Linear polynomials

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.

The governing concept is the zero of a linear polynomial. A zero is the value of x for which the polynomial evaluates to 0. Set p(x) equal to zero: (2/5)x − 6 = 0. Adding 6 to both sides gives (2/5)x = 6. Multiply both sides by the reciprocal 5/2: x = 6 × 5/2 = 15. Substitution verifies the result because p(15) = (2/5)(15) − 6 = 6 − 6 = 0. Therefore, option C is correct. The other values do not make the expression zero and can arise from incorrectly handling the fractional coefficient, especially by multiplying or dividing by 5 without also treating the numerator correctly.

The governing concept is the zero, or root, of a linear polynomial. The zero is the value of x that makes the polynomial equal to 0. Set the expression equal to zero: (3/7)x − 9 = 0. Adding 9 to both sides gives (3/7)x = 9. To remove the coefficient 3/7, multiply both sides by its reciprocal, 7/3. Thus x = 9 × 7/3 = 3 × 7 = 21. Substitution confirms the answer: p(21) = (3/7)(21) − 9 = 9 − 9 = 0. Hence option B is correct. Option D may arise if a solver multiplies by 7 but forgets to divide by 3; options A and C do not make the original polynomial zero.

The governing idea is the zero of a polynomial: it is the input value that makes the polynomial equal to zero. Set p(x) equal to zero and solve: (4/9)x − 8 = 0. Add 8 to both sides, obtaining (4/9)x = 8. Multiply both sides by the reciprocal of 4/9, namely 9/4: x = 8 × 9/4 = 2 × 9 = 18. Verification is important: p(18) = (4/9)(18) − 8 = 8 − 8 = 0. Therefore option A is correct. Values such as 36 or 72 can result from mishandling the fractional coefficient, while 27 does not satisfy the original equation. The substitution check confirms the unique zero because the coefficient of x is nonzero.

Answer: 11x−29, so option A is correct. The zero of a polynomial is the value of x that makes its value equal to zero. For option A, set 11x−29=0. Adding 29 gives 11x=29, and dividing by 11 gives x=29/11, exactly the required zero. Option B gives 29x−11=0, so x=11/29. Option C gives 11x+29=0, so x=−29/11. Option D gives x−11/29=0, so x=11/29. Therefore only A has the stated zero. All four expressions are linear because x has first power and a nonzero coefficient. Common confusion: the zero is found by solving the expression equal to zero, not by simply reading the constant term.