Linear polynomials

First add the polynomials: \(p(x)+q(x)=6x+a+3x-10=9x+a-10\). Since 2 is a zero, the value of this polynomial at \(x=2\) must be zero. Thus, \(9(2)+a-10=0\), or \(18+a-10=0\), which gives \(a=-8\). If \(a=-6\), the value at \(x=2\) is 2, not zero. Exam tip: when a zero is given, substitute it into the polynomial and equate the result to \(0\).

If \(-3\) is a zero of \(p(x)=(c-5)x+6c\), then \(p(-3)=0\). Thus, \(-3(c-5)+6c=0\), which gives \(-3c+15+6c=0\), or \(3c+15=0\). Hence, \(c=-5\). For \(c=5\), the polynomial becomes \(30\), which has no zero. Exam tip: to find a parameter from a given zero \(a\), use \(p(a)=0\).

If \(-4\) is a zero of \(p(x)\), then \(p(-4)=0\). Thus, \((c-2)(-4)+4c=0\). On simplifying, \(-4c+8+4c=0\), which gives \(8=0\), an impossibility. Hence, no value of \(c\) can make \(-4\) a zero of this polynomial. Exam tip: To test a given zero \(a\), always substitute it and use \(p(a)=0\).

Given \(p(x)=rx-9\) and \(p(5)=21\), we get \(5r-9=21\). Hence \(5r=30\) and \(r=6\). Now \(p(-1)=6(-1)-9=-6-9=-15\). Therefore, option B is correct. The value \(-18\) can result from an incorrect sign operation with the constant term. Exam tip: first use the given value of the polynomial to find the unknown coefficient, then substitute the required value of \(x\).

Given \(p(x)=ax+12\), we have \(p(3)=3a+12\) and \(p(7)=7a+12\). Using \(p(3)=p(7)-20\), \(3a+12=7a+12-20=7a-8\). Thus, \(4a=20\), so \(a=5\). Taking \(a=4\) does not satisfy the given condition. Exam tip: Write the polynomial values separately before substituting them into the condition.

Given p(x)=6x-25, we have p(a)=6a-25. Using the condition p(a)=a gives 6a-25=a. Hence, 5a=25 and a=5. For example, if a=6, then p(6)=11, not 6. Exam tip: To find p(a), replace x by a first and then use the given condition to form an equation.

Given \(p(x)=mx+n\). Substituting \(x=0\), we get \(p(0)=n=-6\). Also, \(p(4)=18\) gives \(4m+n=18\), so \(4m-6=18\). Hence \(4m=24\) and \(m=6\). Therefore, \(m+n=6+(-6)=0\). Option 6 is only the value of \(m\), not of \(m+n\). Exam tip: for a linear polynomial, \(p(0)\) directly gives the constant term \(n\).

Given \(p(x)=mx+n\), we have \(p(-3)=-3m+n=2\) and \(p(1)=m+n=18\). Subtracting the first equation from the second gives \(4m=16\), so \(m=4\). The number \(16\) is the difference between the two function values, not the value of \(m\). Exam tip: subtract two equations for a linear polynomial to eliminate the constant term \(n\).

For \(p(x)=x-7\), substituting \(x=7\) gives \(p(7)=7-7=0\). Its degree is 1, so it is a linear polynomial. Although \(x^2-49\) also gives \(p(7)=0\), it has degree 2 and is therefore quadratic, not linear. Exam tip: a linear polynomial always has degree 1.

Given \(p(x)=7x-13\). Replacing \(x\) with \(-x\) gives \(p(-x)=7(-x)-13=-7x-13\). Therefore, \(p(x)+p(-x)=(7x-13)+(-7x-13)=-26\). Option \(0\) is not correct because the constant term \(-13\) occurs in both expressions and adds to \(-26\). Exam tip: while finding \(p(-x)\), substitute \(-x\) for every occurrence of \(x\); the constant term does not change sign.

Given \(p(x)=7x-13\). Replacing \(x\) by \(-x\), we get \(p(-x)=7(-x)-13=-7x-13\). Therefore, \(p(x)-p(-x)=(7x-13)-(-7x-13)=7x-13+7x+13=14x\). Hence, \(14x\) is correct. \(0\) would result for an even polynomial, but the term \(7x\) is odd. Exam tip: while finding \(p(-x)\), replace \(x\) by \(-x\) in every term containing \(x\).

For the zero to be (0), the constant term (5s=0). At (s=0), the coefficient of (x) is (-8), so the polynomial is linear.

We have \(p(x)+q(x)=(6x+a)+(8x-a)=14x\). The terms containing \(a\) cancel, so the sum polynomial does not depend on any value of \(a\). Its zero is \(x=0\), since \(14x=0\) gives \(x=0\). Therefore, no value of \(a\) can make \(-1\) a zero. Exam tip: simplify the sum first; variable terms may cancel before you solve for a zero.

\(q(x)-p(x)=(8x-a)-(6x+a)\). Since there is a minus sign before the second bracket, the signs of both its terms change: \(8x-a-6x-a=2x-2a\). Hence, the correct answer is \(2x-2a\). The option \(2x\) incorrectly omits \(-a-a=-2a\). Exam tip: While subtracting polynomials, keep the second polynomial in brackets and change all its signs.

First, \(p(0)=a\cdot0+6=6\). Hence, \(p(p(0))=p(6)=6a+6\). Given \(6a+6=48\), we get \(6a=42\) and therefore \(a=7\). If \(a=8\), then \(p(6)=54\), not 48. Exam tip: In a composite expression such as \(p(p(0))\), evaluate the inner function first.

Given p(x)=x+a, we get p(4)=4+a. Therefore, p(p(4))=p(4+a)=(4+a)+a=4+2a. Hence, 4+2a=16, so 2a=12 and a=6. If a=8, then p(p(4)) would be 20, so it is not correct. Exam tip: For p(p(4)), first find p(4) and then substitute that result into p again.

Given \(p(-4)=p(2)\), we get \(-4m+n=2m+n\). Thus \(-6m=0\), so \(m=0\). Then \(p(x)=n\), which is a constant polynomial, not a linear polynomial. Hence \(p(x)\) can never be linear under the given condition. Exam tip: \(ax+b\) is linear only when \(a\ne0\).

In \(B(x)=120-8x\), the highest power of \(x\) is 1, so its degree is 1 and it is a linear polynomial. Only 120 is the constant term; the whole polynomial is not constant. Exam tip: use the highest power, not the number of terms.

Given \(p(x)=12x-5\), we get \(p(x+2)=12(x+2)-5=12x+19\) and \(p(x-1)=12(x-1)-5=12x-17\). Hence, \(p(x+2)-p(x-1)=(12x+19)-(12x-17)=36\). Option 24 would correspond to an input difference of 2, whereas \(x+2\) and \(x-1\) differ by 3. Exam tip: for a linear polynomial \(ax+b\), an input change of \(h\) produces an output change of \(ah\).

First evaluate the inner function: \(p(3)=9\times3-26=1\). Then \(p(p(3))=p(1)=9\times1-26=-17\). Hence, the correct answer is \(-17\). Option \(1\) is only the value of \(p(3)\); it must be substituted into \(p\) once more. Exam tip: In a composite function, always evaluate the innermost function first.

Given \(p(x)=13x+a\), we first get \(p(0)=a\). Therefore, \(p(p(0))=p(a)=13a+a=14a\). Since \(14a=182\), \(a=\frac{182}{14}=13\). The value \(14\) arises after adding the constant term \(a\) to the coefficient expression, but it is not the value of \(a\). Exam tip: In composition questions, find the inner function value first and then substitute it into the outer function.

Subtracting the two values gives (5m=25), so (m=5). From (-2m+n=5), (n=15), so (m+n=20), not (15).

Substitute x+1 and x-3 for x separately. \(p(x+1)=14(x+1)-9=14x+5\), while \(p(x-3)=14(x-3)-9=14x-51\). Therefore, \(p(x+1)+p(x-3)=(14x+5)+(14x-51)=28x-46\). The option \(28x-18\) results from adding the constant terms incorrectly. Exam tip: use brackets whenever substituting an expression for x, especially when it contains a minus sign.

On opening the brackets, \(8(2x-1)=16x-8\) and \(-3(5x+4)=-15x-12\). Therefore, \(16x-8-15x-12=x-20\). Hence, the required linear polynomial is \(x-20\). The option \(16x-12\) results from not distributing \(-3\) correctly to both terms in the second bracket. Exam tip: When a negative factor precedes a bracket, multiply every term inside by it.

Since \(a\neq0\), solving \(ax+b=0\) gives \(x=-b/a\). This is only one value, so a linear polynomial has exactly one real zero. Exam tip: always check that \(a\neq0\).