Polynomials in one variable

To find the sum of all coefficients, substitute x = 1 in the polynomial. Thus, p(1) = 8 − 3 + 0 − 4 + 2 = 3, where the coefficient of x² is 0. Therefore, the correct answer is 3. Exam tip: The sum of the coefficients of a polynomial is equal to p(1).

The sum of coefficients is (4+k+0-6+1=k-1), so (k-1=5) and (k=6). The sum of coefficients is found by (p(1)).

The odd-power terms are −5x^5, 4x^3, and x, whose coefficients are −5, 4, and 1. Thus, their sum is −5 + 4 + 1 = 0. The constant term −7 and the coefficients of even-power terms are excluded. Exam tip: The sum of coefficients of odd-power terms can also be found using [p(1) − p(−1)]/2.

The constant term is the term that contains no \(x\). In the product, it is obtained only by multiplying the constant terms: \(3\times5=15\). Therefore, the correct answer is 15. The value 5 is the constant term of only the second factor, not of the complete product. Exam tip: To find the constant term of a product, multiply the constant terms of its factors.

The governing concept is the difference-of-squares identity a^2-b^2=(a-b)(a+b). Let a=2x+5 and b=2x-1. Then a-b=(2x+5)-(2x-1)=6, while a+b=(2x+5)+(2x-1)=4x+4. Therefore the expression equals 6(4x+4)=24x+24, so option B is correct. A direct expansion provides a useful check: (2x+5)^2=4x^2+20x+25 and (2x-1)^2=4x^2-4x+1. Subtracting gives 24x+24 after the x^2 terms cancel. Option A misses a factor of 2 in the x term, C confuses the constant, and D incorrectly uses 6 as the x coefficient.

The conditions give (2a+b=5) and (-a+b=-6), so (a=\frac{11}{3}) and (b=-\frac{7}{3}). Hence (a+b=\frac{4}{3}), so none of the listed choices is correct.

Substituting \(x=1\) gives \(1+a+b=6\), so \(a+b=5\). Substituting \(x=-2\) gives \(-8-2a+b=-9\), so \(-2a+b=-1\). Subtracting the second equation from the first yields \(3a=6\), hence \(a=2\). Therefore, option B is correct. Exam tip: when substituting a negative value in a cubic polynomial, carefully retain the sign of \((-2)^3=-8\).

For a constant polynomial, both (r+4=0) and (r-1=0) are needed, which is impossible together. All variable terms must vanish.

The polynomial factors as \(p(x)=(x-6)(x+2)\). Hence, \(p(6)=(6-6)(6+2)=0\) and \(p(-2)=(-2-6)(-2+2)=0\). Therefore, \(p(6)-p(-2)=0-0=0\). Exam tip: evaluate the polynomial at each value separately before finding the difference.

To find \(p(4)\), substitute 4 for \(x\) in \(p(x)=x^3-64\): \(p(4)=4^3-64=64-64=0\). Hence, 4 is a zero of the polynomial. Option B incorrectly evaluates \(4^3\) as 16. Exam tip: a number is a zero of a polynomial if substitution makes the polynomial’s value equal to zero.

To find \(p(1)\), substitute \(x=1\): \(p(1)=1^4-1^3+1^2-1+1=1-1+1-1+1=1\). Hence, option B is correct. Option A results from incorrectly omitting the final \(+1\) while adding the alternating terms. Exam tip: To evaluate \(p(a)\), substitute \(a\) for every occurrence of \(x\).

Since \(p(0)=c\), we get \(c=-4\). Substituting \(x=1\) and \(p(1)=3\) gives \(5+b-4=3\), so \(b=2\). Option A is the value of \(5+c\), not the value of \(b\). Exam tip: use \(x=0\) first to determine the constant term, then apply the second condition.

In \(3x^2-5x+1\), the highest power of x is 2 and every exponent is a non-negative integer, so it is quadratic. B has degree 3, while C and D are not polynomials because x occurs in a denominator and under a root. Exam tip: check exponents first.

For \(p(a)=0\), we need \(a^2-49=0\), so \(a^2=49\) and \(a=\pm 7\). Among the given options, \(a=7\) is correct; \(a=-7\) is the other zero. Substituting \(0\), \(6\), or \(49\) does not make the polynomial zero. Exam tip: The zeros of \(x^2-c^2\) are \(x=\pm c\).

The polynomial can be written as \(16x^2-24x+9=(4x-3)^2\). For a zero, \(4x-3=0\), giving \(x=\frac{3}{4}\). Substituting option B, \(x=-\frac{3}{4}\), does not make the polynomial zero. In an exam, identifying the perfect-square form is the quickest method.

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.

(p(1)=0) gives (a+b=9) and (p(5)=0) gives (5a+b=-23), so (a=-8). None of the listed choices is correct.

Use the identity (a-b)(a+b)=a^2-b^2. Here, a=x^2 and b=9, so (x^2-9)(x^2+9)=(x^2)^2-9^2=x^4-81. Option B has the wrong sign for 81, while C and D incorrectly introduce an 18x^2 term. Exam tip: whenever (a-b)(a+b) appears, apply the difference-of-squares identity directly.

The polynomial has no term containing \(x^5\), which means its \(x^5\)-term is \(0x^5\). Therefore, the coefficient of \(x^5\) is 0. The other numerical choices are coefficients of \(x^2\), \(x^6\), and \(x^4\), respectively. Exam tip: the coefficient of any missing power in a polynomial is 0.

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.