Polynomials in one variable

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.

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 \(-b/a\).

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.

A non-zero polynomial of degree n has at most n zeroes. Therefore, a degree-4 polynomial can have no more than four real zeroes; it need not have exactly four or even one. Exam tip: distinguish “at most” from “exactly”.

The zeroes of (p(x)) are (2) and (5), so for (p(x+1)=0), (x+1=2) or (x+1=5). Hence the zeroes of (q(x)) are (1) and (4).

For degree (4), the coefficient of (x^6) must be (0) and the coefficient of (x^4) must be non-zero. Both conditions hold for (a=1).

(p(-1)=-k+2+7+5=14-k), so (14-k=16) and (k=-2). Watch signs carefully when substituting a negative value.

The direct answer is option C: the degree is \(4\). Add the polynomials term by term: \((6x^5-4x^2+1)+(-6x^5+3x^4+x-9)\). The fifth-degree terms cancel because \(6x^5-6x^5=0\). The sum becomes \(3x^4-4x^2+x-8\). The highest non-zero power is 4, so the degree is 4. Option A, 2, is wrong because the \(x^4\) term remains. Option B, 3, is wrong because there is no non-zero \(x^3\) term and it overlooks the fourth-degree term. Option C is correct. Option D, 5, is wrong because the fifth-degree terms cancel completely. The degree of a sum may be lower than the degrees of the original polynomials when leading terms cancel. Exam cue: always check cancellation of the highest powers.

In the product of two non-zero polynomials, degrees add, so (6+3=9). In multiplication, look at the highest-power terms.

((3x-2)(2x^2+x-5)=6x^3-x^2-17x+10), so the coefficient of (x^2) is (-1). Combine like terms after expansion.

Substituting \(x=2\), we get \(p(2)=4(2)^3-11(2)^2+6(2)+2=32-44+12+2=2\). Hence, the correct answer is 2. The value 0 may result from mishandling the negative term or the constant term. In an exam, calculate the powers first and carefully preserve the signs while adding the terms.

In \(3x^2-\sqrt{5}x+7\), the powers of \(x\) are \(2,1,0\), all non-negative integers. \(\sqrt{5}\) is a real coefficient. In B and D, the variable occurs in a denominator, while C has power \(\tfrac12\). Exam tip: check powers of the variable first.