Linear polynomials

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.

Option A is x - 5, which can be written as 1x - 5. Therefore, the coefficient of x is 1. In option C, the minus sign before x means that its coefficient is -1, not 1. Exam tip: When no number is written before x, its coefficient is usually taken as 1.

Adding the given polynomials gives \(p(x)+q(x)=(x+2)+(2x+1)\). Combining like terms, \(x+2x=3x\) and \(2+1=3\), so the sum is \(3x+3\). In \(2x+3\), the terms \(x\) and \(2x\) have not been added correctly. Exam tip: While adding polynomials, combine variable terms and constant terms separately.

\(p(x)-q(x)=(5x+4)-(2x+1)\). Removing the brackets gives \(5x+4-2x-1\). Combining like terms, \(5x-2x=3x\) and \(4-1=3\), so the result is \(3x+3\). \(3x+5\) results from an incorrect subtraction of the constant terms. Exam tip: when subtracting a polynomial, change the sign of every term in the second polynomial.

Using the distributive property, \(2(3x-1)=2\times 3x-2\times 1=6x-2\). Therefore, \(6x-2\) is correct. In \(6x+2\), the negative sign of the constant term has been changed incorrectly. Exam tip: multiply the number outside the bracket by every term inside it.

Use the distributive property: multiply \(-3\) by each term inside the bracket. Thus, \((-3)\times x=-3x\) and \((-3)\times 4=-12\), so \(-3(x+4)=-3x-12\). In option B, the sign of the constant term is incorrect; multiplying \(-3\) by \(4\) gives \(-12\), not \(+12\). Exam tip: When multiplying by a negative number, check the sign of every term carefully.

\(4x\) is a linear polynomial with constant term \(0\). Its graph, \(y=4x\), passes through the origin \((0,0)\) because substituting \(x=0\) gives \(y=0\). Although \(4x+1\) is also linear, its constant term is \(1\), so it does not pass through the origin. Exam tip: For \(ax+b\), the graph passes through the origin only when \(b=0\).

The polynomial 11x has only a term containing the variable x; it has no term independent of x. Therefore, its constant term is 0. Here, 11 is the coefficient of x, not the constant term. Exam tip: The term with no variable is the constant term.

Substitute 4 for x: p(4)=-2(4)+8=-8+8=0. Therefore, the correct answer is 0. Option 4 is the input value substituted into the polynomial, not the value of p(4). Exam tip: To evaluate a polynomial, substitute the given number for x and carefully follow the order of multiplication and addition/subtraction.

To find a zero, equate the polynomial to 0: \(7-7x=0\). Thus, \(7x=7\), so \(x=1\). Therefore, 1 is the zero of this polynomial. Although 0 is a possible-looking option, substituting \(x=0\) gives 7, not 0. Exam tip: Verify a zero by substituting it into the polynomial and checking whether the result is 0.