Polynomials in one variable

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.

Use the relation between the zeroes and coefficients of a quadratic. If the equal zeroes are r and r, their product is r^2 = 16, so r = 4 or r = -4. Since the question says the zeroes are negative, r = -4. Their sum is therefore -8. For x^2+px+16, the sum of zeroes equals -p, so -p = -8 and p = 8. Hence option A is correct. Option B results from confusing p with the sum of the zeroes. Options C and D do not satisfy both the product condition and the required negative repeated roots. Equivalently, the discriminant p^2-64 must be zero, giving p = ±8, and negativity selects p = 8.

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.

A number r is a zero of a polynomial p(x) exactly when p(r) = 0. Test each proposed value directly. p(1) = 1 − 2 − 5 + 6 = 0, so 1 is a zero. p(2) = 8 − 8 − 10 + 6 = −4, which is not zero; therefore 2 is not a zero. Also, p(3) = 27 − 18 − 15 + 6 = 0 and p(−2) = −8 − 8 + 10 + 6 = 0, so 3 and −2 are zeroes. The factorisation p(x) = (x − 1)(x − 3)(x + 2) confirms these results. Hence option B is uniquely correct. A nonzero substituted value, such as −4, is sufficient to disqualify a candidate as a zero.

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}).

By Vieta’s formulas, \(\alpha+\beta=8\), since the coefficient of \(x\) is \(-8\). We are also given that \(\alpha-\beta=2\). Adding the two equations gives \(2\alpha=10\), so \(\alpha=5\) and \(\beta=3\). For \(x^2-8x+k\), the product of the zeroes is \(\alpha\beta=k\). Hence, \(k=5\times3=15\). Exam tip: In a monic quadratic \(x^2+bx+c\), the product of the zeroes is \(c\), so 16 is not correct.

For a quadratic polynomial \(ax^2+bx+c\), the sum of its zeroes is \(-\frac{b}{a}\). Here, \(a=k\) and \(b=6\), so \(-\frac{6}{k}=-3\). Thus, \(6=3k\), giving \(k=2\). Option B results from a sign error; if \(k=-2\), the sum would be \(3\). Exam tip: use the coefficient formula \(-\frac{b}{a}\) directly for the sum of zeroes.

By Vieta’s formula, the sum of the zeroes is \(2r+3r=5r\). Thus, \(5r=10\), giving \(r=2\). Their product is \((2r)(3r)=6r^2=6\times 2^2=24\). For the monic quadratic \(x^2-10x+q\), the product of the zeroes equals the constant term \(q\), so \(q=24\). Exam tip: for \(x^2+bx+c\), the product of the zeroes is always \(c\).

Since the zeroes are −1, −2, and −4, the polynomial can be written as (x + 1)(x + 2)(x + 4). Expanding gives x³ + 7x² + 14x + 8. Thus, a = 7 and b = 14, so a + b = 21. Options 7 and 14 represent a and b separately, not their sum. Exam tip: For a cubic polynomial, compare the coefficients of x² and x after forming the product of the corresponding linear factors.

(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)). Here (\alpha\beta=-10) and (\alpha+\beta=3), so the value is (-30).

For a quadratic polynomial ax² + bx + c, the sum of its zeroes is given by alpha + beta = −b/a. Here a = 2 and b = 5, so alpha + beta = −5/2. Direct factorisation confirms this: 2x² + 5x − 3 = (2x − 1)(x + 3), whose zeroes are 1/2 and −3. Their sum is 1/2 − 3 = −5/2. Therefore option A is correct. The constant term −3 is related to the product of the zeroes, not their sum, so option C confuses the two coefficient relationships. Option B has the wrong sign, and option D does not represent either required relation. A leading nonzero constant in the factor form does not change the zeroes.

If the zeroes are 0, 2, and −3, the polynomial is formed as \(x(x-2)(x+3)\). Expanding it gives \(x(x^2+x-6)=x^3+x^2-6x\), so option A is correct. Option B factors as \(x(x-3)(x+2)\), giving different zeroes. Exam tip: form one factor \((x-\text{zero})\) for each zero and then expand.

The governing concept is that a zero of a polynomial is a value that makes its output equal to zero. Factor the polynomial by treating it as a quadratic in x^2: x^4-5x^2+4=(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2). Thus its real zeroes are 1, -1, 2, and -2. Among the listed choices, 1, -1, and 2 are genuine zeroes because one factor becomes zero for each of them. For the remaining choice, p(3)=3^4-5(3^2)+4=81-45+4=40, which is not zero. Therefore 3 is not a zero, so option D is correct. The factorisation also provides a complete check rather than relying on only one substitution.