Linear polynomials

\(6x\) 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.

In (-5x+12), the coefficient of (x) is (-5) and the constant term is (12). Identify both parts separately in a linear polynomial.

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.

Since \(0\cdot x=0\), we get \(p(x)=14\). This is a non-zero constant polynomial, and every non-zero constant polynomial has degree \(0\). Degree \(1\) would apply only if the coefficient of \(x\) were non-zero. Exam tip: simplify a polynomial before identifying its degree.

A zero of a polynomial is a value that makes the polynomial equal to 0. Setting 2x+10=0 gives 2x=-10, so x=-5. Therefore, -5 is the correct answer. If x=5, then 2(5)+10=20, not 0. Exam tip: the zero of a linear polynomial ax+b is -b/a.

On expanding the bracket, \(3(x+2)=3x+6\). Therefore, \(3x+6-2x=(3x-2x)+6=x+6\). Hence, \(x+6\) is the correct polynomial. \(3x+6\) is a close distractor because it ignores the subtraction of \(2x\). Exam tip: expand brackets first, then combine like terms.

For a polynomial to be linear, the coefficient of \(x\) must be non-zero. Here, the coefficient of \(x\) is \(a-3\). On putting \(a=3\), we get \(a-3=0\), so the expression becomes \(8\), a constant polynomial rather than a linear polynomial. For example, when \(a=0\), the coefficient is \(-3\), so it is still linear. Exam tip: in a linear polynomial, the highest power of the variable is 1 and its coefficient must not be zero.