Linear polynomials

Replace the variable \(x\) in the polynomial by the complete expression \(2x\): \(p(2x)=5-3(2x)=5-6x\). Therefore, \(5-6x\) is correct. \(10-6x\) would result from incorrectly multiplying the constant term 5 by 2 as well. Exam tip: To find \(p(kx)\), substitute \(kx\) only for every \(x\).

\(\frac{3}{x}\) can be written as \(3x^{-1}\). A polynomial permits only non-negative integer powers such as 0, 1, and 2, so this expression is not a polynomial. Exam tip: check every exponent before finding the degree.

Given \(p(2)=p(-2)\), we get \(2k+4=-2k+4\). Thus \(4k=0\), so \(k=0\). Then \(p(x)=4\), which is a constant polynomial, not a linear polynomial. Hence, under the given condition, there is no value of \(k\) for which \(p(x)\) remains linear. Exam tip: \(ax+b\) is linear only when \(a\ne0\).

For \(a\ne0\), \(ax+b=0\) gives one value, \(x=-b/a\). Thus a non-constant linear polynomial has exactly one zero. A constant polynomial is excluded. Exam tip: first verify \(a\ne0\).

First simplify the expression: \(q(x)-2p(x)=(5x-1)-2(2x+3)=5x-1-4x-6=x-7\). The highest power of \(x\) in \(x-7\) is 1, with a non-zero coefficient, so its degree is 1. Option 0 would apply only to a constant polynomial; here the \(x\)-term does not cancel. Exam tip: simplify and combine like terms before deciding the degree of a polynomial.

A zero of a polynomial is a value of x that makes the polynomial equal to 0. For \(5x-11=0\), we get \(5x=11\), so \(x=\frac{11}{5}\). Hence, \(5x-11\) has zero \(\frac{11}{5}\). The close distractor \(11x-5\) has zero \(\frac{5}{11}\), which is different. Exam tip: To find the zero of \(ax+b\), solve \(ax+b=0\).

First find p(x+1): p(x+1)=3(x+1)-8=3x-5. Therefore, p(x+1)-p(x)=(3x-5)-(3x-8)=3. Hence, the correct answer is 3. The option 3x is incorrect because the x-terms cancel on subtraction. Exam tip: for a linear polynomial ax+b, p(x+1)-p(x) is always a.

The zero is (-\frac{6}{\lambda}). From (-\frac{6}{\lambda}>-2) and (\lambda>0), we get (\lambda>3).

Since \(a\ne0\), the graph is a non-horizontal straight line with zero \(x=-\frac{b}{a}\). Hence it meets the x-axis exactly once. Exam tip: put \(p(x)=0\) to locate the x-intercept.

Given \(p(x)=x+r\) and \(q(x)=x-r\), \(p(x)+q(x)=(x+r)+(x-r)=x+x+r-r=2x\). The terms \(+r\) and \(-r\) cancel each other. \(2r\) would result only if the \(x\)-terms cancelled, which they do not. Exam tip: While adding polynomials, group like terms together first.

On subtraction, (p(x)-q(x))=(x+r)-(x-r)=2r. Since r is a constant, 2r has no term involving x; therefore, it is a constant polynomial in x. A linear polynomial must contain an x-term, so option A is not correct. Exam tip: while subtracting polynomials, change the sign of every term inside the second bracket. If r=0 specifically, the result is the zero polynomial; normally, r is taken as a non-zero constant here.

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.

For a composite expression, evaluate the inner function first. p(2)=3×2+1=7. Then p(7)=3×7+1=22, so p(p(2))=22. Option 21 results from taking only 3×7 and forgetting to add the constant term 1. Exam tip: in p(p(a)), first find p(a), then substitute that result into p again.

If \(x=1\) is a zero of \(((n-1)x+n+2)\), substituting \(x=1\) must make the expression equal to 0. Thus, \((n-1)(1)+n+2=0\), giving \(2n+1=0\). Hence, \(n=-\frac{1}{2}\). For \(n=1\), the expression becomes \(3\), so it cannot have 1 as a zero. Exam tip: substitute the given zero in the polynomial and set the result equal to 0.

Given \(p(x)=\frac{x}{3}+a\), substitute \(x=6\): \(p(6)=\frac{6}{3}+a=2+a\). Since \(p(6)=5\), we get \(2+a=5\), so \(a=3\). Option 2 is only the value of \(\frac{6}{3}\), not the value of a. Exam tip: when a value of a polynomial is given, substitute the input first and then solve the resulting equation.

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.

First add the polynomials: p(x)+q(x)=2x+a+x-3=3x+a-3. Since 1 is its zero, substituting x=1 gives 3(1)+a-3=0. Therefore, a=0. If a=-1, the value of the polynomial at x=1 would be -1, not zero. Exam tip: when a zero is given, substitute that value of x in the polynomial and set the result equal to 0.

If \(-4\) is a zero of \(p(x)\), then \(p(-4)=0\). Thus, \((c+1)(-4)+2c=0\), or \(-4c-4+2c=0\). This gives \(-2c-4=0\), so \(c=-2\). For \(c=-1\), the polynomial becomes the constant \(-2\), which has no zero. Exam tip: for a given zero \(a\), always use \(p(a)=0\).

Since \(-4\) is a zero of \(p(x)\), we must have \(p(-4)=0\). Thus, \(-4(c+1)+2c=0\Rightarrow -2c-4=0\Rightarrow c=-2\). Therefore, the correct option is \(c=-2\). If \(c=-1\), the constant term becomes \(-2\), so \(-4\) is not a zero. Exam tip: For a given zero \(a\), always use \(p(a)=0\).

Given \(p(x)=rx+5\) and \(p(2)=15\), we get \(2r+5=15\). Hence \(2r=10\), so \(r=5\). Now \(p(-1)=5(-1)+5=-5+5=0\). Therefore, the correct answer is 0. The value \(5\) is only the coefficient \(r\), not the value of \(p(-1)\). Exam tip: first use the given function value to find the unknown coefficient, then substitute the required value of \(x\).