Polynomials in one variable

In (y^2+\frac{4}{y}), (\frac{4}{y}=4y^{-1}), so the power is negative. A variable with a negative power is not a polynomial.

The term (0x^2) is not effective, so the terms are (7x^4), (-3x^3), and (5). Do not count a term with zero coefficient.

Substituting \(x=2\) gives \(p(2)=2(2)^2+3(2)-11=2\times4+6-11=3\). Therefore, option C is correct. Follow the order of operations: evaluate the power and multiplication before addition and subtraction.

Substitute x=-2 in the polynomial: q(-2)=(-2)^3+2(-2)^2-5(-2)-6=-8+8+10-6=4. Therefore, the correct value is 4. An answer such as 2 may result from incorrectly treating -5(-2) as -10. In exams, evaluate the power and sign of each term separately when substituting a negative number.

This polynomial has three distinct terms: \(12x^2\), \(-7x\), and \(4\). Therefore, it is a trinomial. A monomial and a binomial have one and two terms, respectively, while a zero polynomial has all its coefficients equal to zero. In exams, count the terms to classify a polynomial.

A linear polynomial has degree 1, so the coefficient of \(x^2\) must be zero. Thus, \(a+3=0\), giving \(a=-3\). For example, if \(a=0\), the coefficient of \(x^2\) is 3, so the polynomial remains quadratic. Exam tip: To reduce the degree of a polynomial, set the coefficient of its highest-power term to zero.

The polynomial \(p(x)=-18\) can be written as \(-18x^0\). The degree of a polynomial is the highest power of the variable with a non-zero coefficient; here, that power is \(0\). Therefore, the correct answer is 0. Exam tip: every non-zero constant polynomial has degree 0, whereas the degree of the zero polynomial is not defined.

The governing rule is that the degree of a non-zero polynomial is the greatest exponent whose coefficient is non-zero. Although 0x^5 is visibly written, its coefficient is zero, so that term is identically zero and does not affect the polynomial. After removing it, the expression is 7x^3 − 2x + 1. The non-zero powers are 3, 1, and 0, and the greatest of these is 3. Therefore the degree is 3, making option B correct. Option C is the common mistake of selecting the largest exponent printed in the expression without checking its coefficient. Option A corresponds only to the linear term, and option D is a coefficient rather than a possible degree here. The zero coefficient must always be considered before deciding the degree.

A variable is a letter whose value can change. In this polynomial, changing u changes the value of the polynomial, so u is the variable. Here, p is the name of the polynomial, while 13 and 9 are constants. Exam tip: The letter appearing with powers in a polynomial is usually the variable.

The term containing \(x\) is \(cx\), so its coefficient is \(c\). Since this coefficient is given as \(-4\), we get \(c=-4\). The value \(-5\) is the constant term, not the coefficient of \(x\). Exam tip: identify the number multiplying the variable to find its coefficient.

Substituting \(x=4\) in the polynomial gives \(p(4)=4^2-16=16-16=0\). Therefore, the correct answer is 0. Remember that the value of the polynomial is found after applying the complete expression; stopping at \(4^2=16\) would lead to the distractor 16.

Substitute c(x=3c) into the polynomial: c(p(3)=3^2-7(3)+12=9-21+12=0c). Therefore, the correct answer is 0, and 3 is also a zero of the polynomial. Exam tip: to evaluate a polynomial, directly replace c(xc) with the given value before simplifying.

Since \(m\ne0\), the term \(mx\) is present. The highest power of the variable \(x\) in the polynomial is therefore 1, so its degree is 1. Degree 0 applies to a non-zero constant polynomial, not to this expression. Exam tip: identify the highest power of the variable whose coefficient is non-zero.

The leading coefficient is the numerical coefficient of the term with the highest power of the variable. First arrange or inspect the polynomial by powers, identify the term of greatest degree, and then read its coefficient, including its sign. It is not necessarily the largest number in size and it is not the constant term unless the polynomial is constant.

In p(z)=-6z^5+4z^4-z+8, the highest power present is z^5. The coefficient multiplying z^5 is -6, so the leading coefficient is -6 and option A is correct. The number 4 belongs to z^4, -1 belongs to z, and 8 is the constant term. The negative sign must be retained.

The term containing \(x^4\) is \(8x^4\), so its coefficient is 8. Although -5 is a coefficient in the polynomial, it belongs to the \(x^3\) term, not the \(x^4\) term. In exams, match the required power with its term before identifying the coefficient.

To add polynomials, combine terms with the same power of x: (4x² + 3x²) + (−9x + x) + (2 − 7) = 7x² − 8x − 5. Option C has the wrong sign for the x-term, while option D incorrectly resembles a multiplication result rather than a sum. Exam tip: combine only like terms with the same variable and exponent.

The phrase “subtract r(x) from p(x)” means p(x) − r(x). Thus we calculate (6x^2 − x + 3) − (2x^2 + 5x − 8). The minus sign before the second bracket changes every sign inside it, giving 6x^2 − x + 3 − 2x^2 − 5x + 8. Now combine like terms: the x^2 terms give 6x^2 − 2x^2 = 4x^2, the x terms give −x − 5x = −6x, and the constants give 3 + 8 = 11. Hence the result is 4x^2 − 6x + 11, so option A is correct. Option C has the wrong sign for x, while B and D mishandle subtraction as addition or alter signs incorrectly.

Multiplying each term by (-3x) gives (-6x^3+12x^2-15x). Apply signs carefully with a negative multiplier.

Using the distributive property, 8x-598x+29=x^2+2x-5x-10=x^2-3x-10, so option A is correct. Option B results from combining the middle terms incorrectly. In an exam, multiply the first, outer, inner, and last terms, then combine like terms carefully with their signs.

Using the distributive property, multiply every term in the first bracket by every term in the second: \((3x+2)(x-4)=3x\cdot x+3x\cdot(-4)+2\cdot x+2\cdot(-4)=3x^2-12x+2x-8=3x^2-10x-8\). Therefore, option A is correct. In option D, the sign of the constant term is incorrect. In an exam, carefully track the negative signs while expanding brackets.

Substituting \(x=-1\), \(p(-1)=(-1)^3-4(-1)^2+(-1)+6=-1-4-1+6=0\). Therefore, 0 is the correct value, and \(-1\) is a zero of the polynomial. Choosing 2 usually results from mishandling the sign or the square of \(-1\). In an exam, always place a negative substitution inside parentheses.

Since \(p(3)=0\), substitute \(x=3\) into the polynomial: \(p(3)=3^2+3k-15=0\). Thus, \(9+3k-15=0\), so \(3k=6\) and \(k=2\). The value 3 does not make the polynomial equal to zero. Exam tip: directly substitute the given zero of the polynomial for \(x\).

Substituting \(x=3\) in the condition \(p(3)=0\) gives \(3^2-3a+18=0\), or \(27-3a=0\). Hence, \(3a=27\) and \(a=9\). Exam tip: when the value of a polynomial is given, substitute the specified value of \(x\) directly and solve the resulting equation.

If \(2\) is a zero of the polynomial, substituting \(x=2\) must make its value zero: \(2^2-6(2)+m=0\), so \(4-12+m=0\) and \(m=8\). Exam tip: Substitute the given zero into the polynomial and equate the result to zero.

p(1)=3(1)²−10(1)+7=3−10+7=0. Therefore, 1 is a zero of the polynomial because a number a is called a zero when p(a)=0. Options B, C and D are incorrect: the degree is 2, the coefficients include 3 and −10, and the constant term is 7. Exam tip: to test a zero, substitute the given number and check whether the polynomial value is 0.