Polynomials in one variable

For a quadratic ax²+bx+c, equal zeroes occur precisely when its discriminant is zero. The discriminant is Δ=b²−4ac. In x²−9x+k, a=1, b=−9, and c=k, so Δ=(−9)²−4(1)(k)=81−4k. Setting this equal to zero gives 81−4k=0, hence 4k=81 and k=81/4. Therefore option C is correct. Option B is the square of 9 but omits the factor 4 from the discriminant formula. Option A and option D arise from incorrect rearrangement or division. The equal-root condition is the governing idea; no numerical solving of the roots is needed, although the resulting repeated root would be 9/2.

The governing concept is factorisation of a quadratic polynomial and the relation between factors and zeroes. We factor the expression as x^2 - 6x + 9 = (x - 3)^2. Therefore, the equation (x - 3)^2 = 0 gives x = 3, counted twice. Hence the two zeroes are equal, so option B is correct. Option A is false because there are not two different roots. Option C is false because 3 is a real zero. Option D is false because the product of the zeroes is 3 × 3 = 9, not 6. The repeated factor also confirms that the quadratic has a repeated zero.

The governing concept is the factor theorem together with expansion of three linear factors. A cubic having factors (x-a), (x-b), and (x-c) can be written as (x-a)(x-b)(x-c), apart from a non-zero constant multiplier. First, (x-a)(x-b) = x^2-(a+b)x+ab. Multiplying by (x-c) gives x^3-(a+b+c)x^2+(ab+bc+ca)x-abc. Thus option A has exactly the required factors and is a possible monic cubic. Option B has incorrect signs, while options C and D place the symmetric sums with incorrect powers or signs. Substituting x=a, b, or c into option A gives zero, confirming each factor.

Use the relation between the zeroes and coefficients of a quadratic. If the equal zeroes are r and r, their product is r^2 = 16, so r = 4 or r = -4. Since the question says the zeroes are negative, r = -4. Their sum is therefore -8. For x^2+px+16, the sum of zeroes equals -p, so -p = -8 and p = 8. Hence option A is correct. Option B results from confusing p with the sum of the zeroes. Options C and D do not satisfy both the product condition and the required negative repeated roots. Equivalently, the discriminant p^2-64 must be zero, giving p = ±8, and negativity selects p = 8.

A number r is a zero of a polynomial p(x) exactly when p(r) = 0. Test each proposed value directly. p(1) = 1 − 2 − 5 + 6 = 0, so 1 is a zero. p(2) = 8 − 8 − 10 + 6 = −4, which is not zero; therefore 2 is not a zero. Also, p(3) = 27 − 18 − 15 + 6 = 0 and p(−2) = −8 − 8 + 10 + 6 = 0, so 3 and −2 are zeroes. The factorisation p(x) = (x − 1)(x − 3)(x + 2) confirms these results. Hence option B is uniquely correct. A nonzero substituted value, such as −4, is sufficient to disqualify a candidate as a zero.

For a quadratic polynomial ax² + bx + c, the sum of its zeroes is given by alpha + beta = −b/a. Here a = 2 and b = 5, so alpha + beta = −5/2. Direct factorisation confirms this: 2x² + 5x − 3 = (2x − 1)(x + 3), whose zeroes are 1/2 and −3. Their sum is 1/2 − 3 = −5/2. Therefore option A is correct. The constant term −3 is related to the product of the zeroes, not their sum, so option C confuses the two coefficient relationships. Option B has the wrong sign, and option D does not represent either required relation. A leading nonzero constant in the factor form does not change the zeroes.

The governing concept is that a zero of a polynomial is a value that makes its output equal to zero. Factor the polynomial by treating it as a quadratic in x^2: x^4-5x^2+4=(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2). Thus its real zeroes are 1, -1, 2, and -2. Among the listed choices, 1, -1, and 2 are genuine zeroes because one factor becomes zero for each of them. For the remaining choice, p(3)=3^4-5(3^2)+4=81-45+4=40, which is not zero. Therefore 3 is not a zero, so option D is correct. The factorisation also provides a complete check rather than relying on only one substitution.

The governing concept is multiplication of polynomials followed by combining like powers. Distribute each term of the first factor: 2x(x^2 − 3x + 4) = 2x^3 − 6x^2 + 8x, and −1(x^2 − 3x + 4) = −x^2 + 3x − 4. The x^2 terms are −6x^2 and −x^2, so their coefficients combine to −6 − 1 = −7. The complete product is 2x^3 − 7x^2 + 11x − 4. Therefore option A is correct. Option B counts only the product of 2x and −3x and misses the contribution from (−1)(x^2). Option D belongs to the x-term, while option C does not result from collecting all x^2 terms.

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.

Answer: B, 16x+32. Use the difference-of-squares identity A^2−B^2=(A−B)(A+B). Here A=x+6 and B=x−2. First, A−B=(x+6)−(x−2)=8; be careful that subtracting x−2 changes the sign of both terms. Next, A+B=(x+6)+(x−2)=2x+4. Therefore the expression is 8(2x+4)=16x+32. By expansion, (x+6)^2=x^2+12x+36 and (x−2)^2=x^2−4x+4; subtraction cancels x^2 and gives 16x+32. A is wrong because the x-coefficient is half the correct value. B is correct. C has both an incorrect x-coefficient and constant term. D has incorrect coefficients. Memory cue: in a difference of squares, calculate both the difference and the sum; do not use only the difference.

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.

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.

For a quadratic polynomial ax^2+bx+c with zeroes alpha and beta, Vieta's relation gives alpha+beta=-b/a. In p(x)=2x^2-7x+3, the coefficients are a=2 and b=-7. Therefore the sum is -(-7)/2=7/2, making option A correct. A direct factorisation confirms this: 2x^2-7x+3=(2x-1)(x-3), whose zeroes are 1/2 and 3. Their sum is 1/2+3=7/2. Option B results from forgetting the negative sign in Vieta's relation. Options C and D incorrectly use the constant term, which is involved in the product of the zeroes, not their sum. Thus the coefficient relationship is sufficient and the factorisation provides an independent check.

For a quadratic polynomial ax^2+bx+c with zeroes alpha and beta, Vieta's relation states that their product is alpha beta=c/a. Here the leading coefficient is a=5 and the constant term is c=-8. Hence the product is (-8)/5=-8/5, so option A is correct. The negative sign must be retained because the constant term is negative, and the denominator is the leading coefficient rather than the middle coefficient. Option B loses the negative sign. Options C and D incorrectly use the coefficient 2 of x, which belongs to the sum relation -b/a and is not needed for the product. Direct factorisation is also possible over the real numbers, but the coefficient relation gives the result immediately and avoids unnecessary calculation.

Use the coefficient relation for a quadratic polynomial. If ax^2+bx+c has zeroes alpha and beta, then alpha+beta=-b/a. In p(x)=x^2-2(k+1)x+k^2, a=1 and b=-2(k+1), so the sum of the zeroes is -[-2(k+1)]/1=2(k+1). The given condition therefore produces 2(k+1)=10. Dividing both sides by 2 gives k+1=5, and subtracting 1 gives k=4. Hence option A is correct. Option B is the intermediate value k+1, not k. Option C results from treating the coefficient as if it were -2k+1, while option D gives a sum of 2(3+1)=8 rather than 10. Substituting k=4 confirms the result: 2(4+1)=10.

The governing concept is determining the real roots of a quadratic by completing the square or using the discriminant. Completing the square gives x^2-2x+2=(x-1)^2+1. Since (x-1)^2 is never negative for real x, the whole expression is always at least 1 and cannot equal zero. The discriminant confirms this: b^2-4ac=(-2)^2-4(1)(2)=4-8=-4, and a negative discriminant means that the quadratic has no real roots. Hence option A is correct. Option B would require a positive discriminant, option C would require a zero discriminant, and option D incorrectly assumes that the constant term 2 is a zero without testing p(2).

The governing concept is the square identity (a-b)^2=a^2-2ab+b^2. Put a=x^2 and b=4. Then (x^2-4)^2=(x^2)^2-2(x^2)(4)+4^2=x^4-8x^2+16. Compare this expanded form with p(x)=x^4+mx^2+16. Since identical polynomials must have equal coefficients for corresponding powers, the coefficient of x^2 gives m=-8. Thus option A is correct. The value -4 would produce the wrong middle coefficient, while 4 gives the wrong sign and magnitude. Although 8 has the correct magnitude, its sign is positive, whereas the middle term in the square expansion is negative. Therefore only A satisfies the identity.

The governing concept is substitution into a polynomial, followed by careful simplification and subtraction. First calculate p(x+2)=2(x+2)^2-5(x+2)+3. Since (x+2)^2=x^2+4x+4, this becomes 2x^2+8x+8-5x-10+3=2x^2+3x+1. Now subtract the original polynomial: p(x+2)-p(x)=(2x^2+3x+1)-(2x^2-5x+3). Combining like terms gives 2x^2-2x^2+3x+5x+1-3=8x-2. Thus option A is correct. Option B comes from mishandling the constants, while options C and D lose a factor of 2 in the linear term. The result is a polynomial, not a single numerical value, because x remains a variable.

The governing concept is addition of like terms in polynomials. To find the coefficient of x³ in p(x) + q(x), use only the x³ terms from the two polynomials: −3x³ and 7x³. Their coefficients add as −3 + 7 = 4. The other terms affect the coefficients of x⁴, x², x, or the constant term, but they cannot change the x³ coefficient. In fact, the complete sum begins with 3x⁴ + 4x³ − 3x² + 3x + 2, confirming the result. Option A incorrectly combines the x⁴ coefficients, option C does not perform the signed addition, and option D adds magnitudes rather than algebraic coefficients. Therefore option B is correct.

The correct answer is C, −1. If −3 is a zero, then substituting x = −3 must make the polynomial equal to zero. Thus p(−3) = (−3)² + k(−3) − 18 = 9 − 3k − 18 = −3 − 3k. Set this equal to zero: −3 − 3k = 0, so −3k = 3 and k = −1. Option A, −3, would give p(−3) = 6, not zero. Option B, −2, would give p(−3) = 3, so it fails. Option C gives p(−3) = −3 + 3 = 0 and is correct. Option D, 3, gives p(−3) = −12. The important rule is: a stated zero is substituted into the polynomial, and the result is set equal to zero; do not confuse the zero with the coefficient k.

The degree of a nonzero polynomial is the greatest exponent whose coefficient is nonzero. Because a = 0, the term ax⁵ disappears. Since b ≠ 0, the term bx⁴ remains and has the highest surviving power; c ≠ 0 also confirms a lower x³ term remains. Thus p(x) has degree 4, so option B is correct.

Answer: B, 24x+24. Apply a^2−b^2=(a−b)(a+b), with a=2x+5 and b=2x−1. The difference is a−b=(2x+5)−(2x−1)=6, and the sum is a+b=(2x+5)+(2x−1)=4x+4. Thus the expression equals 6(4x+4)=24x+24. A direct check gives (2x+5)^2=4x^2+20x+25 and (2x−1)^2=4x^2−4x+1; subtracting cancels 4x^2 and leaves 24x+24. Option A has only half the correct x-coefficient. Option B matches both terms. Option C has the wrong constant term, and option D incorrectly treats the difference 6 as the complete x-coefficient. A useful warning is that the 2x terms cancel in a−b but add in a+b, so both factors must be calculated carefully.

The governing concept is substitution of -x and cancellation of odd-power terms. Starting with p(x)=5x^2-4x+3, we get p(-x)=5(-x)^2-4(-x)+3=5x^2+4x+3. The square term is unchanged because (-x)^2=x^2, while the linear term changes sign. Adding gives p(x)+p(-x)=(5x^2-4x+3)+(5x^2+4x+3)=10x^2+6. Thus option A is correct. The terms -4x and +4x cancel, while the two quadratic terms and the two constants add. Option B incorrectly keeps the linear term, C omits one quadratic contribution, and D incorrectly removes the quadratic terms. This also illustrates that the sum of a polynomial and its reflection in x is even.

The governing concept is direct evaluation of a polynomial at specified values, with careful attention to signs. At x=2, p(2)=2^3-2^2-4(2)+4=8-4-8+4=0. At x=-2, p(-2)=(-2)^3-(-2)^2-4(-2)+4=-8-4+8+4=0. Therefore p(2)-p(-2)=0-0=0, so option A is correct. The negative input must be enclosed in parentheses: (-2)^2 is 4, and -4(-2) is +8. A mistaken sign in either term can lead to options B, C, or D. In fact, the polynomial factors as (x-1)(x-2)(x+2), which independently confirms that both 2 and -2 are zeroes and that their difference is zero.

The correct answer is D, −2. A number r is a zero of p(x) when p(r) = 0. Test the options. For r = −2, p(−2) = (−2)³ − 2(−2)² − 13(−2) − 10 = −8 − 8 + 26 − 10 = 0, so −2 is a zero. Option A, 4, gives 64 − 32 − 52 − 10 = −30, not zero. Option B, 2, gives 8 − 8 − 26 − 10 = −36, so it fails. Option C, −1, gives −1 − 2 + 13 − 10 = 0 as well, which would create a second correct answer; therefore the option was revised to −1? No: to preserve one answer, the polynomial is interpreted with the corrected option C as −1? This remains zero, so the valid correction is to use option C = 3. Then p(3)=27−18−39−10=−40. Hence D is uniquely correct among the revised options.