Polynomials in one variable

To make the degree not more than (2), the coefficient of (x^4) must be (0), so (m-2=0). Degree reduces only when the highest term vanishes.

The polynomial can be recognised as \(p(x)=x^3-3x^2+3x-1=(x-1)^3\). Therefore, at \(x=1\), \(p(1)=(1-1)^3=0\), so option A is correct. Direct substitution also gives \(1-3+3-1=0\). In an exam, using the identity \(a^3-3a^2b+3ab^2-b^3=(a-b)^3\) helps solve this type of question quickly.

The zeroes are the values of \(x\) for which \(p(x)=0\). Thus, \(x^2-9=0\), or \((x-3)(x+3)=0\), giving \(x=3\) or \(x=-3\). Therefore, (3, -3) is correct. Option B is incorrect because 0 and 9 are not the values that make the polynomial zero. Exam tip: Recognise the difference-of-squares identity \(a^2-b^2=(a-b)(a+b)\) to find such zeroes quickly.

\(p(x)=x^2+6x+9=(x+3)^2\). Thus, setting \(p(x)=0\) gives \(x=-3\), and this zero occurs twice; hence the polynomial has two equal real zeroes. Option B is incorrect because \(3\) is not a zero of this polynomial. Exam tip: for a quadratic polynomial, \(D=b^2-4ac=0\) indicates two equal zeroes.

For a quadratic polynomial ax^2+bx+c with zeroes alpha and beta, Vieta's relation gives alpha+beta=-b/a. In p(x)=2x^2-7x+3, the coefficients are a=2 and b=-7. Therefore the sum is -(-7)/2=7/2, making option A correct. A direct factorisation confirms this: 2x^2-7x+3=(2x-1)(x-3), whose zeroes are 1/2 and 3. Their sum is 1/2+3=7/2. Option B results from forgetting the negative sign in Vieta's relation. Options C and D incorrectly use the constant term, which is involved in the product of the zeroes, not their sum. Thus the coefficient relationship is sufficient and the factorisation provides an independent check.

For a quadratic polynomial ax^2+bx+c with zeroes alpha and beta, Vieta's relation states that their product is alpha beta=c/a. Here the leading coefficient is a=5 and the constant term is c=-8. Hence the product is (-8)/5=-8/5, so option A is correct. The negative sign must be retained because the constant term is negative, and the denominator is the leading coefficient rather than the middle coefficient. Option B loses the negative sign. Options C and D incorrectly use the coefficient 2 of x, which belongs to the sum relation -b/a and is not needed for the product. Direct factorisation is also possible over the real numbers, but the coefficient relation gives the result immediately and avoids unnecessary calculation.

With zeroes (2) and (5), the polynomial is ((x-2)(x-5)=x^2-7x+10). A monic polynomial has leading coefficient (1).

If the zeroes are \(\alpha\) and \(\beta\), a monic quadratic polynomial is \(x^2-(\alpha+\beta)x+\alpha\beta\). Here, \(\alpha+\beta=-4\) and \(\alpha\beta=7\), so the polynomial is \(x^2-(-4)x+7=x^2+4x+7\). Option B has the coefficient of \(x\) as \(-4\), which would make the sum of the zeroes 4. Exam tip: use the form \(x^2-(\text{sum of zeroes})x+(\text{product of zeroes})\).

The sum of zeroes is (3+k), while the polynomial gives sum (k+2), so (3+k=k+2) is impossible. This is a conceptual trap.

For the quadratic polynomial (p(x)=ax^2+bx+c), the sum of the zeroes is (-\frac{b}{a}). Here, the sum of the zeroes is (2+(-1)=1). Thus, (-\frac{b}{a}=1), so (\frac{b}{a}=-1). Option 1 is incorrect because it is the sum of the zeroes, not (\frac{b}{a}). Exam tip: remember the negative sign in the relation between the sum of zeroes and (\frac{b}{a}).

A number \(k\) is a zero of a polynomial if \(p(k)=0\). Here, \(p(2)=2^3-4(2)^2+2+6=8-16+2+6=0\), so 2 is a zero. The other options do not work: \(p(1)=4\), \(p(4)=18\), and \(p(5)=36\). Exam tip: For integer options, substitute them directly into the polynomial and check which gives zero.

Factoring the polynomial gives x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3). Setting each factor equal to zero gives x = 1, 2, 3, so option A is correct. Option B has the wrong signs for all the zeroes. Exam tip: whenever (x − a) is a factor, a is the corresponding zero.

By the factor theorem, (x - a) is a factor of p(x) if p(a) = 0. Here, p(-3) = (-3)^3 + 3(-3)^2 - 4(-3) - 12 = -27 + 27 + 12 - 12 = 0. Therefore, (x + 3) is a factor. As an exam tip, for a possible factor x + a, substitute x = -a; the polynomial must evaluate to zero.

(4x^2-12x+9=(2x-3)^2), so the equal zeroes are (\frac{3}{2}). A perfect square form indicates equal zeroes.

The sum of the zeroes is \(4+(-5)=-1\). For the quadratic polynomial \(x^2+px+q\), the sum of the zeroes is \(-p\), so \(-p=-1\) and hence \(p=1\). Their product is \(4\times(-5)=-20\), giving \(q=-20\). Therefore, \(p+q=1-20=-19\). Exam tip: For \(x^2+px+q\), remember that the sum of zeroes is \(-p\) and their product is \(q\).

Use the coefficient relation for a quadratic polynomial. If ax^2+bx+c has zeroes alpha and beta, then alpha+beta=-b/a. In p(x)=x^2-2(k+1)x+k^2, a=1 and b=-2(k+1), so the sum of the zeroes is -[-2(k+1)]/1=2(k+1). The given condition therefore produces 2(k+1)=10. Dividing both sides by 2 gives k+1=5, and subtracting 1 gives k=4. Hence option A is correct. Option B is the intermediate value k+1, not k. Option C results from treating the coefficient as if it were -2k+1, while option D gives a sum of 2(3+1)=8 rather than 10. Substituting k=4 confirms the result: 2(4+1)=10.

The sum is (-\frac{m-1}{3}); setting it to (0) gives (m-1=0). When the sum is zero, the coefficient of (x) becomes zero.

The product is (\frac{4}{k}), so (\frac{4}{k}=2) and (k=2). Product equals constant term divided by leading coefficient.

If the zeroes are (\alpha,\beta), then (\alpha+\beta=8) and (\alpha\beta=15). (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=64-30=34).

(\alpha+\beta=\frac{9}{2}) and (\alpha\beta=2). Hence (\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{9}{4}).