Linear polynomials

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.

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.

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.

Open the brackets first: \(3(2x-5)=6x-15\) and \(-2(5x+1)=-10x-2\). Therefore, \(6x-15-10x-2=-4x-17\). Hence, the correct polynomial is \(-4x-17\). In \(-4x+17\), the sign of the constant term is incorrect. Exam tip: when a negative multiplier precedes a bracket, apply it to every term inside the bracket.

\(3x+6=3(x+2)\), so its zero is -2 and the highest power of x is 1; hence it is linear. \(x^2+4x+4\) also has zero -2, but it is quadratic. Exam tip: check the degree before selecting the expression.

Since \(a\ne0\), solving \(ax+b=0\) gives the single value \(x=-b/a\). Hence, a non-constant linear polynomial has exactly one zero. Exam tip: if \(a=0\), the expression is not linear.

\(3p(x)-4q(x)=3(5x-4)-4(2x+7)\). Expanding gives \(15x-12-8x-28=7x-40\), so \(7x-40\) is correct. \(7x+16\) results from failing to apply the negative sign to both terms of \(4(2x+7)\). Exam tip: when a minus sign precedes brackets, change the sign of every term inside them.

A zero equal to 3 means that the polynomial must have value 0 when \(x=3\). Thus, \(3(2a-3)-9=0\). Simplifying gives \(6a-18=0\), so \(a=3\). For example, if \(a=1\), the value is \(-12\), so it is not correct. Exam tip: Substitute the given zero into the polynomial and set its value equal to zero.