Polynomials in one variable

Substituting \(x=-3\), \(p(-3)=(-3)^2+6(-3)+9=9-18+9=0\). Therefore, 0 is the correct value, and \(-3\) is a zero of the polynomial. Choosing 9 usually results from omitting the term \(6(-3)\). In an exam, enclose a negative value in parentheses while substituting.

In \(x^2+2x+3\), the powers of \(x\) are 2, 1, and 0, all of which are non-negative integers; hence it is a polynomial. Its highest power is 2, so it is a quadratic polynomial, not a linear one. Option A has degree 1 and is linear, while option C contains \(x^{-1}\) and option D contains \(x^{1/2}\), so neither is a polynomial. Exam tip: in a polynomial, the variable can have only non-negative integer exponents such as 0, 1, 2, and so on.

The zero of a polynomial is the value of \(x\) that makes the polynomial equal to zero. Setting \(9x-27=0\) gives \(9x=27\), so \(x=3\). The value 9 is the coefficient of \(x\), not the zero. Exam tip: set a linear polynomial equal to zero and solve for \(x\).

The zero of a polynomial is the value of \(x\) for which \(p(x)=0\). Thus, \(4x+20=0\), so \(4x=-20\) and \(x=-5\). Therefore, option B is correct. \(-4\) is merely the negative of the coefficient and is not the zero. Exam tip: Set a linear polynomial equal to zero before solving for \(x\).

A zero of a polynomial is the value of \\(x\\) for which \\(p(x)=0\\). Thus, \\(cx+d=0\\) gives \\(cx=-d\\), and since \\(c\ne0\\), \\(x=-\frac{d}{c}\\). Option A misses the negative sign. For exams, remember that the zero of \\(ax+b\\) is \\(-\frac{b}{a}\\).

The direct answer is option A, 0. A coefficient is the number multiplying a particular power of the variable. In the polynomial, the terms are arranged as a cubic term, a quadratic term, a linear term, and a constant: 5x^3, 0x^2, -4x, and 12. The term containing x^2 is explicitly 0x^2, so its coefficient is 0. Therefore option A is correct. Option B, 5, is the coefficient of x^3, not x^2. Option C, -4, is the coefficient of x. Option D, 12, is the constant term and has no x. A missing power is also understood to have coefficient zero. Remember: first locate the exact power asked for, then read the number immediately before it.

The terms containing \(x^5\) and \(x^3\) are present. The constant term \(-2\) can be written as \(-2x^0\), so the power \(x^0\) is also present. There is no \(x^4\) term, which means its coefficient is 0. Exam tip: To identify a missing power, check whether its coefficient appears in the polynomial; a missing term has coefficient zero.

In a polynomial, powers of the variable are non-negative integers like (0,1,2,\ldots). This is the basic identification rule.

Substituting \(x=0\) gives \(p(0)=0^2-8(0)+13=13\), so the correct answer is 13. When \(x=0\), all terms containing \(x\) become zero, leaving only the constant term. Exam tip: \(p(0)\) is directly equal to the constant term of the polynomial.

\(p\left(\frac{2}{3}\right)=9\cdot\frac{4}{9}-8+4=0\). While substituting a fraction, square and multiply carefully.

The polynomial has no term containing \(x^3\), so the missing term is treated as \(0x^3\). Therefore, the coefficient of \(x^3\) is 0. The value -2 is the coefficient of \(x^2\), while 9 is the constant term. Exam tip: The coefficient of an absent power in a polynomial is taken as 0.

The coefficient of (x) is (-6) in the first polynomial and (9) in the second, so the sum is (3). Add coefficients of like powers only.

In (p(x)=x^3+1), the highest power is (3), so it is a cubic polynomial. Decide the type of a polynomial by its degree.

Substituting \(x=0\) gives \(p(0)=a(0)^3+b(0)^2+c(0)+d=d\). Since \(p(0)=-7\), the constant term is \(d=-7\). Option A has the incorrect sign. Exam tip: For any polynomial, \(p(0)\) equals its constant term.

For the degree to be (2), the coefficient of (x^4) must be (0), so (k-4=0) and (k=4). Check the coefficient of the highest power first.

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.

The correct answer is A, 1. A zero is a value r for which p(r) = 0. Factor the polynomial: x³ − 2x² − 5x + 6 = (x − 1)(x − 3)(x + 2). Therefore the zeros are 1, 3 and −2. The positive zeros are 1 and 3, and the smallest positive one is 1. Option A is correct. Option B is not a zero because p(2) = 8 − 8 − 10 + 6 = −4. Option C is a zero, since p(3) = 27 − 18 − 15 + 6 = 0, but it is not the smallest positive zero. Option D gives p(4) = 64 − 32 − 20 + 6 = 18, so it is not a zero. The phrase ‘smallest positive’ is important; without it, both 1 and 3 would be correct and the question would be ambiguous.

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.

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 \((a+b)^2\).

The direct answer is option A: 0. Substitute x=7 into the polynomial: \(p(7)=7^2-10(7)+21\). Calculate each part in order: 7^2=49 and 10(7)=70, so p(7)=49−70+21. Since 49+21=70, the result is 0. Thus 7 is a zero, or root, of the polynomial. Option A, 0, is correct. Option B, 7, is the input value, not the output; it confuses x with p(x). Option C, 14, comes from an arithmetic mistake and is not the value of the expression. Option D, 21, is only the constant term and ignores the other two terms. A useful check is factoring: x^2−10x+21=(x−3)(x−7), so putting x=7 makes one factor zero and the whole product zero. When substituting into a polynomial, use brackets around the value and calculate powers and products before addition or subtraction. Memory cue: a number is a zero when the polynomial’s value at that number is exactly 0.

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.

For a polynomial, the sum of all coefficients is obtained by substituting x=1, because every power of 1 equals 1. Thus p(1)=2(1)³+k(1)²−8(1)+3=2+k−8+3=k−3. The problem states that this sum is zero, so k−3=0 and therefore k=3. Hence option C is correct. Directly adding the coefficients gives the same equation: 2+k−8+3=0. Option A, B, or D would make the coefficient sum respectively −2, −1, or 1 rather than zero. The key idea is that the variable is not assigned an arbitrary value; x=1 is specifically used because it converts every coefficient term into that coefficient itself.

The governing concept is direct evaluation of a polynomial at specified values, followed by multiplication. For x = 1, p(1) = 2(1)³ + (1)² − 5(1) + 2 = 2 + 1 − 5 + 2 = 0. For x = −1, carefully preserve the odd-power sign: p(−1) = 2(−1)³ + (−1)² − 5(−1) + 2 = −2 + 1 + 5 + 2 = 6. Thus p(1)p(−1) = 0 × 6 = 0, making option C correct. Once one factor is zero, the product is zero regardless of the other factor. Options A, B, and D arise from substitution, sign, or multiplication errors. The corrected answer is C, not A.

First calculate the fixed value p(1): p(1)=1²−4(1)+1=1−4+1=−2. Now subtract this value from the general expression: p(x)−p(1)=(x²−4x+1)−(−2)=x²−4x+1+2=x²−4x+3. Therefore option A is correct. The negative sign before p(1) is important: subtracting −2 is the same as adding 2. Option C is merely the original polynomial and ignores the subtraction. Option B would result from subtracting 2 rather than subtracting −2, and option D changes the sign of the linear term without any algebraic basis. The original option list and key were inconsistent; option A has been corrected to the actual result.