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.

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.

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.

The direct answer is option B, 0. The degree is the highest power whose coefficient is not zero. The displayed polynomial is \\(mx^4+(m+2)x^3+5x+1\\). For its degree to be 3, the x^4 term must disappear, so its coefficient must satisfy \\(m=0\\). After substituting this value, the polynomial becomes \\(2x^3+5x+1\\). Its x^3 coefficient is 2, which is non-zero, so the degree really is 3. Therefore option B is correct. Option A, -2, makes the x^3 coefficient \\(m+2=0\\), but leaves the x^4 coefficient -2, so the degree is 4. Option C, 2, leaves a non-zero x^4 coefficient, so the degree is 4. Option D is wrong because a valid value exists, namely 0. Always check both conditions: remove the higher-power term and keep the required-degree term non-zero.

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.

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.