Linear polynomials

A zero of a polynomial is the value of the variable that makes the polynomial equal to zero. For a linear polynomial, we find it by equating the expression to zero and solving for the variable. This is different from merely looking at the constant term; both terms must be considered together.

For option C, set \(7x+3=0\). Subtracting 3 gives \(7x=-3\), and dividing by 7 gives \(x=-\frac{3}{7}\). Therefore option C has the required zero. Option A would give \(x=\frac{3}{7}\), while option B gives \(-\frac{7}{3}\); hence neither has the stated value.

The governing concept is repeated function evaluation, also called composition of a function with itself. In p(p(3)), evaluate the inner function first. Using p(x)=2x−5, p(3)=2(3)−5=6−5=1. Now use this output as the input for the second application: p(p(3))=p(1)=2(1)−5=2−5=−3. Thus option A is correct. A common distractor is option C, which is the value of p(3) after only one application; it does not complete p(p(3)). The other values result from arithmetic errors or from substituting the original input again instead of the intermediate result. The order of evaluation is essential because the outer p acts on the value produced by the inner p.

The governing concept is determining the degree of the simplified difference of two polynomials. A linear polynomial must have degree exactly 1, meaning that after subtraction a nonzero x-term must remain. In option B, p(x) − q(x) = (2x + 3) − (x − 5) = 2x + 3 − x + 5 = x + 8, which has degree 1. Therefore, option B is correct. In option A, the x-terms cancel and the result is 3, a constant. In option C, the two identical polynomials subtract to give 0. In option D, 5 − 2 = 3, also a constant. Thus the other options do not produce a linear polynomial after simplification.

A polynomial’s zero is found by setting the polynomial equal to zero and solving for x. For option D, 9x + 7 = 0. Subtracting 7 from both sides gives 9x = −7, and dividing by 9 gives x = −7/9. Hence option D is correct. Checking the distractors clarifies the possible errors: 9x − 7 = 0 gives x = 7/9, so its sign is positive. Both 7x + 9 = 0 and x + 9/7 = 0 give x = −9/7, which reverses the required fraction. Therefore only 9x + 7 has the stated zero. The coefficient and constant must be used in the equation exactly as written; changing their order changes the zero.

The governing concept is evaluating a linear polynomial at given inputs and solving the resulting simultaneous equations. From p(−1) = 4, substitute x = −1: −m + n = 4. From p(3) = 20, substitute x = 3: 3m + n = 20. Subtract the first equation from the second: (3m + n) − (−m + n) = 20 − 4, so 4m = 16 and m = 4. Substituting into −m + n = 4 gives −4 + n = 4, hence n = 8. Therefore m + n = 4 + 8 = 12, so option C is correct. Checking gives p(−1) = −4 + 8 = 4 and p(3) = 12 + 8 = 20. Other options arise from sign or elimination errors.

A polynomial of the form ax+b is linear when the coefficient a of x is nonzero. In the given expression, that coefficient is r²−5r+6. Therefore, the expression will stop being linear precisely when r²−5r+6=0. Factorising gives r²−5r+6=(r−2)(r−3). Hence (r−2)(r−3)=0, so r=2 or r=3. For either value, the x-term disappears and the expression becomes the constant polynomial 4, whose degree is 0 rather than 1. Thus option A is correct. Option B does not satisfy the coefficient equation, option C gives only one valid value, and option D is false because two values exist.

For an expression ax+b to be a linear polynomial, the coefficient a of x must be nonzero. Here the coefficient is r²−9r+20. The expression fails to be linear when this coefficient equals zero. Factorising gives r²−9r+20=(r−4)(r−5). Thus (r−4)(r−5)=0, so r=4 or r=5. At either value, the x-term vanishes and the expression becomes the constant polynomial 6, which has degree 0 rather than degree 1. Therefore option A is correct. Option B contains values that do not solve the coefficient equation. Option C gives neither of the roots, and option D is false because two valid values exist.

To decide whether the product is linear, find the highest power of x after multiplication. Expanding gives (6x−1)(x+8)=6x²+48x−x−8=6x²+47x−8. The leading term is 6x², and its coefficient is nonzero. Therefore the resulting polynomial has degree 2. A linear polynomial must have degree exactly 1, so this product is quadratic and is not linear. Option C gives the correct reason. Option A incorrectly claims degree 1. Options B and D describe a constant polynomial or the zero polynomial, but the expanded expression has a nonzero x² term and variable terms, so neither description can apply.

Use the definition p(x)=mx+n and substitute the two given input-output pairs. From p(−3)=4, we obtain −3m+n=4. From p(2)=29, we obtain 2m+n=29. Subtracting the first equation from the second eliminates n: (2m+n)−(−3m+n)=29−4, so 5m=25 and m=5. Substitute m=5 into −3m+n=4: −15+n=4, hence n=19. Therefore m+n=5+19=24, so option B is correct. The other choices do not satisfy both original conditions simultaneously; for instance, the value 29 is one given output, not the requested sum of the coefficients.