Polynomials in one variable

A non-zero polynomial of degree n has at most n zeroes. Therefore, a degree-4 polynomial can have no more than four real zeroes; it need not have exactly four or even one. Exam tip: distinguish “at most” from “exactly”.

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 zeroes of (p(x)) are (2) and (5), so for (p(x+1)=0), (x+1=2) or (x+1=5). Hence the zeroes of (q(x)) are (1) and (4).

The conditions give (2a+b=2) and (3a+b=-5), so (a=-7), (b=16), and (a+b=9). When two values are given, form two equations.

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.

For degree (4), the coefficient of (x^6) must be (0) and the coefficient of (x^4) must be non-zero. Both conditions hold for (a=1).

(p(-1)=-k+2+7+5=14-k), so (14-k=16) and (k=-2). Watch signs carefully when substituting a negative value.

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 (x^5)-terms cancel and the highest remaining power is (4). Recheck the degree of the polynomial after addition.

In the product of two non-zero polynomials, degrees add, so (6+3=9). In multiplication, look at the highest-power terms.

((3x-2)(2x^2+x-5)=6x^3-x^2-17x+10), so the coefficient of (x^2) is (-1). Combine like terms after expansion.

The governing concept is the zero of a polynomial: if r is a zero of p(x), then p(r) = 0. Since -3 is given as a zero, substitute x = -3 into the polynomial: p(-3) = (-3)^2 + k(-3) - 18 = 9 - 3k - 18 = -3 - 3k. Setting this equal to zero gives -3 - 3k = 0, so -3k = 3 and k = -1. However, -1 is not present among the listed options. Thus the supplied answer key and options are inconsistent; option A cannot be mathematically justified. The item must be revised by adding -1 as an option or correcting the polynomial.

Substituting \(x=2\), we get \(p(2)=4(2)^3-11(2)^2+6(2)+2=32-44+12+2=2\). Hence, the correct answer is 2. The value 0 may result from mishandling the negative term or the constant term. In an exam, calculate the powers first and carefully preserve the signs while adding the terms.

In \(3x^2-\sqrt{5}x+7\), the powers of \(x\) are \(2,1,0\), all non-negative integers. \(\sqrt{5}\) is a real coefficient. In B and D, the variable occurs in a denominator, while C has power \(\tfrac12\). Exam tip: check powers of the variable first.

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.

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.

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 \(b\).

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.

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.