Linear polynomials

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 \(n\).

In option D, \(p(x)=x-4\). On substituting \(x=4\), we get \(p(4)=4-4=0\). Its degree is 1, so it is a linear polynomial. Option C also gives \(p(4)=0\), but its degree is 2; therefore, it is quadratic, not linear. Exam tip: a linear polynomial always has degree 1.

Given \(p(x)=5x-12\). To find \(p(-x)\), substitute \(-x\) for \(x\): \(p(-x)=5(-x)-12=-5x-12\). Therefore, \(p(x)+p(-x)=(5x-12)+(-5x-12)=-24\). \(10x-24\) would result only if the \(x\)-term remained \(5x\) in both expressions. Exam tip: while finding \(p(-x)\), substitute \(-x\) carefully in every term containing \(x\).

Given \(p(x)=5x-12\). Replacing \(x\) with \(-x\) gives \(p(-x)=5(-x)-12=-5x-12\). Therefore, \(p(x)-p(-x)=(5x-12)-(-5x-12)=5x-12+5x+12=10x\). Hence, \(10x\) is correct. \(0\) would result only if the two expressions were identical. Exam tip: while finding \(p(-x)\), substitute \(-x\) for every occurrence of \(x\).

For the zero to be (0), the constant term (3s=0). At (s=0), the coefficient of (x) is (-5), so the polynomial is linear.

We have p(x)+q(x)=(4x+a)+(6x-a)=10x. The terms a and -a cancel, so the sum polynomial is unchanged for every value of a. The only zero of 10x is x=0, since 10x=0 gives x=0. Hence, no value of a can make x=-2 a zero. Exam tip: Simplify the sum first; cancellation may show that a parameter has no effect on the zero.

\(q(x)-p(x)=(6x-a)-(4x+a)\). Since there is a minus sign before the second bracket, the signs of both its terms change: \(6x-a-4x-a=2x-2a\). Hence, the correct answer is \(2x-2a\). Writing only \(2x\) is incorrect because the terms containing \(a\) must also be subtracted. Exam tip: while subtracting polynomials, keep the second polynomial in brackets and distribute the negative sign carefully.

Given \(p(x)=ax+4\), we first get \(p(0)=4\). Hence \(p(p(0))=p(4)=4a+4\). So, \(4a+4=28\), which gives \(4a=24\) and \(a=6\). Therefore, option C is correct. For example, if \(a=7\), then \(p(4)=32\), not 28. Exam tip: In a composite expression such as \(p(p(0))\), evaluate the inner function first.

Given \(p(x)=x+a\), we get \(p(2)=2+a\). Therefore, \(p(p(2))=p(2+a)=(2+a)+a=2+2a\). Hence \(2+2a=10\), so \(2a=8\) and \(a=4\). Option 8 may result from the mistake of concluding \(a=8\) from \(2a=8\). Exam tip: For a composite function, evaluate the inner function first and substitute its result into the outer function.

Given \(p(-2)=p(1)\), we get \(-2m+n=m+n\). Thus \(-3m=0\), so \(m=0\). Then \(p(x)=n\), which is a constant polynomial, not a linear polynomial. A linear polynomial requires \(m\ne0\), so option C contradicts the given condition. Exam tip: \(ax+b\) is linear only when \(a\ne0\).

A zero must make \(p(x)=0\). Substituting \(x=5\) gives \(m(5-5)=0\), so 5 is the zero for every non-zero \(m\). Exam tip: if a factor is \((x-a)\), test \(x=a\) first.

Given \(p(x)=8x-3\), we get \(p(x+2)=8(x+2)-3=8x+13\) and \(p(x-1)=8(x-1)-3=8x-11\). Therefore, \(p(x+2)-p(x-1)=(8x+13)-(8x-11)=24\). Option 16 would result from using only \(8\times2\), but the difference between the inputs is \((x+2)-(x-1)=3\). Exam tip: for a linear polynomial \(ax+b\), use \(p(u)-p(v)=a(u-v)\) to calculate quickly.

First evaluate the inner function: \(p(3)=5\times3-14=1\). Then \(p(1)=5\times1-14=-9\). Therefore, \(p(p(3))=-9\). Option \(1\) is only the value of \(p(3)\), not the final value. Exam tip: In a composite function, always evaluate the innermost function first.

Here, \(p(0)=6(0)+a=a\). Therefore, \(p(p(0))=p(a)=6a+a=7a\). Since \(p(p(0))=42\), we get \(7a=42\). Hence, \(a=6\). Option 7 is the coefficient in \(7a\), not the value of \(a\). Exam tip: In a composite function, evaluate the inner function first.

Given \(p(x)=9x-5\), \(p(x+1)=9(x+1)-5=9x+4\) and \(p(x-2)=9(x-2)-5=9x-23\). Therefore, \(p(x+1)+p(x-2)=(9x+4)+(9x-23)=18x-19\). Option A results from an error while adding the constant terms. Exam tip: substitute using brackets first, expand, and then combine like terms.