+5. Hence, k+5=0, giving k=-5. Exam tip: for a factor of the form (x-a), substitute x=a; therefore, (x+1)=(x-(-1)) requires substituting x=-1.

By the remainder theorem, the remainder when p(x) is divided by x - a is p(a). Here, a = 1, so p(1) = 1^3 - 5(1)^2 + 8(1) - 4 = 1 - 5 + 8 - 4 = 0. Therefore, the remainder is 0. Exam tip: For a divisor of the form x - a, substitute a directly into the polynomial instead of performing long division.

By the Remainder Theorem, the remainder when \(p(x)\) is divided by \(x-a\) is \(p(a)\). Since \(x+2=x-(-2)\), substitute \(x=-2\): \(p(-2)=3(-2)^3-2(-2)^2+(-2)+7=-24-8-2+7=-27\). Therefore, option A is correct. Exam tip: Rewrite the divisor as \(x-a\) before substituting, so the sign of \(a\) is not missed.

By the Remainder Theorem, when a polynomial \(p(x)\) is divided by \(x-a\), the remainder is \(p(a)\). Here, \(a=4\) and the given remainder is \(7\), so \(p(4)=7\). The value \(0\) would apply if \(x-4\) were a factor of \(p(x)\). Exam tip: For a divisor of the form \(x-a\), substitute \(x=a\) in the polynomial.

The direct answer is option C, \\(x^2+1\\). A linear factor is a polynomial of degree 1, such as \\(x-a\\). Factor the expression as \\(x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)\\). Over the complex numbers, \\(x^2+1=(x-i)(x+i)\\), so \\(x-1\\), \\(x+1\\), and \\(x-i\\) are linear factors. Option C itself has degree 2, so it is not linear, even though it is a factor. Option A is linear because its degree is 1 and it is a factor. Option B is also linear and a factor. Option D is linear over the complex numbers and is a factor because i is a root of \\(x^2+1\\), so it divides \\(x^4-1\\). The important distinction is between being a factor and being a linear factor. Do not call every factor linear merely because it cannot be factored over the reals.

Substituting \(x=1\) gives \(p(1)=1^2-2(1)+5=1-2+5=4\). Since \(p(1)\neq 0\), \(1\) is not a zero of the polynomial. Option B incorrectly treats the value as zero, while option D ignores the contribution of the term \(-2x\). Exam tip: a number is a zero of a polynomial only when substitution makes the polynomial’s value equal to zero.

The monic quadratic with zeroes \(\alpha\) and \(\beta\) is \(x^2-(\alpha+\beta)x+\alpha\beta\). Substituting the given values gives \(x^2-5x+6\), so option A is correct. Option B has a positive coefficient of \(x\), but the coefficient must be the negative of the sum of the zeroes. Exam tip: for a monic quadratic, the coefficient of \(x\) is the negative of the sum of the zeroes, and the constant term is their product.

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}). Here (\frac{3}{2}\div-\frac{5}{2}=-\frac{3}{5}).

By the Factor Theorem, if \(p(a)=0\), then \(x-a\) is a factor. Here, \(2(3)^3-3(3)^2-11(3)+6=0\), so \(x-3\) is correct, not \(x+3\). In exams, remember to reverse the sign in the factor.

For a quadratic polynomial \(ax^2+bx+c\), the sum of its zeroes is \(\alpha+\beta=-\frac{b}{a}\). Here, \(a=3\) and \(b=2\), so \(\alpha+\beta=-\frac{2}{3}\). Therefore, \((\alpha+eta)^2=\left(-\frac{2}{3}\right)^2=\frac{4}{9}\). Option B is incorrect because squaring removes the negative sign. Exam tip: while using \(-\frac{b}{a}\), carefully retain the negative sign before squaring.

Substituting the given values gives \(p(1)=1^2-4(1)+k=k-3\) and \(p(3)=3^2-4(3)+k=k-3\). Thus the two values are equal for every real \(k\), so there is no further restriction on \(k\). Therefore, \(k\) can be any real value. Exam tip: Evaluate the polynomial at both specified inputs before trying to solve for the parameter.

By the Factor Theorem, if P(a) = 0, then (x − a) is a factor of the polynomial. Here, P(1) = 2 − 9 + 13 − 6 = 0, so (x − 1) is a factor. For such questions, quickly test the zero associated with each option.

The governing concept is transformation of the zeroes of a polynomial. First factor the original polynomial: x² + 5x + 6 = (x + 2)(x + 3), so alpha = −2 and beta = −3. Adding 1 to each zero gives new zeroes −1 and −2. The monic quadratic having these zeroes is (x − (−1))(x − (−2)) = (x + 1)(x + 2) = x² + 3x + 2. Hence option A is correct. This can also be verified from the sum and product: the new sum is −1 + (−2) = −3, giving coefficient +3, and the product is 2. Option B is the original polynomial; C shifts in the wrong direction, and D has the wrong sign for the linear term.

(\alpha+\beta=4) and (\alpha\beta=1). (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=64-12=52).

Let the zeroes be (t) and (3t), so (3t^2=9) gives (t=\sqrt{3}). The sum is (4\sqrt{3}), hence (k=-4\sqrt{3}).