Linear polynomials

Using the distributive property, multiply \(-4\) by each term inside the bracket: \(-4\times x=-4x\) and \(-4\times 3=-12\). Therefore, \(-4(x+3)=-4x-12\). In \(-4x+12\), the sign of the constant term is incorrect. Exam tip: When multiplying by a negative number, check the sign of every term carefully.

The polynomial 23x contains only a term with x; it has no term without the variable x. Hence, its constant term is 0. Here, 23 is the coefficient of x, not the constant term. Exam tip: identify the term that has no variable to find the constant term.

Given \(p(x)=-5x+25\). Substituting \(x=5\), \(p(5)=-5\times5+25=-25+25=0\). Therefore, the correct answer is 0. Option 5 is not the value of the polynomial; it is the \(x\)-value at which the polynomial becomes zero. Exam tip: To find a polynomial’s value, substitute the given \(x\)-value carefully and keep track of signs.

To find a zero, equate the polynomial to 0: \(20-5x=0\). Thus, \(5x=20\), so \(x=4\). Checking, \(20-5(4)=0\); hence 4 is the zero. If 5 is substituted, the value is \(-5\), not 0. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\).

In the polynomial \(30-11v\), \(v\) is the letter whose value can change, so it is the variable. \(30\) is the constant term, while \(-11\) is the coefficient of \(v\); neither is a variable. Exam tip: Identify the letter whose value can vary.

The governing concept is the degree of a polynomial. In p(x)=ax+b, the expression is linear only if the highest power of x that remains has exponent 1. Therefore the coefficient a must be nonzero. If a=0, the x-term disappears and p(x)=b becomes a constant polynomial of degree 0, not a linear polynomial. The coefficient does not need to equal 1: values such as a=2, a=−3, or a=1/2 all give linear polynomials when they are nonzero. There is also no requirement that a=b, because the coefficient of x and the constant term are independent parameters. Hence option B is the only correct choice; the other options confuse a necessary nonzero condition with unnecessary special values.

In a linear polynomial, the term without a variable is called the constant term. In \(3x-8\), the variable-free term is \(-8\), so option A is correct. In \(-8x+3\), \(-8\) is the coefficient of \(x\), while the constant term is \(3\). Exam tip: identify the term that has no variable to find the constant term.

To find the zero of the linear polynomial \(x-8\), set \(x-8=0\). This gives \(x=8\), so its zero is 8. In contrast, \(x+8\) has zero \(-8\). Exam tip: Set a polynomial equal to 0 and solve for \(x\) to find its zero.

A zero of a polynomial is a value that makes the polynomial equal to 0. Substituting \(x=-10\) in \(x+10\) gives \(-10+10=0\), so its zero is \(-10\). The close distractor \(x-10\) has zero \(10\), not \(-10\). Exam tip: the zero of \(x+a\) is \(-a\), while the zero of \(x-a\) is \(a\).

\(6x\) and \(7x\) are like terms because both have the variable \(x\) with exponent 1. Adding their coefficients gives \(6+7=13\), so the expression becomes \(13x-3\). \(42x-3\) is incorrect because the coefficients are added, not multiplied. Exam tip: Combine only like terms.

Combine the like terms \(17x\) and \(-8x\): \(17x-8x=9x\). The constant term \(+5\) remains unchanged, so the simplified expression is \(9x+5\). In \(9x-5\), the sign of the constant term has been changed incorrectly. Exam tip: Add or subtract only terms having the same variable and exponent.

To find the value of the function, substitute 4 for x: \(p(4)=5\times4+6=20+6=26\). Therefore, the correct answer is 26. The value 20 is only \(5\times4\); the constant term 6 must also be added. Exam tip: While evaluating a polynomial, substitute the given value carefully in every term.

Given \(p(x)=6x-7\), substitute \(x=5\): \(p(5)=6\times5-7=30-7=23\). Option 30 is only the product \(6\times5\); the subtraction of 7 has not been done. Exam tip: To find the value of a polynomial, substitute the given value for every \(x\) and follow the correct order of operations.

For a linear polynomial \(p(x)=ax+b\), substituting \(x=0\) gives \(p(0)=b\), the constant term. In option A, \(p(x)=7x+4\), so \(p(0)=4\). In option C, 7 is the constant term; 4 is the coefficient of \(x\), so it is not correct. Exam tip: To find \(p(0)\), put \(x=0\) and identify the constant term.

Since \(0\cdot x=0\), we get \(p(x)=0-6=-6\). The coefficient of \(x\) is zero, so there is no non-zero variable term. Its degree is \(0\), making it a constant polynomial. A linear polynomial has the form \(ax+b\), where \(a\ne0\), so this is not linear. It is also not a zero polynomial because its value is \(-6\), not \(0\). Exam tip: simplify a polynomial before deciding its degree.

Given \(p(x)=8x-9\), substitute 2 for \(x\): \(p(2)=8\times2-9=16-9=7\). Hence, 7 is the correct answer. The value 16 is only \(8\times2\); subtracting 9 is still necessary. Exam tip: To evaluate a polynomial, replace every \(x\) with the given value and then simplify carefully.

The direct answer is option B, \(-5x+12\). In a linear polynomial written as \(ax+b\), \(a\) is the coefficient of \(x\), and \(b\) is the constant term because it has no variable. We need coefficient \(-5\) and constant \(12\). Option B has \(-5x+12\), so the coefficient is \(-5\) and the constant is \(12\). Option A, \(5x+12\), has the correct constant but coefficient \(+5\), not \(-5\). Option C, \(-5x-12\), has the correct coefficient but constant \(-12\), not 12. Option D, \(12x-5\), reverses the requested roles: its coefficient is 12 and its constant is -5. The minus sign belongs to the number immediately attached to the term. Memory cue: in \(ax+b\), the number beside \(x\) is the coefficient and the number alone is the constant.

To find the zero, equate the polynomial to 0: \(9x-27=0\). Thus, \(9x=27\), so \(x=3\). Hence, substituting 3 makes the polynomial equal to 0. Substituting \(-3\) gives \(9(-3)-27=-54\), so it is not a zero. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\), where \(a\ne0\).

For \((r+11)x+4\) to be a linear polynomial, the coefficient of \(x\), namely \(r+11\), must be non-zero. When \(r=-11\), \(r+11=0\), so the polynomial becomes \(4\), which has degree 0 and is not linear. For all the other given values, the coefficient of \(x\) is non-zero. Exam tip: In parameter-based linear-polynomial questions, set the coefficient of \(x\) equal to zero first.

While adding polynomials, add like terms. Here, \(4x+3x=7x\) and \(-9+2=-7\), so \(p(x)+q(x)=7x-7\). Option \(7x+11\) results from incorrectly adding the constant terms. Exam tip: add the \(x\)-terms and constant terms separately.

Multiplying 6 by each term inside the bracket gives 6(2x - 1) = 12x - 6. Then, subtracting 5x gives 12x - 6 - 5x = 7x - 6. Hence, the correct answer is 7x - 6. In 7x + 6, the sign of the constant term is incorrect. Exam tip: While removing brackets, carefully track multiplication and minus signs.

Substitute x=1 in the polynomial: p(1)=11-3(1)=11-3=8. Therefore, 8 is the correct answer. Option 11 is only the constant term; 3 must be subtracted after substituting x=1. Exam tip: To evaluate a polynomial, replace the variable with the given value and then simplify carefully.

A linear polynomial has degree 1 and can be written as ax+b, where a is nonzero. In 6x, the coefficient of x is 6, so the leading coefficient is 6, and there is no constant term, meaning b=0. Thus option A satisfies both conditions. x+6 has the wrong leading coefficient, 6 is constant, and 6x+1 has constant term 1.

\(q(x)-p(x)=(5x-3)-(2x+7)\). On removing the brackets, the sign of each term in \(2x+7\) changes: \(5x-3-2x-7=3x-10\). Therefore, the correct answer is \(3x-10\). The option \(3x+10\) results from a sign error while subtracting the constant terms. Exam tip: When subtracting polynomials, write the second polynomial in brackets and change the signs of all its terms.

For \((h-1)x+10\) to be linear, the coefficient of \(x\) must be non-zero. Thus, \(h-1\ne0\), so \(h\ne1\). If \(h=1\), the expression becomes \(10\), which is a constant polynomial, not a linear polynomial. Exam tip: In a linear polynomial, the highest power of the variable is 1 and its coefficient must be non-zero.