Linear polynomials

First evaluate the inner function: \(p(1)=4(1)+1=5\). Then \(p(5)=4(5)+1=21\). Therefore, \(p(p(1))=21\). Option 20 results from taking only \(4\times5\) and missing the constant term \(+1\). Exam tip: In a composite function, always evaluate the innermost input first.

A zero equal to 1 means that the polynomial must have value 0 when \(x=1\). Thus, \((n+2)(1)+3n-1=0\), which gives \(4n+1=0\). Hence, \(n=-\frac{1}{4}\). For instance, with \(n=-1\), the value of the polynomial is \(-4\), so it is not correct. Exam tip: When a zero is given, substitute that value for \(x\) and set the expression equal to 0.

Given \(p(x)=\frac{x}{4}+a\). Substituting \(x=8\) gives \(p(8)=\frac{8}{4}+a=2+a\). Since \(p(8)=7\), we get \(2+a=7\), so \(a=5\). Option 7 is the given value of \(p(8)\), not the value of \(a\). Exam tip: First substitute the given value of \(x\) into the polynomial, then solve the resulting equation.

\(p(x)+q(x)=(3x+a)+(2x-5)=5x+a-5\). Since its zero is \(1\), put \(x=1\): \(5(1)+a-5=0\). Hence, \(a=0\). If \(a=1\), the polynomial has value \(1\), not zero, at \(x=1\). Exam tip: if \(r\) is a zero of a polynomial, substituting \(x=r\) must make the polynomial equal to \(0\).

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.