Polynomials in one variable

Correct answer: A. 0

Explanation: Substituting \(x=-3\), we get \(p(-3)=(-3)^4-10(-3)^2+9=81-90+9=0\). Since the powers are even, \((-3)^4=81\) and \((-3)^2=9\). Option B may result from omitting the middle term \(-10(-3)^2\). In an exam, evaluate the powers first, then perform multiplication and addition or subtraction.

Correct answer: A. -13

Explanation: Since p(-2)=0 and p(5)=0, the zeros of the polynomial are -2 and 5. Therefore, p(x)=(x+2)(x-5)=x²-3x-10. Comparing coefficients gives a=-3 and b=-10, so a+b=-3+(-10)=-13. Remember that the sum of the zeros is 3, whereas a is its negative.

Correct answer: A. -4

Explanation: Since the polynomial is monic and its zeros are \(-4\) and \(1\), we can write \(p(x)=(x+4)(x-1)\). Expanding gives \(p(x)=x^2+3x-4\), so the constant term is \(b=-4\). The value \(4\) is incorrect because the constant term equals the product of the zeros, not their sum. Exam tip: For \(x^2+ax+b\), the product of the zeros is directly equal to \(b\).

Correct answer: C. (x^{\frac{3}{2}}+x+1)

Explanation: In (x^{\frac{3}{2}}), the power of the variable is fractional, so it is not a polynomial. In a polynomial, powers are non-negative integers.

Correct answer: C. 3

Explanation: 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).

Correct answer: C. (6)

Explanation: 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)).

Correct answer: B. 0

Explanation: 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.

Correct answer: D. 15

Explanation: 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.

Correct answer: B. (2)

Explanation: 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.

Correct answer: B. 2

Explanation: 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 \((-2)^3=-8\).

Correct answer: D. No such value

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

Correct answer: A. 0

Explanation: 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.

Correct answer: A. \(4^3-64=64-64=0\)

Explanation: 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.

Correct answer: B. 1

Explanation: 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 \(x\).

Correct answer: B. 2

Explanation: 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.

Correct answer: A. \(3x^2-5x+1\)

Explanation: 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.

Correct answer: C. \(a=7\)

Explanation: 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 \(x=\pm c\).

Correct answer: A. \(\frac{3}{4}\)

Explanation: 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.

Correct answer: B. (-6)

Explanation: (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.

Correct answer: A. x^4-81

Explanation: 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.

Correct answer: A. 0

Explanation: 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.

Correct answer: B. 0

Explanation: \(p(2)=3(2)^3-12(2)=24-24=0\), while \(p(-2)=3(-2)^3-12(-2)=-24+24=0\). Hence \(p(-2)+p(2)=0+0=0\), so option B is correct. Remember that the cube of a negative number is negative, whereas \(-12(-2)\) is positive. In an exam, evaluate both values separately and check the signs carefully.

Correct answer: C. \(x=6\)

Explanation: Completing the square gives \(p(x)=x^2-12x+40=(x-6)^2+4\). Since \((x-6)^2\geq 0\), its least value is 0 when \(x=6\). Therefore, the polynomial has minimum value 4 at \(x=6\). As an exam tip, writing a quadratic in the form \(a(x-h)^2+k\) immediately gives the vertex's x-coordinate as \(h\) when \(a>0\).

Correct answer: B. 4

Explanation: Completing the square gives \(p(x)=x^2+14x+53=(x+7)^2+4\). Since \((x+7)^2\geq 0\), we have \(p(x)\geq 4\), and equality occurs at \(x=-7\). Therefore, the minimum value is 4, so option B is correct. The nearby distractor 2 does not follow from the correct completed-square form. Exam tip: when a quadratic is written as \((x-h)^2+k\) with a positive leading coefficient, its minimum value is \(k\).

Correct answer: A. -10

Explanation: A perfect-square form of \(x^2+kx+25\) must be \((x\pm5)^2\). Since \(k<0\), use \((x-5)^2=x^2-10x+25\), giving \(k=-10\). The form \((x+5)^2\) gives \(k=10\), but it does not satisfy the stated condition. Exam tip: in \((x+a)^2=x^2+2ax+a^2\), the coefficient of the middle term is \(2a\).