Linear polynomials

\(p(x)-q(x)=(8x+3)-(3x+2)\). Removing the brackets gives \(8x+3-3x-2\), so \((8x-3x)+(3-2)=5x+1\). Therefore, the correct answer is \(5x+1\). \(11x+5\) would be obtained by adding the two polynomials, not subtracting them. Exam tip: while subtracting polynomials, change the sign of every term in the second polynomial.

Use the distributive property: \(3(2x-5)=3\times 2x+3\times(-5)=6x-15\). Therefore, the correct linear polynomial is \(6x-15\). In \(6x+15\), the sign is incorrect because \(3\times(-5)=-15\). Exam tip: multiply the number outside the bracket by every term inside it.

Using the distributive property, multiply

dot(-2) by each term inside the bracket:



dot(-2)

times x=-2x and

dot(-2)

times(-6)=+12. Therefore,



dot(-2(x-6))=-2x+12. In option C, the sign of the constant term is incorrect. Exam tip: the product of two negative numbers is positive.

A linear polynomial of the form \(ax\) has a non-zero coefficient of \(x\) and no constant term. In \(7x\), \(a=7\) and the constant term is \(0\), so it is of the form \(ax\). Although \(7x+1\) is also linear, it has the constant term \(1\). Exam tip: In \(ax+b\), the polynomial is of the form \(ax\) only when \(b=0\).

The polynomial \(15x\) has only a variable term and no term without \(x\). Hence, its constant term is \(0\). Here, \(15\) is the coefficient of \(x\), not the constant term. Exam tip: The term containing no variable is the constant term.

Substituting 4 for x gives p(4)=-3(4)+12=-12+12=0. Hence, the correct answer is 0. The number 12 is only the constant term; it is not the value of p(4). Exam tip: To find the value of a polynomial, substitute the given value for every occurrence of x and calculate carefully.

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

In the polynomial \(21-4t\), \(t\) is the letter whose value can change, so it is the variable. Here, \(21\) is the constant term and \(-4\) is the coefficient of \(t\), not a variable. Exam tip: Identify the letter whose value may vary in a polynomial.

A linear polynomial has degree 1. In (p(x)=ax+b), (a) is the coefficient of (x), so (a) must be non-zero for the polynomial to retain degree 1. If (a=0), the expression reduces to the constant term (b) and is not a degree-1 polynomial. Exam tip: the coefficient of the highest-degree term of a polynomial cannot be zero.

A linear polynomial has the form \(ax+b\), where \(b\) is the constant term. In \(5x+6\), the term without the variable \(x\) is \(6\), so its constant term is \(6\). In \(5x-6\), the constant term is \(-6\), not \(6\). Exam tip: To identify the constant term, look for the term containing no variable.

To find the zero of \(x-6\), set the polynomial equal to zero: \(x-6=0\). Hence, \(x=6\), so \(x-6\) is the correct option. The zero of \(x+6\) is \(-6\), not 6. Exam tip: To find the zero of a linear polynomial, equate it to 0 and solve for \(x\).

A zero of a polynomial is a value that makes the polynomial equal to 0. Substituting \(x=-7\) in \(x+7\) gives \(-7+7=0\), so its zero is \(-7\). In contrast, \(x-7\) has zero \(7\), so it is not correct. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\).

\(4x\) and \(5x\) are like terms, so their coefficients are added: \(4x+5x=9x\). The constant term \(-8\) remains unchanged. Therefore, the simplified polynomial is \(9x-8\). In \(9x+8\), the sign of the constant term has been changed incorrectly. Exam tip: Add or subtract only terms with the same variable and the same power.

13x and -6x are like terms, so combine their coefficients: 13x-6x=(13-6)x=7x. The constant term +2 remains unchanged. Therefore, the simplified form is 7x+2. Option 19x+2 incorrectly adds 13x and 6x without considering the minus sign. Exam tip: While combining like terms, always include the signs of the coefficients.

In \(8x\), the highest power of \(x\) is \(1\), so it is a linear polynomial. It has no independent constant term, hence its constant term is \(0\). Although \(8x+1\) 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.

Given \(p(x)=4x+3\). To find \(p(3)\), substitute 3 for \(x\): \(p(3)=4\times3+3=12+3=15\). Hence, the correct answer is 15. The value 12 is only \(4\times3\); the constant term 3 must also be added. Exam tip: While evaluating a polynomial, substitute the given value in every term carefully.

Substituting 4 for x gives p(4)=5×4−2=20−2=18. Therefore, 18 is correct. The value 20 is only 5×4; the constant term 2 still has to be subtracted. Exam tip: To find the value of a polynomial, substitute the given number for the variable and follow the correct order of operations.

For a linear polynomial \(p(x)=ax+b\), substituting \(x=0\) gives \(p(0)=b\). Thus, \(p(0)\) is the constant term. In option A, \(p(x)=4x-3\), so \(p(0)=4(0)-3=-3\). In option C, \(-3\) is the coefficient of \(x\), while the constant term is \(4\), so it is not correct. Exam tip: To find \(p(0)\), directly substitute \(x=0\).

In the linear polynomial \(ax+b\), \(x\) is the variable. The term \(ax\) contains the variable, whereas \(b\) does not contain any variable. Therefore, \(b\) is the constant term. Here, \(a\) is a coefficient, not the constant term. Exam tip: The term with no variable is the constant term.

\(p(x)=0\cdot x+5=5\) because \(0\cdot x=0\). Thus, no \(x\)-term remains after simplification, and the polynomial has degree \(0\). A linear polynomial must have degree \(1\), so this is a constant polynomial. It is not the zero polynomial because its value is \(5\). Exam tip: remove terms with zero coefficients before finding a polynomial’s degree.

A linear polynomial has highest power 1 and is of the form \(ax+b\), where \(a\ne0\). Thus, \(7x-4\) is linear. In contrast, \(x^2+7\) has degree 2. Exam tip: check both the highest exponent and whether a variable appears in a denominator.

In (-4x+9), the coefficient of (x) is (-4) and the constant term is (9). Identify the coefficient and constant separately.

To find the zero, set the polynomial equal to zero: \(8x-16=0\). Thus, \(8x=16\), so \(x=2\). On checking, \(8(2)-16=0\); hence, \(2\) is the correct zero. Substituting \(-2\) gives \(-32\), not zero. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-b/a\).

In a linear polynomial, the coefficient of x must be non-zero. On putting r=9, (r-9)x-3=(9-9)x-3=-3, which is a constant polynomial of degree 0, not a linear polynomial. For the other values, the coefficient of x is non-zero. Exam tip: In parameter-based linear polynomials, set the coefficient of x equal to zero to find when it ceases to be linear.

While adding polynomials, combine like terms. Here, \(3x+2x=5x\) and \(-7+5=-2\). Therefore, \(p(x)+q(x)=5x-2\), so option A is correct. In option C, the constant terms have been added incorrectly. Exam tip: add the \(x\)-terms and the constant terms separately.