Polynomials in one variable

((\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).

The zeroes are (-2,0,3), so the polynomial is (x(x+2)(x-3)=x^3-x^2-6x). Intersections with the (x)-axis give zeroes.

Since \(x=1\) is a zero, \((x-1)\) must be a factor of \(p(x)\). Dividing \(x^3+x^2-10x+8\) by \((x-1)\) gives the quotient \(x^2+2x-8\) with remainder \(0\). Verification: \((x-1)(x^2+2x-8)=x^3+x^2-10x+8\). Therefore, option A is correct. Exam tip: if \(a\) is a zero, divide the polynomial by \((x-a)\) to obtain the remaining factor.

By the Remainder Theorem, the remainder when p(x) is divided by x-a is p(a). Here the divisor is x-1, so a=1. Therefore, p(1)=2(1)^3+3(1)^2-8(1)+3=2+3-8+3=0. Hence, the correct remainder is 0. Exam tip: For a divisor of the form x-a, evaluate p(a) directly instead of carrying out polynomial division.

Substituting \(x=3\), \(p(3)=3^3-2(3)^2-5(3)+6=27-18-15+6=0\). Hence, the correct answer is 0. Since \(p(3)=0\), 3 is also a zero of the polynomial. Exam tip: substitute the given value for every occurrence of \(x\), including its powers.

The zero polynomial has no non-zero term, so its degree is not defined. A non-zero constant polynomial has degree (0).

A constant polynomial has no variable term, so (7) is a constant polynomial. A non-zero constant polynomial has degree (0).

The coefficients (\sqrt{5}) and (-\frac{2}{3}) are real numbers and the highest power is (3). Hence it is a cubic polynomial.

The governing concept is the degree of a nonzero polynomial. A linear polynomial has degree exactly 1, meaning the highest power of the variable with a nonzero coefficient is x^1. In option A, 5x-9 has highest exponent 1 and therefore is linear. Option B, x^2-9, has degree 2 and is quadratic. Option C is a nonzero constant polynomial, whose degree is 0. Option D, x^3+x, has highest exponent 3 and is cubic, even though it also contains a linear term. Thus option A is the only correct answer. The classification depends on the greatest exponent, not simply on whether the expression contains x or has only a few terms. The nonzero coefficient of x in 5x-9 confirms its degree is exactly one.

A zero is the value of \(x\) for which \(p(x)=0\). Setting \(ax+b=0\) gives \(ax=-b\), and hence \(x=-\frac{b}{a}\). Option B misses the negative sign, while options C and D interchange the numerator and denominator. For exams, remember that the zero of \(ax+b\) is \(-\frac{b}{a}\).

A zero of a polynomial is the value of the variable for which the polynomial equals zero. Setting \(p(x)=0\) gives \(4x-12=0\), so \(4x=12\) and \(x=3\). For the closest distractor, \(p(-3)=-24\), so −3 is not a zero. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-b/a\).

The governing concept is the definition of a zero of a polynomial: a number r is a zero when p(r)=0. The polynomial can be factored as p(x)=7x^3-2x=x(7x^2-2). Thus x is a factor of every term, and substituting x=0 gives p(0)=0(7(0)^2-2)=0. Equivalently, the constant term is zero, which is why the graph passes through the origin. Option A states the direct algebraic reason. Option B is incorrect because 7 is the coefficient of x^3, not the constant term. The degree being 3 and the presence of a negative coefficient do not by themselves make zero a root. Therefore A is the only unambiguous answer.

For a zero, \(x^2+1=0\), so \(x^2=-1\) must hold. For every real \(x\), \(x^2\geq 0\), hence \(x^2+1\geq 1>0\), and the expression can never be zero. Therefore, there are no real zeroes. Option C is incorrect because \(p(0)=1\), not 0. As an exam tip, when an equation gives \(x^2=-1\), check the domain: its solutions are complex, not real.

For zeroes 8-19 and 849, the polynomial is formed as 8x+198x-49. Expanding it gives \(x^2-3x-4\), so option A is correct. Option B has zeroes 1 and -4, so it is not correct. Exam tip: for zeroes \(\alpha\) and \(\beta\), use the monic quadratic form \(x^2-(\alpha+\beta)x+\alpha\beta\).

For a quadratic polynomial \(x^2+bx+c\), the sum of the zeroes is \(-b\) and their product is \(c\). Here, \(b=-a\) and \(c=a\), so both the sum and product are \(a\). Hence the condition is automatically satisfied for every non-zero real value of \(a\). The values 1, −1, and 2 are only particular examples, not the complete answer. Exam tip: compare \(-b\) and \(c\) directly when using the relationships between zeroes and coefficients.

For a quadratic polynomial to have equal zeroes, its discriminant must be zero. Thus, using (b^2-4ac=0), we get (k^2-4\cdot2\cdot8=0), so (k^2=64). Therefore, option A is correct. Exam tip: For equal zeroes of a quadratic, set the discriminant to zero.

Let the zeroes be (t) and (2t), then (2t^2=16) gives (t=2\sqrt{2}). The sum is (6\sqrt{2}), so (-k=6\sqrt{2}) and (k=-6\sqrt{2}).

For a quadratic polynomial \(ax^2+bx+c\), the sum of its zeroes is \(-\frac{b}{a}\). Here, the sum is \(-\frac{-10}{1}=10\). Therefore, if one zero is \(4\), the other zero is \(10-4=6\). The value \(4\) is only the given zero, not the other one. Exam tip: use the sum-of-zeroes formula directly when one zero is known.

For a quadratic polynomial \(ax^2+bx+c\), the product of its zeroes is \(\frac{c}{a}\). Here, the product is \(\frac{-18}{1}=-18\). Since one zero is \(3\), the other zero is \(\frac{-18}{3}=-6\). Hence, option A is correct. Exam tip: when one zero is given, use the product of zeroes directly rather than the sum unless needed.