Linear polynomials

From (-3(c-2)+3c=0), we get (6=0), so there is no solution. Therefore none of the listed values should be correct.

Since \(-2\) is a zero of the polynomial, \(p(-2)=0\). Thus, \((c-2)(-2)+3c=0\). Simplifying gives \(-2c+4+3c=0\), or \(c+4=0\). Therefore, \(c=-4\). If \(c=2\), the coefficient of \(x\) becomes zero, so the given expression would not remain a linear polynomial. Exam tip: Substitute the given zero in the polynomial and equate the result to zero.

Given \(p(x)=rx-4\) and \(p(3)=14\), we get \(3r-4=14\). Thus, \(3r=18\) and \(r=6\). Now, \(p(-1)=6(-1)-4=-6-4=-10\). Therefore, \(-10\) is the correct answer. An answer such as \(-8\) can result from an error while substituting \(-1\) or handling the negative sign. Exam tip: first find the unknown coefficient using the given condition, then substitute the required value of \(x\).

Given \(p(x)=ax+8\), we have \(p(1)=a+8\) and \(p(4)=4a+8\). Using \(p(1)=p(4)-12\), we get \(a+8=(4a+8)-12\), or \(a+8=4a-4\). Hence \(12=3a\), so \(a=4\). For example, \(a=3\) does not satisfy the given condition. Exam tip: first evaluate the polynomial at the specified values, then substitute them into the given relation.

Given \(p(x)=3x-10\), we have \(p(a)=3a-10\). Using the condition \(p(a)=a\), we get \(3a-10=a\). Hence, \(2a=10\), so \(a=5\). If 10 is chosen, then \(p(10)=20\), which is not equal to 10. Exam tip: In \(p(a)=a\), first substitute \(a\) for \(x\), then equate the two expressions.

Given \(p(x)=mx+n\). Substituting \(x=0\) gives \(p(0)=n=-2\). Also, \(p(4)=10\) gives \(4m+n=10\). Using \(n=-2\), we get \(4m-2=10\), so \(m=3\). Therefore, \(m+n=3+(-2)=1\). Option 2 can result from incorrectly handling the sign of \(n\). Exam tip: for a linear polynomial, \(p(0)\) is always its constant term \(n\).

Given \(p(x)=mx+n\), we have \(p(2)=2m+n=9\) and \(p(5)=5m+n=21\). Subtracting the first equation from the second gives \(3m=12\), so \(m=4\). The constant term \(n\) cancels because it is the same in both equations. Exam tip: When two values of a linear polynomial are given, subtract the equations to find \(m\) first.

For \(p(x)=x+3\), substituting \(x=-3\) gives \(p(-3)=-3+3=0\). Its degree is 1, so it is a linear polynomial. In option D, \(p(-3)=0\), but its degree is 2; therefore, it is quadratic, not linear. Exam tip: A linear polynomial always has degree 1.

Given \(p(x)=4x+9\), substitute \(-x\) for \(x\): \(p(-x)=4(-x)+9=-4x+9\). Therefore, \(p(x)+p(-x)=(4x+9)+(-4x+9)=18\). The expression \(8x+18\) would result only if the sign of \(x\) were not changed in \(p(-x)\). Exam tip: while finding \(p(-x)\), replace every occurrence of \(x\) with \(-x\) before simplifying.

Given \(p(x)=4x+9\), substitute \(-x\) for \(x\) to get \(p(-x)=-4x+9\). Hence, \(p(x)-p(-x)=(4x+9)-(-4x+9)=4x+9+4x-9=8x\). It is not \(18\), because the constant terms \(+9\) and \(-9\) cancel. Exam tip: while finding \(p(-x)\), replace every occurrence of \(x\) with \(-x\).

For the zero to be (0), the constant term (2s=0). At (s=0), the coefficient of (x) is (4), so the polynomial is linear.

Adding the polynomials gives (p(x)+q(x))=(3x+a)+(5x-a)=8x. The terms containing (a) cancel, so the sum does not depend on (a). The only zero of (8x) is (0), since (8x=0) gives (x=0). Hence, no value of (a) can make (2) a zero. Exam tip: Simplify the polynomial first and check whether variable parameters cancel before applying the zero condition.

\(q(x)-p(x)=(5x-a)-(3x+a)\). Since there is a minus sign before the second bracket, the signs of both its terms change: \(5x-a-3x-a=2x-2a\). Hence, the correct answer is \(2x-2a\). Writing only \(2x\) is incorrect because the constant terms also give \(-a-a=-2a\). Exam tip: while subtracting polynomials, keep the second polynomial in brackets and change every sign inside it.

Given \(p(x)=ax-2\), we get \(p(0)=-2\). Therefore, \(p(p(0))=p(-2)=-2a-2\). Using \(-2a-2=10\), we obtain \(-2a=12\), so \(a=-6\). Hence, option D is correct. If \(a=6\), then \(p(p(0))=-14\), not 10. Exam tip: In a composite-function question, evaluate the inner function first and substitute its value into the outer function.

Given \(p(x)=x+a\), first \(p(1)=1+a\). Therefore, \(p(p(1))=p(1+a)=(1+a)+a=1+2a\). Hence \(1+2a=7\), so \(2a=6\) and \(a=3\). If \(a=4\), then \(p(p(1))=1+2(4)=9\), not 7. Exam tip: In a composite function, evaluate the inner function first and substitute its result into the outer function.

A linear polynomial has degree exactly 1, so the coefficient \(a\) of \(x\) must be non-zero. If \(a=0\), then \(p(x)=b\) is a constant polynomial. Exam tip: first check the coefficient of the highest-power term.

Given \(p(x)=6x-1\), we get \(p(x+3)=6(x+3)-1=6x+17\) and \(p(x-2)=6(x-2)-1=6x-13\). Therefore, \(p(x+3)-p(x-2)=(6x+17)-(6x-13)=30\). Option 18 is only the change for an input increase of 3; here the inputs differ by \((x+3)-(x-2)=5\), so the change is \(6\times5=30\). Exam tip: for a linear polynomial \(ax+b\), use \(p(u)-p(v)=a(u-v)\).

First evaluate the inner function: \(p(3)=4\times3-9=3\). Therefore, \(p(p(3))=p(3)=3\). Hence, the correct answer is 3. The value 9 is only the constant term and is not the value of the function here. Exam tip: In a composite function, always evaluate the inner function first.

Given \(p(x)=5x+2\), \(p(x+1)=5(x+1)+2=5x+7\) and \(p(x-1)=5(x-1)+2=5x-3\). Therefore, \(p(x+1)+p(x-1)=(5x+7)+(5x-3)=10x+4\). The option \(10x\) would result only if the constant terms added to zero, but here \(7+(-3)=4\). Exam tip: substitute \(x+1\) and \(x-1\) carefully in the complete polynomial before adding.

For it not to be linear, the coefficient of (x) must be (0). (r^2-2r=r(r-2)=0) gives (r=0) or (r=2).

Given \(p(x)=ax+3\), first find \(p(0)=a(0)+3=3\). Therefore, \(p(p(0))=p(3)=3a+3\). Using \(3a+3=18\), we get \(3a=15\), so \(a=5\). If \(a=6\), then \(p(p(0))=21\), not 18. Exam tip: In a composite expression such as \(p(p(0))\), evaluate the inner function first.

From the two equations, (4m=12), so (m=3). From (2m+n=13), (n=7) and (m-n=-4), which is not in the options.

On expanding the brackets, \(7(2x-5)=14x-35\) and \(-5(3x+4)=-15x-20\). Therefore, \(14x-35-15x-20=-x-55\). Hence, the correct polynomial is \(-x-55\). In \(-x+55\), the sign of the constant term is incorrect. Exam tip: when a minus sign is multiplied with a bracket, change the sign of every term inside it.

In \(P(x)=\sqrt{2}x-5\), the highest power of \(x\) is 1 and \(\sqrt{2}\) is a real coefficient, so it is linear. Exam tip: determine degree from the exponent, not from the number of terms.

For a linear polynomial, the coefficient of \(x^2\) must be zero and the coefficient of \(x\) must be non-zero. From \(a^2-9=0\), \(a=\pm3\). At \(a=-3\), the \(x\)-coefficient is also 0, so only \(a=3\) works. Exam tip: check both conditions.