Polynomials in one variable

(p(1)=a+b+c) and (p(-1)=a-b+c); subtracting gives (2b=0). In exams, use addition or subtraction for symmetric inputs.

By factor theorem (p(2)=0), so (8+4k-8-4=0) and (k=1). In exams, substitute the given zero directly.

By remainder theorem (p(-1)=0), so (-2-5-m+6=0), giving (-1-m=0) and (m=-1). Always check signs carefully.

The highest power is (x^4), so the degree is (4). Terms with zero coefficients do not affect the degree.

(p(2)=12-14+5=3) and (p(1)=3-7+5=1), so the difference is (2). Match every option after calculation to avoid traps.

For it to be quadratic, the coefficient of (x^3) must be (0), while the coefficient of (x^2) is non-zero. Focus on the leading non-zero term.

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.

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 \(q\).

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}).

((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=25-24=1). This identity avoids finding the zeroes separately.

The new sum is (\alpha+\beta+2=8) and product is (\alpha\beta+\alpha+\beta+1=15). Thus the polynomial is (x^2-8x+15).