Polynomials in one variable

Substitute \(x=2\) into the polynomial: \(p(2)=2(2)^2+2b-10=4\). Thus, \(8+2b-10=4\), so \(2b-2=4\) and \(b=3\). Therefore, option C is correct. Exam tip: when a value such as \(p(a)\) is given, substitute \(x=a\) directly in the polynomial.

The polynomial \(x^3-1\) has two terms, \(x^3\) and \(-1\), so it is a binomial. Its highest exponent is 3, so it is a cubic binomial. Option B is incorrect because the polynomial has neither three terms nor degree 2. Exam tip: classify a polynomial first by its highest degree and then by its number of terms.

For a polynomial, \(p(0)\) equals its constant term because \(p(0)=0^2+5(0)+c=c\). Here, the constant term is \(c=-6\), so \(p(0)=-6\). The value 5 is the coefficient of \(x\), not the constant term. Exam tip: To find \(p(0)\), substitute \(x=0\) in the polynomial.

In a polynomial, the powers of the variable must be non-negative integers, while the coefficients may be real numbers. In option C, the powers of \(t\) are 3, 1, and 0, so it is a polynomial. Option A contains \(\sqrt{t}=t^{1/2}\), option B contains the negative power \(t^{-2}\), and option D has the variable in the denominator; hence, they are not polynomials. Exam tip: check the powers of the variable first—fractional or negative powers make an expression non-polynomial.

Substituting \\(x=2\\) gives \\(p(2)=2^2-2(2)+3=3\\). Substituting \\(x=-1\\) gives \\(p(-1)=(-1)^2-2(-1)+3=6\\). Therefore, \\(p(2)+p(-1)=3+6=9\\). Exam tip: when substituting \\(x=-1\\), remember that \\(-2(-1)=+2\\).

Since \(a\ne 0\), the term \(ax\) is definitely present, so the highest power of \(x\) in \(ax+b\) is \(1\). Therefore, it is a linear polynomial. It cannot be constant because the coefficient of \(x\) is non-zero, and its degree is not \(2\). Exam tip: a polynomial whose highest power of the variable is \(1\) is called a linear polynomial.

The term \\(x^3\\) is absent from the polynomial, so its coefficient is 0. The coefficient of \\(x\\) is -5. Therefore, their sum is \\(0+(-5)=-5\\). Exam tip: The coefficient of any missing power in a polynomial is taken as 0.

\(p\left(\frac{1}{3}\right)=3\left(\frac{1}{3}\right)^2+2\left(\frac{1}{3}\right)-1=0\). Use brackets while substituting fractions.

In descending order, the term with the highest power of x is written first, followed by terms with decreasing powers. The powers here are 2, 1, and 0, so the correct form is \\(2x^2-3x+8\\). Options B and C incorrectly change the sign of the x-term, while D changes signs and does not preserve the polynomial. Exam tip: identify each term’s power before arranging the terms from greatest to least.

In \(x^3-2x+1\), the powers of x are 3, 1 and 0, all non-negative integers, so it is a polynomial in one variable. \(1/x=x^{-1}\) and \(\sqrt{x}=x^{1/2}\) are not polynomials. Exam tip: reject expressions with negative or fractional powers.

The zero polynomial has no non-zero term, so there is no highest exponent to determine its degree; hence, its degree is not defined. In contrast, a non-zero constant polynomial has degree 0, so option A is incorrect. Exam tip: distinguish the zero polynomial from a non-zero constant polynomial.

To find \(p(1)\), substitute \(x=1\) in the polynomial: \(p(1)=1^3-2(1)^2+1=1-2+1=0\). Hence, the correct answer is 0. The closest distractor, 1, results from not combining the terms correctly; the term \(-2x^2\) contributes \(-2\), not \(+2\). In an exam, substitute the given value directly to check the value of a polynomial or its zero.

Substituting \(x=-1\), \(p(-1)=2(-1)^2+3(-1)+4=2-3+4=3\). Also, \(p(0)=2(0)^2+3(0)+4=4\). Therefore, \(p(-1)-p(0)=3-4=-1\), so option A is correct. Choosing 0 would incorrectly assume that the two polynomial values are equal. Exam tip: the square of a negative number is positive, but the linear term \(3x\) becomes \(-3\) when \(x=-1\).

(\sqrt{5}) is a real coefficient and the powers of (x) are whole numbers. So it is a polynomial in (x).

Substituting \(x=3\) in \(p(x)=0\) gives \(3^2-3a+12=0\), so \(21-3a=0\). Hence, \(3a=21\) and \(a=7\), making option C correct. Exam tip: If \(r\) is a zero of a polynomial, substitute \(x=r\) directly into the polynomial equation.

For a non-zero polynomial in one variable, the degree is the greatest exponent of the variable whose coefficient is non-zero. In 2x^3 − 5x^2 + x − 4, the exponents are 3, 2, 1, and 0. The greatest exponent is therefore 3. The constant term is the term that contains no variable; here it is −4, which may also be viewed as −4x^0. Thus the correct pair is degree 3 and constant term −4, so option A is correct. Option B incorrectly uses the exponent of the second term and also selects the coefficient 1. Option C confuses the coefficient −5 with a constant term. Option D treats the coefficient 2 as though it were an exponent.

The degree of a nonzero polynomial is the greatest exponent of its variable that has a nonzero coefficient. In \(p(x)=11x^6-4x^3+x-9\), the exponents present are 6, 3, 1, and 0. The largest of these is 6, and its coefficient 11 is nonzero. Therefore the degree of the polynomial is 6, so option C is correct.

The degree is determined by the exponent, not by the numerical size of the coefficient. Thus the coefficient 11 does not make the degree 11. Likewise, the terms \(-4x^3\) and \(x\) have lower powers and do not control the degree. Always identify the highest power of the variable after ignoring zero coefficients.

The term containing x² is −8x², so its coefficient is −8. The number 13 is the coefficient of x, while −21 is the constant term. In an exam, always include the positive or negative sign when writing a coefficient.

The constant term is the term that does not contain the variable \(t\). In the given polynomial, \(4t^5\), \(-6t^2\), and \(t\) contain \(t\), whereas 15 does not; therefore, the constant term is 15. Exam tip: identify the term whose variable has exponent zero or is absent.

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.