Linear polynomials

In the polynomial 8z - 15, the term 8z can be written as 8z^1, so the power of z is 1. Hence, the correct answer is 1, and the polynomial is linear in z. Option 0 applies to the constant term -15, not to z. Exam tip: If no exponent is written with a variable, its exponent is 1.

In \(3x^2-1\), the highest power of \(x\) is 2, so its degree is 2. Hence, it is a quadratic polynomial, not a linear polynomial. A linear polynomial has degree 1; therefore, \(7x+4\), \(x-10\), and \(12-5x\) are linear polynomials. Exam tip: identify the degree by finding the highest power of the variable.

The general form of a linear polynomial is (ax+b) where (a\neq0). If (a=0), it will not remain linear.

The degree of a polynomial is the highest power of its variable with a non-zero coefficient. Here, x in 9x has power 1, while the constant term −2 has power 0. Therefore, the degree of p(x) is 1. Degree 0 would apply to a non-zero constant polynomial only. Exam tip: identify the greatest exponent of x to find the degree.

In a polynomial, the coefficient of a variable is the number multiplying that variable. Here, the x-term is -7x, which means -7 × x. Therefore, the coefficient of x is -7. The closest distractor is 7, but omitting the negative sign makes the answer incorrect. Exam tip: Always include the sign attached to the term when identifying a coefficient.

A constant term is a term with no variable. In \(x-11\), \(x\) contains the variable, whereas \(-11\) has no variable. Therefore, the constant term is \(-11\). The constant term would be \(0\) only if no separate constant number were present. Exam tip: identify the term without a variable to find the constant term quickly.

The polynomial 5x-3 contains an x term and the constant term -3, but no x^2 term. It can be written as 0x^2+5x-3, so the coefficient of x^2 is 0. Note that 5 is the coefficient of x, not of x^2. Exam tip: If a term of a particular power is absent, its coefficient is 0.

In 14x-5, the highest power of x is 1, so the leading term is 14x. The coefficient of this leading term is 14; therefore, the correct answer is 14. Here, -5 is the constant term, not the leading coefficient. Exam tip: identify the term with the highest power first, then write its numerical coefficient.

In \(4x-9\), the highest power of the variable \(x\) is 1. Hence, its degree is 1 and it is a linear polynomial. \(x^2-4\) is quadratic, \(x^3+2x\) is cubic, and \(15\) is a constant polynomial with degree 0. Exam tip: identify the degree of a polynomial by finding the highest exponent of its variable.

\(p(x)=x+16=1x+16\). The highest power of \(x\) is 1, so it is a linear polynomial and has the general form \(ax+b\), where \(a\ne 0\). The form \(ax^2+b\) has highest power 2, so it represents a quadratic polynomial. Exam tip: identify a polynomial’s degree by checking the highest exponent of the variable.

Since 0x equals 0,
(0x-8)=-8
. The expression
-8
has no variable term, so its degree is 0 and it is a non-zero constant polynomial. A linear polynomial must have its highest power of the variable equal to 1, which is not the case here. Exam tip: simplify terms with zero coefficients before finding the degree of a polynomial.

In a linear polynomial, the highest power of x must be 1, so the coefficient of x cannot be zero. For m=-5, m+5=0 and the expression becomes 0x+7=7, which is a constant polynomial, not a linear polynomial. For the other given values, the coefficient of x is non-zero. Exam tip: In a parameter-based linear polynomial, set the coefficient of x equal to zero to find when it stops being linear.

In \((k-6)x+2\), the coefficient of \(x\) is \(k-6\). For a polynomial to be linear, the coefficient of \(x\) must be non-zero; hence \(k-6\ne 0\), or \(k\ne 6\). At \(k=6\), the expression becomes \(2\), which is a constant polynomial, not a linear polynomial. Exam tip: for a parameter-based linear polynomial, check that the coefficient of the variable is not zero.

A zero of a polynomial is a value of x that makes the polynomial equal to 0. Setting 4x-12=0 gives 4x=12, so x=3. Substituting x=-3 gives 4(-3)-12=-24, so -3 is not a zero. Exam tip: the zero of a linear polynomial ax+b is -b/a, where a≠0.

A zero of a polynomial is a value that makes the polynomial equal to 0. Setting x - 14 = 0 gives x = 14. Therefore, 14 is the correct answer. Substituting -14 gives -14 - 14 = -28, so it is not a zero. Exam tip: To find a zero, set the polynomial equal to 0 and solve for x.

To find p(2), substitute 2 for x in the polynomial: p(2)=3(2)+4=6+4=10. Therefore, the correct answer is 10. The value 6 is only 3×2; the constant term 4 must also be added. Exam tip: To evaluate a polynomial, replace the variable with the given number and perform all operations in order.

Given \(p(x)=7x-9\). Substituting \(x=0\), we get \(p(0)=7\times 0-9=-9\). Therefore, \(-9\) is the correct answer. Option \(9\) misses the negative sign. Exam tip: when \(x=0\) is substituted in a polynomial, its value is the constant term.

To find a zero, set the polynomial equal to 0: \(15-3x=0\). Thus, \(3x=15\), so \(x=5\). On checking, \(15-3(5)=0\); therefore, 5 is the correct zero. Substituting \(-5\) gives \(30\), not 0. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\).

In the polynomial \(5s+12\), \(s\) is the only variable. Its highest power is 1, so the expression is a linear polynomial in \(s\). The variables \(x\), \(y\), and \(z\) do not occur in the expression. Exam tip: identify the letter whose power appears, explicitly or implicitly, in the expression.

A linear polynomial has degree 1. In \(ax+b\), \(a\) is the coefficient of \(x\), so \(a\neq0\) is required. If \(a=0\), the expression reduces to \(b\), a constant polynomial rather than a linear polynomial. The value of \(a\) does not have to be 1; it may be any non-zero number. Exam tip: Determine a polynomial’s degree from the highest power with a non-zero coefficient.