Linear polynomials

The governing concept is classification by the degree of a polynomial. The degree of a non-zero polynomial is the greatest exponent of its variable having a non-zero coefficient. A polynomial of degree 0 is called a constant polynomial, one of degree 1 is called a linear polynomial, one of degree 2 is quadratic, and one of degree 3 is cubic. Since the highest power stated in the question is 1, the polynomial has degree 1 and is therefore called a linear polynomial. Hence option B is correct. Option A describes degree 0, option C describes degree 2, and option D describes degree 3. The classification depends on the greatest exponent, not on the number of terms.

A linear polynomial has highest exponent 1 and is of the form \(ax+b\), where \(a\neq 0\). In \(x^2+1\), the highest exponent of \(x\) is 2, so it is a quadratic polynomial, not a linear polynomial. Although \(\frac{x}{4}-6\) has a fractional coefficient, its highest exponent of \(x\) is 1, so it is linear. Exam tip: To find the degree of a polynomial, identify the greatest exponent of its variable.

In \(4x-9\), the highest power of \(x\) is \(1\), so its degree is 1. A polynomial of degree 1 is called a linear polynomial. A quadratic polynomial would have highest power 2. Exam tip: identify the type of a polynomial by checking the highest power of its variable.

The governing definition is that a linear polynomial in one variable has the form ax + b, where a is nonzero and b is a constant. The highest power of x that actually occurs is therefore x^1, so its degree is 1. For example, 4x − 7 is linear because the variable appears to the first power. A constant such as 5 has degree 0, while x^2 + 1 is quadratic and has degree 2. The number of terms does not determine degree; the greatest nonzero exponent does. Thus option C is correct. Option A describes a nonzero constant polynomial, and options B and D describe polynomials of higher degrees rather than a linear polynomial.

In \(7y+2\), the term \(7y\) can be written as \(7y^1\). Therefore, the highest power of \(y\) is \(1\), so the expression is a linear polynomial. The \(2\) is a constant term and does not affect the power of \(y\). Exam tip: Find the highest exponent of the variable in the terms containing that variable.

In \(2x^2+3\), the highest power of \(x\) is 2, so its degree is 2. Therefore, it is a quadratic polynomial, not a linear polynomial. A linear polynomial has degree 1, as in \(5x-1\), \(x+8\), and \(9-4x\). Exam tip: Identify the degree by checking the highest exponent of the variable.

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

In \(F(x)=18x+50\), the highest power of \(x\) is \(1\), so it is a linear polynomial. The constant term 50 does not prevent it from being a polynomial. Exam tip: check the highest exponent.

In the polynomial (12-5x), the term containing x is -5x. The number multiplying x is -5, so the coefficient of x is -5. The number 5 alone is not the coefficient because the negative sign is part of it. Exam tip: Always include the sign while identifying a coefficient.

A constant term is a term that has no variable. In \(x+4\), \(x\) contains the variable, whereas \(4\) has no variable. Therefore, the constant term is \(4\). Although \(0\) can be a constant, it is not the constant term in this polynomial. Exam tip: identify the term without \(x\), \(y\), or any other variable.

The polynomial \(2x+7\) has an \(x\)-term and the constant term \(7\), but no \(x^2\)-term. It can be written as \(0x^2+2x+7\), so the coefficient of \(x^2\) is \(0\). The number \(2\) 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 the polynomial 9x+1, the highest power of x is 1, so the leading term is 9x. Its numerical coefficient is 9; therefore, the leading coefficient is 9. The number 1 is the constant term, not the leading coefficient. Exam tip: identify the term with the highest power of the variable, then take its numerical coefficient.

A linear polynomial has the highest power of its variable equal to 1. In \(3x+10\), the highest power of \(x\) is 1, so it is a linear polynomial. \(x^2-2\) is quadratic and \(x^3+1\) is cubic, while \(7\) is a constant polynomial. Exam tip: To identify the degree of a polynomial, look for the greatest exponent of the variable.

Here, \(p(x)=x-13=1x+(-13)\). The highest power of \(x\) is 1, so it is a linear polynomial of the form \(ax+b\), where \(a=1\) and \(b=-13\). The form \(ax^2+b\) represents a quadratic polynomial because it has \(x\) raised to the power 2. Exam tip: identify a polynomial’s degree by looking at the highest power of the variable.

Since 0x equals 0, (0x+6) simplifies to 6. The polynomial 6 has no x-term, so its degree is 0 and it is a constant polynomial. A linear polynomial has degree 1 and must have a non-zero coefficient of x. Exam tip: simplify the expression first, then identify the polynomial type from its highest power.

For \((m-2)x+5\) to be linear, the coefficient of \(x\), namely \(m-2\), must be non-zero. On putting \(m=2\), we get \(m-2=0\), so the expression becomes \(5\), a constant polynomial rather than a linear polynomial. For the other given values, the coefficient of \(x\) is non-zero. Exam tip: a linear polynomial must have highest power of the variable equal to 1.

For a polynomial to be linear, the coefficient of x must be non-zero. Here, the coefficient of x is k+4. Therefore, k+4\ne0, so k\ne-4. At k=-4, the expression becomes -7, which is a constant polynomial, not a linear polynomial. Exam tip: In a polynomial with a parameter, first check whether the coefficient of the highest-degree term is zero.

A zero of a polynomial is a value of x that makes the polynomial equal to 0. Setting 3x-6=0 gives 3x=6, so x=2. Substituting x=-2 gives 3(-2)-6=-12, so -2 is not a zero. Exam tip: The zero of a linear polynomial ax+b is -b/a.

A zero of a polynomial is a value that makes the polynomial equal to 0. Here, x+9=0 gives x=-9. Checking: (-9)+9=0, so -9 is the correct zero. Substituting 9 gives 9+9=18, so it is not a zero. Exam tip: for ax+b, the zero is x=-b/a.

Given \(p(x)=2x+3\), substitute 1 for \(x\): \(p(1)=2\times1+3=5\). Therefore, the correct answer is 5. The value 4 would result from incorrectly forgetting to add the constant term 3. Exam tip: To evaluate a polynomial, replace the variable with the given number and perform each operation carefully.

Given \(p(x)=5x-4\), substitute \(x=0\): \(p(0)=5\times 0-4=-4\). Therefore, \(-4\) is correct. Option 0 is incorrect because only the term \(5x\) becomes zero; the constant term \(-4\) remains. Exam tip: when finding \(p(0)\), the constant term of a polynomial is left.

To find the zero, set the polynomial equal to 0: \(10-2x=0\). Thus, \(2x=10\), so \(x=5\). On checking, \(10-2(5)=0\); therefore, 5 is the correct zero. Substituting \(-5\) gives \(20\), not 0. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\), where \(a\ne0\).

A linear polynomial has highest power 1 of the variable. In \(4x+2\), the highest power of \(x\) is 1, so it is linear. \(4x^2+2\) is quadratic and \(4x^3+2\) is cubic, while \(4\) is a constant polynomial. Exam tip: identify the degree by looking for the highest exponent of the variable.

In \(3t-8\), the only variable present is \(t\). Its highest power is 1, so it is a linear polynomial in \(t\). The variables \(x\), \(y\), and \(z\) do not occur in the expression, so the polynomial is not described as linear in them. Exam tip: a linear polynomial has highest degree 1 in its variable.

The polynomial \(ax+b\) is linear only if the coefficient of \(x\), namely \(a\), is non-zero. Then the highest power of \(x\) is 1, so the polynomial has degree 1. If \(a=0\), the expression becomes \(b\), which is a constant polynomial, not a linear polynomial. Exam tip: while finding a polynomial’s degree, ensure that the coefficient of the highest-power term is non-zero.