Polynomials in one variable

The leading coefficient of a polynomial is the coefficient of the term with the highest power of the variable. In option B, the highest-degree term is \(-2x^3\), so its leading coefficient is \(-2\). The leading coefficients of the other options are 5, 1, and 3. Exam tip: identify the highest power first, then read its coefficient.

For the degree to be (2), the coefficient of (x^3) must be (0), so (m-2=0) and (m=2). In exams, check the coefficient of the highest power first.

(p(0)=-4) and (p(1)=2-5+7-4=0), so (p(0)+p(1)=-4). Always substitute each value separately before adding.

(p(1)=1-3+3-1=0), so (1) is a zero of this polynomial. Substituting options can quickly identify the zero.

The governing concept is the zero condition for a polynomial: if p(r) = 0, then substituting x = r into the polynomial must give zero. Here r = −3, so calculate p(−3) = k(−3)² + 6(−3) + 9. Since (−3)² = 9, this becomes 9k − 18 + 9 = 9k − 9. The given condition requires 9k − 9 = 0, so 9k = 9 and k = 1. Therefore option A is correct. The sign of the first term is positive because the negative number is squared. Substituting k = 2, 3, or 4 would give p(−3) equal to 9, 18, or 27, respectively, so none of those values satisfies the zero condition.

The statement is incorrect. \(3x^2-5x+7\) has only one variable, \(x\), with powers 2, 1 and 0, all non-negative integers. The number of terms does not decide the number of variables. Exam tip: a constant has power 0.

If a number is a zero of a polynomial, substituting that number into the polynomial gives zero. Here x=2 is a zero, so we must use \\(p(2)=0\\). Substitution produces a simple linear equation in b, and solving it gives \\(b=-1\\). Therefore the supplied answer A is correct.

Calculate carefully: \\(p(2)=3(2)^2+b(2)-10=12+2b-10=2+2b\\). Since 2 is a zero, \\(2+2b=0\\). Hence \\(2b=-2\\) and \\(b=-1\\). Checking, \\(3(4)-2-10=12-2-10=0\\), so the value satisfies the zero condition. The important idea is to set the polynomial equal to zero, not merely substitute and stop.

The degree of a polynomial is the highest power of the variable with a non-zero coefficient. Here, the coefficient of \(x^2\) is \(a\). For the polynomial to have degree 1, this coefficient must be zero, so \(a=0\); the polynomial then becomes \(3x+5\), whose degree is 1. Exam tip: A term with a zero coefficient is omitted when determining the degree of a polynomial.

In the given polynomial, the coefficient of \\(x^2\\) is \\(-2\\), and the constant term is \\(-9\\). Therefore, their sum is \\((-2)+(-9)=-11\\). Exam tip: the constant term is the term containing no variable.

Substituting \(x=2\) in the condition \(q(2)=0\) gives \(2^2-2k+6=0\), so \(10-2k=0\). Hence, \(2k=10\) and \(k=5\), making option C correct. Exam tip: If \(a\) is a zero of a polynomial, substitute \(x=a\) and set the polynomial equal to zero.

(\frac{4}{x}=4x^{-1}) has a negative power of the variable. In a polynomial, powers of the variable must be whole numbers.

In the polynomial \(2x^3+mx^2-x+4\), the coefficient of \(x^2\) is directly \(m\). Since this coefficient is given as \(-5\), we get \(m=-5\). Exam tip: identify the coefficient attached to the required power of the variable.

The polynomial has no term containing \(x^3\), so it can be written as \(0x^3\) when all powers are considered. Therefore, the coefficient of \(x^3\) is 0. Options A and B are the coefficients of \(x^5\) and \(x^4\), while option D is the coefficient of \(x^2\). Exam tip: the coefficient of any missing power in a polynomial is 0.

A number is a zero of a polynomial if the value of the polynomial at that number is zero. \(f(2)=2^2+2(2)-8=4+4-8=0\), so 2 is a zero of the polynomial. For comparison, \(f(4)=16\), so 4 is not a zero. Exam tip: substitute each option into the polynomial and select the one that gives 0.

Substituting \(x=2\) gives \(p(2)=3(2)^2-2(2)+1=9\), while substituting \(x=1\) gives \(p(1)=3(1)^2-2(1)+1=2\). Hence, \(p(2)-p(1)=9-2=7\), so option C is correct. Option 6 results from incorrectly taking \(p(1)\) as 3. Exam tip: substitute each value carefully and simplify the squared term before subtracting.

The governing concept is the leading term of a polynomial. When a polynomial is arranged in descending powers, its leading term is the complete term having the greatest power of the variable. The coefficient and its sign are both part of that term. In 2x^2 + 5x - 3, the powers of x are 2, 1, and 0. Since 2 is the greatest exponent, the first term, 2x^2, is the leading term. Therefore option A is correct. The term 5x has lower degree 1, while -3 is the constant term with degree 0. Option D is incomplete because x^2 gives only the variable part and leaves out the coefficient 2, so it is not the complete leading term.

For the polynomial to have degree 2, the coefficient of \(x^3\) must be zero so that the cubic term disappears and \(4x^2\) becomes the highest-degree term. Thus, \(a-2=0\), giving \(a=2\). For example, if \(a=4\), the coefficient of \(x^3\) is 2, so the degree remains 3. Exam tip: To reduce a polynomial’s degree, set the coefficient of its highest-power term equal to zero.

In the polynomial \(3x^2-4\), the term containing \(x\) is absent. The coefficient of an absent term is taken as \(0\), so option B is correct. In option C, the coefficient of \(x\) is \(1\), not zero. Exam tip: treat every missing term in a polynomial as having coefficient zero.

Substitute \(x=-2\) into the polynomial: \(r(-2)=(-2)^3-4(-2)=-8+8=0\). Therefore, the correct answer is 0. The distractor −4 may result from mishandling the sign in \(-4(-2)\) or in the final addition. In an exam, always use brackets when substituting a negative value.

In the polynomial \(5x^2-3x+7\), the highest power of the variable \(x\) is 2, and its leading coefficient, 5, is non-zero. Hence, its degree is 2 and it is a quadratic polynomial. It would be linear if the highest power were 1 and cubic if it were 3. Exam tip: The degree of a polynomial is the highest exponent of its variable with a non-zero coefficient.