Polynomials in one variable

The degree of a nonzero polynomial is the greatest exponent of x whose coefficient is nonzero. Since a=0, the cubic term ax³ disappears completely. The next term is bx², and b≠0 guarantees that this term remains present. Therefore the highest surviving power is x², so degree(p)=2 and option C is correct. The value of c or d does not affect the degree once a nonzero x² term is already present. Option D incorrectly retains the vanished cubic term. Option B would be possible only if b were also zero and c were nonzero, while option A would describe a nonzero constant polynomial, not the general expression under the stated condition. The condition b≠0 is essential because it prevents further reduction of the degree.

The sum of all coefficients of a polynomial is found by evaluating it at \(x=1\). Here, \(p(1)=5(1)^3-2(1)^2+1-4=5-2+1-4=0\). Therefore, the correct answer is 0. Exam tip: use \(p(1)\) for the sum of coefficients, not \(p(0)\), which gives only the constant term.

For a polynomial, the sum of all coefficients is obtained by substituting x=1, because every power of 1 equals 1. Thus p(1)=2(1)³+k(1)²−8(1)+3=2+k−8+3=k−3. The problem states that this sum is zero, so k−3=0 and therefore k=3. Hence option C is correct. Directly adding the coefficients gives the same equation: 2+k−8+3=0. Option A, B, or D would make the coefficient sum respectively −2, −1, or 1 rather than zero. The key idea is that the variable is not assigned an arbitrary value; x=1 is specifically used because it converts every coefficient term into that coefficient itself.

The powers in \(3x^4\), \(-5x^2\), and \(-7=-7x^0\) are 4, 2, and 0, respectively, and all are even. Only \(2x=2x^1\) has an odd power, so the required sum is 2. Exam tip: the constant term has power 0, which is even, not odd.

To find the constant term, multiply the constant terms of the two factors. The constant terms are -3 in (2x-3) and 1 in (x^2+x+1), so the constant term is (-3)·1 = -3. Option C, 1, is only the constant term of the second factor, not of the product. Exam tip: In a product of polynomials, the constant term is obtained by multiplying the constant terms of all factors.

\(x^2(x^2+1)=x^4+x^2\), so the highest power of the variable is 4 and all powers are non-negative integers. Option C has degree 5, while B and D involve fractional and negative powers. Exam tip: check exponents first.

Using \(f(1)=4\), we get \(1+a+b=4\), so \(a+b=3\). Using \(f(-1)=-2\), we get \(-1-a+b=-2\), so \(-a+b=-1\). Subtracting the second equation from the first gives \(2a=4\), hence \(a=2\). Therefore, option B is correct. Exam tip: when values at \(1\) and \(-1\) are given, substitute them directly and eliminate the common constant term.

Substituting \(x=0\) gives \(p(0)=s=5\). Then, substituting \(x=1\), \(p(1)=1+r+5=9\), so \(r=3\). Therefore, option C is correct. Exam tip: when evaluating a polynomial at zero, all terms containing \(x\) vanish, leaving only the constant term.

For a constant polynomial, both (m+1=0) and (m-2=0) are needed, which is impossible together. All variable terms must vanish.

Substituting the values gives \(p(3)=3^2+2(3)-8=7\) and \(p(-2)=(-2)^2+2(-2)-8=-8\). Therefore, \(p(3)-p(-2)=7-(-8)=15\). Exam tip: when subtracting \(p(-2)\), subtract the negative value correctly, which becomes addition.

To find \(p(2)\), substitute 2 for \(x\): \(p(2)=2^3-8=8-8=0\). Hence, 2 is a zero of the polynomial. Option B incorrectly uses \(3^2\) instead of \(2^3\), while option C changes the exponent from 3 to 2. In an exam, remember that a number is a zero of a polynomial exactly when substitution gives a value of 0.

Substituting -1 for x gives p(-1)=(-1)^3+(-1)^2+(-1)+1=-1+1-1+1=0. Therefore, option B is correct. Exam tip: always use parentheses when substituting a negative number, because ignoring them can cause a sign error and lead to an answer such as -2.

Since \(p(0)=c=5\), substituting \(x=1\) gives \(p(1)=2+b+5=1\). Hence, \(b+7=1\) and \(b=-6\). The distractor \(-5\) results from an arithmetic error while solving this equation. Exam tip: for a polynomial, the constant term is obtained directly by evaluating it at \(x=0\).

The polynomial \(p(x)=x^2-5x+6=(x-2)(x-3)\). Therefore, \(p(2)=0\) and \(p(3)=0\), since 2 and 3 are its zeroes. Hence, \(p(2)+p(3)=0+0=0\). Exam tip: If a number is a zero of a polynomial, the value of the polynomial at that number is directly 0.

For (p(a)=0ig), we require (a^2-9=0ig), so (a^2=9ig) and (a=\pm3ig). Since the question asks for the positive value, the correct answer is 3. Note that (a=-3ig) is also a zero of the polynomial, but it is not positive. Exam tip: Factor the expression as (x^2-9=(x-3)(x+3)ig) to find its zeros quickly.

The polynomial can be factorised as \(p(x)=(2x-3)^2\). Thus, \(p(x)=0\) when \(2x-3=0\), giving \(x=\frac{3}{2}\). Indeed, \(p\left(\frac{3}{2}\right)=9-18+9=0\), whereas the other options do not make the polynomial zero. Exam tip: For a quadratic polynomial, try factorisation before substituting every option.

The conditions give (a+b=-7) and (4a-2b=2), so (a=-2), (b=-5), and (a-b=3). The correct value is (3).

(p(1)=1+a+b+4=0) and (p(-1)=-1+a-b+4=0); adding gives (2a+8=0), so (a=-4). Add signs carefully.

By the Factor Theorem, if x − 2 is a factor of p(x), then p(2) = 0. Therefore, 2³ + k(2²) − 4(2) − 12 = 0, giving 8 + 4k − 8 − 12 = 0. Thus, 4k − 12 = 0 and k = 3. Option B results from a sign error. Exam tip: whenever x − a is a factor, substitute x = a and set the polynomial equal to zero.

The polynomial has no visible x²-term, which means its x²-term is 0x². Therefore, the coefficient of x² is 0. Here, −3 is the coefficient of x³ and 5 is the coefficient of x. Exam tip: The coefficient of a missing term in a polynomial is taken as 0.