Linear polynomials

A linear polynomial has the form \(ax+b\), where \(a\neq0\) and powers of \(x\) are non-negative integers. Here \(\frac{4}{x}=4x^{-1}\), so it is not a polynomial. Exam tip: if a variable is in the denominator, rewrite it using a negative exponent first.

First simplify the expression: \(q(x)-2p(x)=(9x-3)-2(4x+1)=9x-3-8x-2=x-5\). The highest power of \(x\) is 1, so its degree is 1. Option 0 is incorrect because \(x-5\) is not a constant polynomial. Exam tip: After adding or subtracting polynomials, determine the degree from the highest non-zero power in the simplified result.

A zero of a polynomial is a value that makes the polynomial equal to 0. For \(8x-17=0\), we get \(8x=17\), so \(x=\frac{17}{8}\). Hence, \(8x-17\) is the correct polynomial. Both \(17x-8\) and \(x-\frac{8}{17}\) have zero \(\frac{8}{17}\), not \(\frac{17}{8}\). Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\).

Given \(p(x)=7x-10\), we get \(p(x+1)=7(x+1)-10=7x-3\). Therefore, \(p(x+1)-p(x)=(7x-3)-(7x-10)=7\). The option \(7x\) is incorrect because the terms containing \(x\) cancel on subtraction. Exam tip: for a linear polynomial \(ax+b\), \(p(x+1)-p(x)=a\), the coefficient of \(x\).

The zero is (-\frac{12}{\lambda}). From (-\frac{12}{\lambda}>-3) and (\lambda>0), we get (\lambda>4).

A linear polynomial has zero \(x=-b/a\). Since \(a\ne0\), this is one definite real number, so its graph meets the x-axis exactly once. Exam tip: a non-constant linear polynomial has exactly one zero.

\(p(x)+q(x)=(x+2s)+(x-3s)\). Combining like terms gives \(x+x=2x\) and \(2s-3s=-s\). Therefore, the sum is \(2x-s\). In \(2x+s\), the coefficients of \(s\) have been added with the wrong sign. Exam tip: while adding polynomials, combine coefficients of like terms carefully, including their signs.

On subtracting, \(p(x)-q(x)=(x+2s)-(x-3s)=x+2s-x+3s=5s\). There is no term containing \(x\), so it is a constant polynomial with respect to \(x\). It becomes the zero polynomial only in the special case \(s=0\); normally, \(s\) is treated as a constant parameter. Exam tip: simplify and cancel like \(x\)-terms before identifying the degree of a polynomial.

First evaluate the inner function: \(p(3)=4\times3-11=1\). Now substitute this value into \(p\) again: \(p(1)=4\times1-11=-7\). Therefore, \(p(p(3))=-7\). The value \(1\) is only \(p(3)\), not the final answer. Exam tip: In a composite function, always evaluate the innermost function first.

First evaluate the inner function: p(0)=5(0)+2=2. Then p(p(0))=p(2)=5(2)+2=12. Therefore, the correct answer is 12. The value 10 results from taking only 5×2 and missing the constant term +2. Exam tip: In a composite function, always evaluate the inner function first.

If \(x=1\) is a zero of the polynomial, substituting \(x=1\) must make its value 0. Thus, \((n-3)(1)+2n+1=0\), so \(3n-2=0\). Hence, \(n=\frac{2}{3}\). For example, when \(n=1\), the polynomial has value 1 at \(x=1\), not 0. Exam tip: To find a parameter for a given zero, substitute that zero and equate the polynomial to 0.

Given \(p(x)=\frac{x}{5}+a\). Substituting \(x=15\), we get \(p(15)=\frac{15}{5}+a=3+a\). Since \(p(15)=11\), \(3+a=11\), so \(a=8\). If \(a=6\), then \(p(15)=9\), not 11. Exam tip: To use a given polynomial value, substitute the specified value of \(x\) first and then simplify.

Adding the polynomials gives (p(x)+q(x)=4x+a+x-6=5x+a-6). Since 2 is a zero, substitute x=2: (5×2+a-6=0), so (a+4=0). Hence, (a=-4). Option 4 results from a sign error. Exam tip: when a zero is given, substitute it into the polynomial and set the result equal to 0.

If \(-2\) is a zero of \(p(x)\), then \(p(-2)=0\). Thus, \((c+4)(-2)+2c=0\). On simplifying, \(-2c-8+2c=-8=0\), which is impossible. Hence, there is no value of \(c\). In particular, \(c=-4\) makes the polynomial equal to the constant \(-8\), which has no zero. Exam tip: to test whether \(a\) is a zero, always substitute it and use \(p(a)=0\).

If \(-1\) is a zero of the polynomial, then \(p(-1)=0\). Thus, \((c+4)(-1)+2c=0\), or \(-c-4+2c=0\). Hence, \(c-4=0\), giving \(c=4\). If \(c=-4\), the polynomial becomes \(-8\), which has no zero. Exam tip: When a zero is given, substitute it into the polynomial and set the result equal to zero.

Given \(p(x)=rx+3\) and \(p(4)=23\), we get \(4r+3=23\). Hence \(4r=20\) and \(r=5\). Now \(p(-2)=5(-2)+3=-10+3=-7\). Therefore, \(-7\) is correct. \(-10\) is only the product \(5\times(-2)\); the constant term \(+3\) must also be added. Exam tip: first use the given function value to find the unknown coefficient, then substitute the required input.

Given \(p(x)=ax-9\), we have \(p(2)=2a-9\) and \(p(6)=6a-9\). Using \(p(2)=p(6)-20\), \(2a-9=(6a-9)-20\), so \(2a-9=6a-29\). Hence \(20=4a\), giving \(a=5\). Substituting \(4\) does not satisfy the given condition. Exam tip: first substitute each given value of \(x\) into the polynomial, then form the equation.

Given p(x)=4x-15, we have p(a)=4a-15. Using the condition p(a)=a gives 4a-15=a. Hence, 3a=15 and a=5. If 4 is chosen, then p(4)=1, which is not equal to 4. In exams, replace x with a first and then apply the given condition.

Given \(p(x)=mx+n\). Substituting \(x=0\) gives \(p(0)=n=5\), so \(n=5\). Next, substituting \(x=3\) gives \(3m+n=14\). Using \(n=5\), we get \(3m+5=14\), hence \(m=3\). Therefore, \(m+n=3+5=8\). Option 5 is only the value of \(n\), not of \(m+n\). Exam tip: in a linear polynomial, putting \(x=0\) directly gives the constant term.

Given \(p(x)=mx+n\), we have \(p(2)=2m+n=11\) and \(p(-1)=-m+n=2\). Subtracting the equations gives \(3m=9\), so \(m=3\). Option \(9\) is only the difference \(11-2\) between the function values, not the value of \(m\). Exam tip: When two values of a linear polynomial are given, form equations and subtract them to eliminate the constant term \(n\).