Polynomials in one variable

The sum of the coefficients is \(7+(-4)+(-2)+(-1)=0\). For any polynomial, the sum of its coefficients is obtained by calculating \(p(1)\); here, \(p(1)=7-4-2-1=0\). Therefore, option A is correct. Exam tip: substitute \(x=1\) to find the sum of coefficients, whereas \(x=0\) gives the constant term.

For a polynomial, the sum of all coefficients equals p(1), because substituting x=1 changes every power of x to 1. Here p(1)=3(1)^3+k(1)^2-7(1)+2=3+k-7+2=k-2. The given condition says k-2=5, so k=7. Therefore option C is correct. Directly adding the coefficients gives the same equation: 3+k-7+2=5. Option A incorrectly treats the given total as k, while options B and D result from arithmetic errors when combining the constant terms and the -7 coefficient. The p(1) rule is especially useful because it avoids unnecessary expansion or root calculations.

The odd-power terms are 5x^5, 2x^3, and -x. Their coefficients are 5, 2, and -1, so their sum is 5 + 2 - 1 = 6. Option 7 results from mishandling the negative coefficient of -x. In an exam, remember that the term x has power 1, which is also odd.

To obtain the constant term, multiply the constant terms of the two factors. They are \\-2\\) and \\-1\\), so the constant term is \\((-2)(-1)=2\\). Choosing \\-3\\) results from an incorrect sign calculation. Exam tip: For a product of polynomials, multiply the constant terms of the factors directly.

The governing identity is the difference of squares: A^2 − B^2 = (A − B)(A + B). Let A = x + 6 and B = x − 2. Then A − B = (x + 6) − (x − 2) = 8, and A + B = (x + 6) + (x − 2) = 2x + 4. Therefore the expression becomes 8(2x + 4) = 16x + 32, so option B is correct. Direct expansion confirms this: (x + 6)^2 = x^2 + 12x + 36 and (x − 2)^2 = x^2 − 4x + 4. Subtracting cancels x^2 and leaves 16x + 32. Option A misses a factor of 2 in the x-term, while C and D have incorrect coefficients.

Substituting \(x=1\) gives \(1+a+b=6\), or \(a+b=5\). Substituting \(x=-1\) gives \(-1-a+b=-4\), or \(-a+b=-3\). Subtracting the second equation from the first yields \(2a=8\), so \(a=4\). In an exam, take care that \((-1)^3=-1\) and \(a(-1)=-a\).

Since \(p(0)=s\), we get \(s=-3\). Substituting \(x=1\) and \(s=-3\) gives \(p(1)=2+r-3=4\), so \(r=5\). Hence, option C is correct. Exam tip: use the value at \(x=0\) first to determine the constant term, then apply the second condition.

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

First, \(p(4)=4^2+3(4)-10=16+12-10=18\) and \(p(-2)=(-2)^2+3(-2)-10=4-6-10=-12\). Therefore, \(p(4)-p(-2)=18-(-12)=30\). In such questions, be careful to substitute the negative value with parentheses and remember that subtracting a negative becomes addition.

Substituting -3 for x gives p(-3)=(-3)^3+27=-27+27=0. Therefore, -3 is a zero of the polynomial. Option C is incorrect because it ignores the negative sign; for an odd power, (-3)^3=-27, not 27. Exam tip: To verify a zero, substitute the given value directly into the polynomial and check whether the result is 0.

To find \(p(1)\), substitute 1 for every occurrence of \(x\): \(p(1)=1^3-1^2+1-1=1-1+1-1=0\). Therefore, the correct answer is 0. Exam tip: when evaluating \(p(a)\), replace every \(x\) consistently by \(a\), including all powers.

Since \(p(0)=c\), we get \(c=7\). Substituting \(x=1\) gives \(p(1)=3+b+c=2\), so \(3+b+7=2\) and hence \(b=-8\). Option B, \(-7\), results from omitting the contribution of the term \(3x^2\). Exam tip: substituting \(x=0\) in a polynomial directly gives its constant term.

The polynomial factors as
(p(x)=(x-3)(x-4))
. Hence,
(p(3)=(3-3)(3-4)=0)
and
(p(4)=(4-3)(4-4)=0)
. Therefore,
(p(3)+p(4)=0+0=0)
. Exam tip: If
(x-a)
is a factor of a polynomial, then
(p(a)=0)
.

From \(p(a)=0\), we get \(a^2-25=0\), so \(a^2=25\) and \(a=\pm5\). Therefore, the positive value of \(a\) is \(5\). Remember that the polynomial has two zeroes, \(5\) and \(-5\); the question specifically asks for the positive value.

Factoring the polynomial gives \(9x^2+12x+4=(3x+2)^2\). For a zero, \(p(x)=0\), so \((3x+2)^2=0\), which implies \(3x+2=0\) and hence \(x=-\frac{2}{3}\). The closest distractor, \(\frac{2}{3}\), gives \(p(x)=16\), not zero. Exam tip: check whether a trinomial can be written as a perfect square such as \((a+b)^2\).

From (p(1)=0), we directly get (a+b=-13). The second condition checks consistency, but the first is enough for the asked sum.

(p(1)=0) gives (a+b=5) and (p(2)=0) gives (2a+b=-1), so (a=-6). None of the listed choices is correct.

Use the difference-of-squares identity (a-b)(a+b)=a^2-b^2. Here, a=x^2 and b=4, so (x^2-4)(x^2+4)=(x^2)^2-4^2=x^4-16. Option B has the wrong sign, while C and D are the expansions of (x^2-4)^2 and (x^2+4)^2, respectively. Exam tip: in the product of a sum and a difference with the same terms, the middle terms cancel.

The polynomial has no x^3-term, so it can be written as 0x^3. Therefore, the coefficient of x^3 is 0. The values -2, 3, and 7 are the coefficients of x^4, x^5, and x, respectively. Exam tip: the coefficient of a missing power in a polynomial is taken as 0.

For degree (3), the coefficient of (x^4) must be (0) and the coefficient of (x^3) must be non-zero. For (m=0), (m+2=2).