Linear polynomials

For a composite expression, evaluate the inner function first. p(2)=3×2+1=7. Then p(7)=3×7+1=22, so p(p(2))=22. Option 21 results from taking only 3×7 and forgetting to add the constant term 1. Exam tip: in p(p(a)), first find p(a), then substitute that result into p again.

If \(x=1\) is a zero of \(((n-1)x+n+2)\), substituting \(x=1\) must make the expression equal to 0. Thus, \((n-1)(1)+n+2=0\), giving \(2n+1=0\). Hence, \(n=-\frac{1}{2}\). For \(n=1\), the expression becomes \(3\), so it cannot have 1 as a zero. Exam tip: substitute the given zero in the polynomial and set the result equal to 0.

Given \(p(x)=\frac{x}{3}+a\), substitute \(x=6\): \(p(6)=\frac{6}{3}+a=2+a\). Since \(p(6)=5\), we get \(2+a=5\), so \(a=3\). Option 2 is only the value of \(\frac{6}{3}\), not the value of a. Exam tip: when a value of a polynomial is given, substitute the input first and then solve the resulting equation.

First add the polynomials: p(x)+q(x)=2x+a+x-3=3x+a-3. Since 1 is its zero, substituting x=1 gives 3(1)+a-3=0. Therefore, a=0. If a=-1, the value of the polynomial at x=1 would be -1, not zero. Exam tip: when a zero is given, substitute that value of x in the polynomial and set the result equal to 0.

If \(-4\) is a zero of \(p(x)\), then \(p(-4)=0\). Thus, \((c+1)(-4)+2c=0\), or \(-4c-4+2c=0\). This gives \(-2c-4=0\), so \(c=-2\). For \(c=-1\), the polynomial becomes the constant \(-2\), which has no zero. Exam tip: for a given zero \(a\), always use \(p(a)=0\).

Since \(-4\) is a zero of \(p(x)\), we must have \(p(-4)=0\). Thus, \(-4(c+1)+2c=0\Rightarrow -2c-4=0\Rightarrow c=-2\). Therefore, the correct option is \(c=-2\). If \(c=-1\), the constant term becomes \(-2\), so \(-4\) is not a zero. Exam tip: For a given zero \(a\), always use \(p(a)=0\).

Given \(p(x)=rx+5\) and \(p(2)=15\), we get \(2r+5=15\). Hence \(2r=10\), so \(r=5\). Now \(p(-1)=5(-1)+5=-5+5=0\). Therefore, the correct answer is 0. The value \(5\) is only the coefficient \(r\), not the value of \(p(-1)\). Exam tip: first use the given function value to find the unknown coefficient, then substitute the required value of \(x\).

Given \(p(x)=ax-6\), we have \(p(2)=2a-6\) and \(p(5)=5a-6\). Substituting these in \(p(2)=p(5)-9\) gives \(2a-6=(5a-6)-9\). Hence, \(2a=5a-9\), so \(3a=9\) and \(a=3\). For example, \(a=2\) does not satisfy the given condition. Exam tip: first evaluate the polynomial at each stated value of x, then form the equation carefully.

In option B, the coefficient of x is 0. Hence, (0x+8) = 8, which is a constant polynomial of degree 0, so it cannot be called a linear polynomial. In contrast, the coefficient of x is non-zero in options A, C, and D; therefore, they are linear polynomials. Exam tip: For a polynomial to be linear, the coefficient of x must be non-zero.

Given \(p(x)=2x-7\), we have \(p(a)=2a-7\). Using the condition \(p(a)=a\), we get \(2a-7=a\). Hence, \(a=7\). Option 5 is not correct because \(p(5)=3\), which is not equal to 5. Exam tip: \(p(a)=a\) means that \(a\) is a fixed point of the polynomial; substitute \(a\) for \(x\) and solve the resulting equation.

Given \(p(x)=mx+n\). Substituting \(x=0\), we get \(p(0)=n=4\), so \(n=4\). Now, substituting \(x=2\) gives \(2m+n=10\). Using \(n=4\), \(2m+4=10\), hence \(m=3\). Therefore, \(m+n=3+4=7\). Option 4 is only the value of \(n\), not of \(m+n\). Exam tip: Substitute the most convenient given values of \(x\) first to form equations for the coefficients.

Given \(p(x)=mx+n\), we have \(p(1)=m+n=5\) and \(p(3)=3m+n=13\). Subtracting the first equation from the second gives \(2m=8\), so \(m=4\). \(8\) is the difference \(p(3)-p(1)\), not the value of \(m\). Exam tip: Subtract two value equations of a linear polynomial to eliminate the constant term \(n\).

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

Given \(p(x)=3x-4\). Replacing \(x\) by \(-x\) gives \(p(-x)=3(-x)-4=-3x-4\). Therefore, \(p(x)+p(-x)=(3x-4)+(-3x-4)=-8\). \(0\) is a close distractor because only the terms containing \(x\) cancel; the two constant terms, \(-4\) and \(-4\), add to \(-8\). Exam tip: while finding \(p(-x)\), replace every occurrence of \(x\) with \(-x\).

Given \(p(x)=3x-4\). Replacing \(x\) with \(-x\) gives \(p(-x)=3(-x)-4=-3x-4\). Therefore, \(p(x)-p(-x)=(3x-4)-(-3x-4)=3x-4+3x+4=6x\). Hence, \(6x\) is correct. \(-8\) may result from incorrectly subtracting only the constant terms. Exam tip: while finding \(p(-x)\), replace \(x\) by \(-x\) in every term containing \(x\).

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

Adding the given polynomials gives \(p(x)+q(x)=(2x+a)+(4x-a)=6x\). The terms containing \(a\) cancel, so the sum polynomial is the same for every value of \(a\). The zero of \(6x\) is \(x=0\), since \(6(0)=0\). Hence, its zero cannot be \(-1\), and no value of \(a\) is possible. Exam tip: First simplify the polynomial, then set it equal to zero to find its zero.

\(p(x)-q(x)=(2x+a)-(4x-a)\). Since a minus sign precedes the second bracket, the signs of all its terms change: \(2x+a-4x+a=-2x+2a\). Therefore, the correct answer is \(-2x+2a\). \(2x\) cannot be obtained because \(a-(-a)=2a\), not zero. Exam tip: While subtracting polynomials, write the second polynomial in brackets and change the sign of every term.

First, find \(p(0)=a(0)+1=1\). Hence, \(p(p(0))=p(1)=a(1)+1=a+1\). Since \(p(p(0))=7\), we get \(a+1=7\), so \(a=6\). If \(a=7\), then \(p(1)=8\), which does not satisfy the given condition. Exam tip: In a composite function, evaluate the inner function first.

Given \(p(x)=x+a\), we get \(p(0)=a\). Therefore, \(p(p(0))=p(a)=a+a=2a\). Since \(p(p(0))=0\), \(2a=0\), so \(a=0\). If \(a=-1\), then \(p(p(0))=-2\), so it is not correct. Exam tip: In a composite function, evaluate the inner function first and substitute its value into the outer function.

For \(p(x)=mx+n\), we have \(p(1)=m+n\) and \(p(2)=2m+n\). Given \(p(1)=p(2)\), \(m+n=2m+n\), which gives \(m=0\). However, a linear polynomial must have a non-zero coefficient of \(x\). When \(m=0\), \(p(x)=n\) is a constant polynomial, not a linear polynomial. Hence, it can never be linear. Exam tip: always check that the coefficient of \(x\) is non-zero for a linear polynomial.

Putting \(p(x)=0\) gives \(ax+b=0\), so \(x=-b/a\). Since \(a\ne0\), this is one unique value; therefore two or infinitely many zeroes are impossible. Exam tip: a non-zero polynomial of degree 1 is linear.

Here, p(x+2)=7(x+2)-2=7x+12 and p(x-1)=7(x-1)-2=7x-9. Therefore, p(x+2)-p(x-1)=(7x+12)-(7x-9)=21. Option 14 may seem related to the shift by 2, but the inputs x+2 and x-1 differ by 3, so the result is 7×3=21. Exam tip: for a linear polynomial ax+b, p(x+r)-p(x+s)=a(r-s).

First evaluate the inner expression: \(p(3)=2\times3-5=1\). Then \(p(1)=2\times1-5=-3\). Hence, \(p(p(3))=-3\), so option C is correct. Option B is only the value of \(p(3)\), not the final composite value. Exam tip: In \(p(p(a))\), find the inner \(p(a)\) first and substitute that result into \(p\) again.

Open the brackets first: \(6(3x-2)=18x-12\) and \(-4(4x+5)=-16x-20\). Therefore, \(18x-12-16x-20=2x-32\). Hence, the correct polynomial is \(2x-32\). The option \(2x+8\) can result from not changing the signs of both terms while multiplying by \(-4\). Exam tip: when a negative factor is outside a bracket, apply it to every term inside the bracket.