Polynomials in one variable

The governing idea is substitution and the behavior of even and odd powers under x to -x. Replacing x by -x gives p(-x)=4(-x)^2-3(-x)+2=4x^2+3x+2. The quadratic term remains positive because (-x)^2=x^2, whereas the linear term changes sign. Adding the two expressions gives p(x)+p(-x)=(4x^2-3x+2)+(4x^2+3x+2)=8x^2+4, since -3x and +3x cancel. Therefore option A is correct. Option B fails to cancel the opposite linear terms, option C omits one 4x^2 contribution, and option D has the wrong degree and coefficients. The result also illustrates that adding f(x) and f(-x) removes the odd part of a polynomial.

p(2)=2(2)³−8(2)=16−16=0 and p(−2)=2(−2)³−8(−2)=−16+16=0. Therefore, p(−2)+p(2)=0+0=0, so option B is correct. Choices −16 and 16 result from mishandling the signs of the cubic and linear terms. In such questions, substitute negative values with proper brackets.

For \((x-1)^2(x+3)\), the zeroes are 1 and -3. The factor \((x-1)^2\) shows that 1 has multiplicity 2, while -3 occurs once. Option B has three distinct zeroes. Exam tip: use the exponent of each factor to identify multiplicity.

Completing the square gives \(p(x)=x^2+10x+29=(x+5)^2+4\). Since \((x+5)^2\geq 0\), we have \(p(x)\geq 4\), and equality occurs at \(x=-5\). Therefore, the minimum value is 4. The value 29 is only the constant term, not the minimum. Exam tip: Rewrite a quadratic in the form \((x-a)^2+k\) to identify its minimum quickly.

A perfect-square polynomial with leading coefficient 1 has the form \((x+a)^2=x^2+2ax+a^2\). Since the constant term is \(16=4^2\), we get \(p(x)=(x-4)^2=x^2-8x+16\), so \(k=-8\). The condition \(k<0\) rules out \(k=8\). Exam tip: For \(x^2+bx+c^2\) to be a perfect square, the middle coefficient must be \(\pm 2c\).

The polynomial matches the identity \(a^3+3a^2b+3ab^2+b^3=(a+b)^3\), so \(p(x)=(x+1)^3\). Hence, \(p(-2)=(-2+1)^3=(-1)^3=-1\). The value 0 would result from incorrectly substituting \(x=-1\). Exam tip: look for the perfect-cube identity before expanding or calculating term by term.

\(p(1)=1-2-3+4=0\) and \(p(-1)=(-1)^3-2(-1)^2-3(-1)+4=-1-2+3+4=4\). Therefore, \(p(1)\times p(-1)=0\times4=0\), so option B is correct. Option C is only the value of \(p(-1)\), not the required product. Exam tip: When substituting a negative number, carefully apply the powers and signs to every term.

The governing concept is evaluation of a polynomial at specified values, with special care for signs when the input is negative. Substituting x=2 gives p(2)=2^3+2(2^2)-4(2)-8=8+8-8-8=0. Substituting x=-2 gives p(-2)=(-2)^3+2(-2)^2-4(-2)-8=-8+8+8-8=0. Consequently p(2)-p(-2)=0-0=0, so option A is correct. A structural check is also possible: p(x)=x^3+2x^2-4x-8=(x+2)^2(x-2), so both 2 and -2 are zeroes. Option C usually comes from mishandling the negative input or from subtracting an incorrect value, while B and D have no support after correct evaluation. Both methods confirm the same answer.

Since \(f(0)=b\), we get \(b=-8\). Substituting \(x=4\) in the second condition gives \(16+4a-8=0\), so \(4a+8=0\) and hence \(a=-2\). Exam tip: use the condition at \(x=0\) first to find the constant term, then use the other condition to determine the unknown coefficient.

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.

The governing concept is the definition of a zero of a polynomial: a number r is a zero of p(x) exactly when p(r)=0. Substitute x=0 into every option. For A, p(0)=0^2+5=5, so it is not zero. For B, p(0)=3(0)^3-2(0)=0, so it satisfies the required condition. For C, p(0)=0^4+1=1, and for D, p(0)=2(0)^2-7(0)+9=9. Thus only option B is correct. The factor theorem gives the same result because 3x^3-2x=x(3x^2-2), so x is a factor and x=0 is a zero. The other polynomials have nonzero constant terms, which prevents their value at zero from being zero.

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.

(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.