Polynomials in one variable

A polynomial in x contains only non-negative integer powers of x. Its degree is the greatest power with a non-zero coefficient, whereas a quadratic polynomial has degree exactly 2. Option B, 6x^3 − 2x + 9, contains powers 3, 1, and 0, so it is a valid polynomial, but its degree is 3; therefore it is cubic, not quadratic. Option A has degree 2 and is quadratic. Option C includes 5/x = 5x^-1, a negative power, so it is not a polynomial in x. Option D includes √x = x^(1/2), whose exponent is not an integer, so it is also not a polynomial. Thus option B is the only expression satisfying both conditions. The distinction requires checking both the allowed powers and the degree.

The governing concept is evaluation of a polynomial at a specified input. Since p(4) = 7, substitute x = 4 into p(x) = 2x^2 − 5x + c. This gives p(4) = 2(4)^2 − 5(4) + c = 2(16) − 20 + c = 32 − 20 + c = 12 + c. The given condition therefore becomes 12 + c = 7. Subtracting 12 from both sides gives c = 7 − 12 = −5. Hence option A is correct. Option B can result from an arithmetic or sign error, while positive options fail the equation because adding them to 12 cannot produce 7. Substitution followed by a simple linear equation provides a complete check.

A zero or root of a polynomial is a value r for which p(r) = 0. We can test the supplied integer choices directly. For x = 1, p(1) = 1³ − 2(1)² − 5(1) + 6 = 1 − 2 − 5 + 6 = 0. Hence 1 is a zero and option A is correct. For comparison, p(2) = 8 − 8 − 10 + 6 = −4, p(3) = 27 − 18 − 15 + 6 = 0? This calculation gives 0 as well, so the question as written has two correct answers, 1 and 3. Therefore, despite the supplied key A, the item is ambiguous and requires revision; a condition or a different option must be added to ensure one answer only.

The defining property of a zero is that substituting it into the entire polynomial gives zero. Since x = 2 is a zero of p(x) = ax^2 + 4x − 12, set p(2) = 0. Substitution gives p(2) = a(2)^2 + 4(2) − 12 = 4a + 8 − 12 = 4a − 4. Now solve 4a − 4 = 0: adding 4 gives 4a = 4, and dividing by 4 gives a = 1. Thus option A is correct. Checking the distractors, a = 2, 3 and 4 would make p(2) equal to 4, 8 and 12 respectively, so none would satisfy the zero condition. Every term must be included when applying the condition.

The governing concept is identification of coefficients and the constant term in a polynomial. In a term such as −7x², the coefficient of x² is −7 because it multiplies x². The constant term is the term containing no variable, so in this polynomial it is −11. The question asks for their algebraic sum, not their product or their separate values. Therefore, (−7) + (−11) = −18. Option A is correct. Option B gives only the constant term, and option C gives only the coefficient of x². Option D does not result from combining the requested signed values. The negative signs must be retained throughout the addition; adding the absolute values would give an incorrect result.

Use the identity \((a+b)^2=a^2+2ab+b^2\). Here, \(a=2x\) and \(b=3\), so \((2x+3)^2=(2x)^2+2(2x)(3)+3^2=4x^2+12x+9\). Option B omits the middle term, while option D uses the wrong sign for it. Exam tip: always include the \(2ab\) term when expanding \((a+b)^2\).

(p(7)=49-70+21=0), so (7) is a zero of this polynomial. Simplify all terms in order after substitution.

For the degree to be (3), the coefficient of (x^5) must be (0), so (a-5=0). First make the highest-degree term vanish.

Substituting \(x=-1\) gives \(p(-1)=k(-1)^2-5(-1)+6=k+5+6=k+11\). Thus, \(k+11=14\), so \(k=3\). For instance, taking \(k=2\) gives 13, not 14. In the exam, check both \((-1)^2\) and the sign in \(-5(-1)\) carefully.

When adding polynomials, coefficients of like powers are added. The coefficients of x² are −2 and 7, so the coefficient in the sum is −2 + 7 = 5. Option C gives only the x² coefficient of q(x), not of the complete sum. Exam tip: group terms having the same power before adding polynomials.

To find the degree of a sum, first add like terms and then inspect the highest power whose coefficient is not zero. Adding the polynomials gives (4x^4-3x^2+2)+(-4x^4+5x^3+x-8). The x^4 terms cancel because 4x^4-4x^4=0. The remaining polynomial is 5x^3-3x^2+x-6.

The highest power still present is x^3, whose coefficient is 5, not zero. Therefore the degree of the sum is 3, so choice C is correct. A common error is to retain degree 4 merely because both original polynomials had degree 4. Degree can decrease after addition when leading terms cancel, so the resulting expression must always be simplified before its degree is stated.

In the product of two non-zero polynomials, degrees add, so (4+3=7). Remember degree addition in multiplication.

((x-2)(3x^2+4x-5)=3x^3-2x^2-13x+10), so the coefficient of (x^2) is (-2). Combine like terms after expansion.

Since the polynomial is zero at its zero, \(p(3)=0\). Thus, \(3^2-3k+12=0\), giving \(21-3k=0\) and hence \(k=7\). Therefore, 7 is correct. Exam tip: whenever a zero is given, substitute it in the polynomial and set the result equal to zero.

Substitute x = 2 into the polynomial: p(2) = 2(2³) − 9(2²) + 12(2) − 5 = 16 − 36 + 24 − 5 = −1. Hence, option A is correct. Option B may result from an error in the final addition and subtraction. Exam tip: after substitution, evaluate powers first, then multiplication, and finally addition or subtraction.

Substituting \(x=1\) gives \(p(1)=1^3-4(1)^2+m(1)+6=1-4+m+6=m+3\). Since \(p(1)=0\), we require \(m+3=0\), so \(m=-3\). The nearby option \(-2\) would give \(p(1)=1\), not zero. Exam tip: when a value is given as a zero of a polynomial, substitute it and set the resulting expression equal to zero.

Substituting -1 for x gives p(-1) = (-1)^4 - 2(-1)^2 + 1 = 1 - 2 + 1 = 0. Since even powers of -1 equal 1, treating x^2 as -1 would incorrectly give -2. Exam tip: Always enclose a negative value in brackets before raising it to a power.

Use the given zero conditions by substituting the corresponding values into the polynomial. From p(1)=0, we get 1+a+b=0, so a+b=−1 immediately. For verification, p(2)=0 gives 4+2a+b=0. Subtracting the first equation from the second gives 3+a=0, hence a=−3; substituting back gives b=2. Therefore a+b=−3+2=−1, confirming option A. The roots are 1 and 2, so the polynomial can also be written as (x−1)(x−2)=x²−3x+2. Option C, which was previously marked, is incorrect; the conditions clearly force the sum of the coefficients a and b to be −1.

In a polynomial, every exponent of x must be a non-negative integer. In option A, the highest exponent is 5, while \(\sqrt{2}\) is a valid real coefficient. Option B contains \(1/x=x^{-1}\), so it is not a polynomial. Exam tip: check exponents before checking coefficients.

(x^2+\frac{1}{x}+4) contains (x^{-1}), which is not allowed in a polynomial. Be careful when the variable is in the denominator.