Polynomials in one variable

The zero of a polynomial is the value of \(x\) for which \(p(x)=0\). Thus, \(7x-14=0\), so \(7x=14\) and \(x=2\). Therefore, option B is correct. Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\); here, \(-\frac{-14}{7}=2\).

To find the zero, set p(x) equal to 0: 3x + 12 = 0. Thus, 3x = -12 and x = -4. Therefore, -4 is correct. Option 4 results from missing the negative sign. Exam tip: for ax + b, the zero is -b/a.

The zero of a polynomial is the value of \(x\) for which \(p(x)=0\). Thus, \(ax+b=0\) gives \(ax=-b\), so \(x=-\frac{b}{a}\). Option A has the wrong sign. In an exam, remember that moving the constant term to the other side changes its sign.

The governing concept is direct identification of coefficients. In a polynomial written in descending powers, the coefficient of x is the number multiplying the first-power term x. Here the expression is p(x) = 2x^2 + 0x + 3, so the x-term is explicitly 0x. Hence its coefficient is 0, and option A is correct. The number 2 is the coefficient of x^2, while 3 is the constant term because it does not contain x. Option D, 5, is not obtained from any coefficient in the given polynomial; adding coefficients is not what the question asks. The zero coefficient also explains why the x-term contributes nothing.

The polynomial contains the terms \(x^4\), \(2x^2\), and the constant term \(1=x^0\), but it has no \(x^3\) term. Therefore, among the given options, B is correct. The coefficient of a missing term is taken as 0; in an exam, first list the powers that are explicitly present.

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

A polynomial in one variable contains only one variable, raised only to non-negative integer powers. Option C, \(x^2+y+1\), contains two different variables, \(x\) and \(y\), so it is not a polynomial in one variable. Options A, B and D contain only one variable each—\(x\), \(t\) and \(z\), respectively—and are therefore polynomials in one variable. Exam tip: count the distinct variables, regardless of their names.

Substituting \(x=0\), we get \(p(0)=0^2-5(0)+6=6\). The terms containing \(x\) become zero, leaving only the constant term, 6. Therefore, option D is correct. Exam tip: For any polynomial, \(p(0)\) equals its constant term.

\(p\left(\frac{3}{2}\right)=4\cdot\frac{9}{4}-18+9=0\). While substituting a fraction, square it carefully first.

The polynomial has no x-term, so the missing term is understood as 0x. Therefore, the coefficient of x is 0. Here, 2 is the coefficient of x^3, 3 is the coefficient of x^2, and -5 is the constant term. Exam tip: The coefficient of an absent power is taken as 0.

The coefficient of (x) is (-2) in the first polynomial and (4) in the second, so the sum is (2). Add coefficients of like powers only.

The governing concept is classification by degree. The degree of a non-zero polynomial is the greatest exponent of the variable whose coefficient is non-zero. In p(x) = x² + 1, the coefficient of x² is 1, so the highest exponent is 2. The term 1 is a constant term with degree 0. Consequently, p(x) has degree 2 and is called a quadratic polynomial, making option B correct. It is not linear, because a linear polynomial has degree 1. It is certainly a polynomial because the variable has a non-negative integer exponent and there are finitely many terms. Option D incorrectly treats the constant term or another lower degree as the degree.

Substituting \(x=0\) gives \(p(0)=2(0)^2+3=0+3=3\). Thus, the correct answer is 3. The term containing \(x^2\) becomes zero, leaving the constant term. Exam tip: To find \(p(0)\), replace \(x\) with 0; for a polynomial written in standard form, the result is its constant term.

In option A, the greatest exponent of \(x\) is 4 and its coefficient is non-zero, so its degree is 4. The degrees of options B, C and D are 3, 2 and 1, respectively. Exam tip: The degree of a polynomial is the greatest exponent of the variable with a non-zero coefficient.

The term containing \(x^3\) is \(5x^3\), so the coefficient of \(x^3\) is 5. The numbers 2, -1, and 8 are associated with the \(x^2\) term, the \(x\) term, and the constant term, respectively. Exam tip: identify the numerical multiplier attached to the required power of the variable.

Substituting \(x=-4\), we get \(f(-4)=(-4)^2+4(-4)=16-16=0\). Therefore, the correct answer is 0. The value 16 results from calculating only \((-4)^2\) and ignoring the term \(4x=-16\). Exam tip: the square of a negative number is positive, but the linear term \(4x\) remains negative here.

The constant term is the term that does not contain the variable \(x\). In the given polynomial, \(12x^2\) and \(-9x\) contain \(x\), whereas 4 does not; therefore, the correct answer is 4. Exam tip: Identify the term independent of the variable to find the constant term.

To find the zero, set \(p(x)=0\): \(x+9=0\), which gives \(x=-9\). Therefore, -9 is correct. The distractor 9 has the wrong sign because \(9+9\neq0\). Exam tip: the zero of \(x+a\) is always \(-a\).

Terms are separated by plus or minus signs. The terms here are \\(4x^2\\), \\(-6x\\), and \\(1\\), so there are 3 terms and the polynomial is a trinomial. In an exam, count the negative sign with its term; \\(-6x\\) is one term, not two.

Substituting 1 for \(x\) gives \(g(1)=3(1)^2-1=3-1=2\). Therefore, option B is correct. Remember to evaluate \(x^2\) before performing the subtraction; taking only \(3x^2\) would incorrectly give 3.