Polynomials in one variable

Substituting \(x=0\) gives \(p(0)=0^4+a\cdot0^2+b=b\). Hence the condition \(p(0)=0\) necessarily implies \(b=0\). It does not require \(a=0\); for example, \(a=1,b=0\) is possible. Also, \(p(1)=1+a\) is not always zero, and the polynomial generally has degree 4, not 2. Exam tip: evaluating a polynomial at zero isolates its constant term.

To find \(p(x-1)\), substitute the entire expression \((x-1)\) for \(x\): \(p(x-1)=(x-1)^2+4(x-1)+6=x^2-2x+1+4x-4+6=x^2+2x+3\). Therefore, option A is correct. Option B has an incorrect constant-term calculation, while option D results from mishandling the expansion of \(4(x-1)\). Exam tip: whenever the replacement contains a minus sign, keep the complete expression inside parentheses.

To find \(p(2x)\), substitute \(2x\) for every occurrence of \(x\): \(p(2x)=3(2x)^2-2(2x)+5=12x^2-4x+5\). Hence, option A is correct. In option B, the term \(3(2x)^2\) has incorrectly been treated as \(6x^2\). Exam tip: when squaring \(2x\), square the entire expression, so \((2x)^2=4x^2\).

First evaluate \(p(1)\): \(p(1)=1^2-6(1)+5=1-6+5=0\). Therefore, \(p(x)-p(1)=p(x)-0=p(x)=x^2-6x+5\), so option A is correct. Option B incorrectly subtracts 1 from the constant term. Exam tip: when evaluating a polynomial, substitute the given value directly into the original polynomial.

The conditions give (2a+b=2) and (3a+b=-5), so (a=-7), (b=16), and (a+b=9). When two values are given, form two equations.

The governing concept is the zero condition p(r)=0. Because x=1 is a zero, substitute 1 into the polynomial: p(1)=1^3+a(1)^2+b(1)+18=1+a+b+18=0. Hence a+b+19=0, so a+b=-19. Therefore option A is correct. Notice that the second zero, x=-2, is not needed for the requested sum; it would be needed if the individual values of a and b were required. A common error is to omit the constant 18, producing -1 less or another nearby distractor, or to mishandle 1^3. The condition x=-2 can serve as a consistency check, but the first zero alone determines the asked quantity. Thus the answer is uniquely A.

The governing concept is that the degree of a nonzero polynomial is the greatest exponent with a nonzero coefficient. To make the degree equal to 4, the coefficient of x^5 must be zero, so m=0. We must then verify that the x^4 coefficient remains nonzero: m-4=0-4=-4, which is indeed nonzero. Therefore the x^4 term remains and the degree is exactly 4. Option A is correct. If m=4, the x^4 coefficient would vanish, but the x^5 term would have coefficient 4, giving degree 5. If m=2, the x^5 coefficient would still be nonzero, also giving degree 5. Thus both the removal of the fifth-degree term and survival of the fourth-degree term must be checked.

Let α and β be the zeroes of p(x). Since p(x)=x^2-6x+8 is monic, Vieta’s relations give α+β=6 and αβ=8. The required zeroes are α^2 and β^2. Their sum is α^2+β^2=(α+β)^2-2αβ=6^2-2(8)=36-16=20. Their product is α^2β^2=(αβ)^2=8^2=64. A monic quadratic with zeroes having sum S and product P is x^2-Sx+P. Substituting S=20 and P=64 gives x^2-20x+64, so option A is correct. Option B is merely the original polynomial. Option C uses an incorrect transformed sum and product, while D has product 64 but the wrong sum. The calculation can also be checked from the actual zeroes 2 and 4, whose squares are 4 and 16; their polynomial is (x-4)(x-16)=x^2-20x+64.