Linear polynomials

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.

In \(5y-2\), the highest power of the variable \(y\) is 1, and the coefficient of \(y\) is non-zero. Hence, it is a linear polynomial in \(y\). \(9\) is a constant polynomial of degree 0, while \(y^2+3\) and \(y^3-y\) have degrees 2 and 3 respectively. Exam tip: A linear polynomial must have its highest variable power exactly 1.

In the polynomial 13 - 2z, the value of z can change, so z is the variable. Here, 13 is the constant term and -2 is the coefficient of z; neither is a variable. Exam tip: The letter whose value can vary is called the variable.

Because \(0x^2=0\), this term does not contribute to the degree of the polynomial. The expression becomes \(7-3x\), whose highest power of \(x\) is 1; therefore, it is a linear polynomial. Merely writing an \(x^2\) term does not make a polynomial quadratic—the coefficient must be non-zero. Exam tip: remove terms with zero coefficients before determining the degree.

A linear polynomial has the form ax + b, where b is the constant term. In 2x - 3, the term without x is -3, so its constant term is -3. In 2x + 3, the constant term is +3, not -3. Exam tip: To identify the constant term, look for the term that has no variable.

A zero of a polynomial is a value that makes the polynomial equal to 0. Substituting \(x=3\) in \(x-3\) gives \(3-3=0\), so its zero is 3. In contrast, \(x+3\) has zero \(-3\), not 3. Exam tip: Substitute the given value in the polynomial and check whether the result is 0.

To find the zero of \(x+4\), set \(x+4=0\). This gives \(x=-4\), so \(x+4\) is the correct option. The zero of \(x-4\) is \(4\), not \(-4\). Exam tip: Set a polynomial equal to 0 and solve for \(x\) to find its zero.

\(2x\) and \(3x\) are like terms, so their coefficients are added: \(2x+3x=(2+3)x=5x\). The constant term \(-1\) remains unchanged; hence the simplified polynomial is \(5x-1\). \(6x-1\) would result only if the coefficients added to 6. Exam tip: combine only terms with the same variable and the same exponent.

Here, 9x and -4x are like terms, so their coefficients are combined: 9x-4x=(9-4)x=5x. The constant term 6 remains unchanged; hence the simplified expression is 5x+6. In 5x-6, the sign of the constant term has been incorrectly changed. Exam tip: Combine only terms with the same variable and exponent.

In \(6x\), the highest power of \(x\) is \(1\), so it is a linear polynomial. It has no constant term and can be written as \(6x+0\). Although \(6x+1\) is also linear, its constant term is \(1\). Exam tip: a linear polynomial has the form \(ax+b\), where \(a\ne0\), and \(b\) is the constant term.

Substitute 2 for x in the polynomial: \(p(2)=3\times2+2=6+2=8\). Therefore, the correct answer is 8. The value 6 represents only \(3\times2\); the constant term 2 must also be added. Exam tip: To find the value of a polynomial, replace every x with the given number and then simplify carefully.

Substitute 3 for x: p(3)=4(3)-1=12-1=11. Therefore, 11 is the correct option. The closest distractor, 12, results from calculating 4×3 but forgetting to subtract 1. Exam tip: To find the value of a polynomial, substitute the given number for the variable and follow the correct order of operations.

A linear polynomial has degree 1 and is of the form ax+b, where a ≠ 0. In -3x+4, the highest power of x is 1 and its leading coefficient is -3, which is negative. x^2-3 is quadratic, while 7 is a constant polynomial. Exam tip: To find the leading coefficient, look at the coefficient of the term with the highest power.

For a linear polynomial \(p(x)=ax+b\), substituting \(x=0\) gives \(p(0)=b\), the constant term. In option A, \(p(x)=5x+2\), so \(p(0)=5(0)+2=2\). Option C has constant term 5, so it does not satisfy the condition. Exam tip: To find \(p(0)\), put 0 in place of \(x\).

In the linear polynomial ax+b, ax is the term containing x, so it is the x-term. The term b is the constant term because it does not contain x. Exam tip: identify a variable term by writing the complete term along with the coefficient of x.

The direct answer is option C: no, it is the zero polynomial. Since 0·x=0 and the constant term is also 0, p(x)=0·x+0 is simply p(x)=0. A linear polynomial must have degree exactly 1, usually written as ax+b with a≠0. Here the coefficient of x is 0, so there is no actual x term. The zero polynomial is special: its degree is not defined in the usual school convention, so it cannot be called linear. Option A is wrong because merely writing x does not make a term non-zero. Option B is wrong because the degree is not 1; the whole polynomial is zero. Option C is correct because the expression is the zero polynomial. Option D is wrong because a zero constant does not create a linear term. Do not confuse the form ax+b with the condition a≠0. Exam cue: if every coefficient is zero, identify the zero polynomial first; never assign it degree 1.

Given \(p(x)=2x-5\), substitute 5 for \(x\): \(p(5)=2\times5-5=10-5=5\). Therefore, the correct answer is 5. The value 10 is only \(2\times5\); the subtraction of 5 has not yet been done. Exam tip: To evaluate a polynomial, replace every \(x\) with the given number and then simplify.