Linear polynomials

First add the polynomials: \(p(x)+q(x)=5x+a+2x-9=7x+a-9\). Since \(2\) is a zero of this polynomial, substitute \(x=2\): \(7(2)+a-9=0\). Thus, \(14+a-9=0\), so \(a+5=0\) and \(a=-5\). If \(a=-3\), the polynomial does not give zero at \(x=2\). Exam tip: when a zero is given, substitute that value in the polynomial and equate the result to \(0\).

If \(-2\) is a zero of \(p(x)\), then \(p(-2)=0\). Thus, \((c-3)(-2)+4c=0\), which gives \(-2c+6+4c=0\), or \(2c+6=0\). Hence, \(c=-3\). For \(c=3\), the polynomial becomes \(12\), which has no zero. Exam tip: for a given zero \(a\), directly use \(p(a)=0\).

If \(-2\) is a zero of \(p(x)\), then \(p(-2)=0\). Thus, \(-2(c-1)+2c=0\). Simplifying gives \(-2c+2+2c=0\), or \(2=0\), which is impossible. Hence, no value of \(c\) can make \(-2\) a zero of this polynomial. Exam tip: For a given zero \(a\), always substitute it and use \(p(a)=0\).

Given \(p(x)=rx-6\) and \(p(4)=18\), we have \(4r-6=18\). Thus \(4r=24\), so \(r=6\). Now \(p(-2)=6(-2)-6=-12-6=-18\). Therefore, \(-18\) is the correct answer. Getting \(-12\) is a common error caused by forgetting the constant term \(-6\). Exam tip: first use the given function value to find the coefficient, then substitute the required input.

Given \(p(x)=ax+10\), we have \(p(2)=2a+10\) and \(p(5)=5a+10\). Using \(p(2)=p(5)-15\), \(2a+10=5a+10-15\), so \(2a=5a-15\). Hence \(3a=15\), giving \(a=5\). For example, \(a=3\) does not satisfy the given condition. Exam tip: first substitute the stated values of \(x\) into the polynomial, then form and solve the equation.

A polynomial’s degree is determined by its highest exponent, not by its number of terms. Since \(x^2\) has exponent 2, this is a quadratic polynomial, not a linear one. Exam tip: identify the highest power first.

Given \(p(x)=mx+n\). Substituting \(x=0\) gives \(p(0)=n=-4\). Next, substituting \(x=3\) gives \(3m+n=11\). Using \(n=-4\), we get \(3m-4=11\), so \(m=5\). Hence, \(m+n=5+(-4)=1\). Option \(5\) is only the value of \(m\), not of \(m+n\). Exam tip: In such questions, put \(x=0\) first to find the constant term \(n\) quickly.

Given \(p(x)=mx+n\), we have \(p(2)=2m+n=17\) and \(p(-2)=-2m+n=1\). Subtracting the second equation from the first gives \(4m=16\), so \(m=4\). Note that \(16\) is the difference \(p(2)-p(-2)\), not the value of \(m\). Exam tip: subtracting two function-value equations eliminates the constant term \(n\).

For \(p(x)=x+5\), substituting \(x=-5\) gives \(p(-5)=-5+5=0\). Its highest power of \(x\) is 1, so it is a linear polynomial. Although \(x^2-25\) also gives 0 at \(-5\), its degree is 2, so it is quadratic, not linear. Exam tip: a linear polynomial always has degree 1.

Given p(x)=6x+11, substitute -x for x: p(-x)=6(-x)+11=-6x+11. Therefore, p(x)+p(-x)=(6x+11)+(-6x+11)=22. Hence, option B is correct. Zero is a close distractor, but only the x-terms cancel; the constant terms give 11+11=22. Exam tip: in p(x)+p(-x), odd-power terms cancel, while even-power and constant terms are doubled.

Given \(p(x)=6x+11\). Replacing \(x\) with \(-x\) gives \(p(-x)=-6x+11\). Therefore, \(p(x)-p(-x)=(6x+11)-(-6x+11)=12x\). The constant term \(11\) cancels on subtraction; \(22\) would result from addition, not subtraction. Exam tip: while finding \(p(-x)\), carefully change the sign of every term containing \(x\).

For the zero to be (0), the constant term (4s=0). At (s=0), the coefficient of (x) is (7), so the polynomial is linear.

We get p(x)+q(x)=(5x+a)+(7x-a)=12x. The terms a and -a cancel, so the sum polynomial does not depend on any value of a. The only zero of 12x is 0, since 12x=0 gives x=0. Hence, no value of a can make 1 a zero. Exam tip: While adding polynomials, first check whether opposite constant terms cancel.

\(q(x)-p(x)=(7x-a)-(5x+a)\). Since a minus sign occurs before the second bracket, the signs of all its terms change: \(7x-a-5x-a=2x-2a\). Therefore, the correct answer is \(2x-2a\). \(2x\) would result only if the terms containing \(a\) cancelled, but here they give \(-a-a=-2a\). Exam tip: while subtracting polynomials, use brackets and change the sign of every term in the polynomial being subtracted.

Given \(p(x)=ax+5\), we first get \(p(0)=5\). Hence \(p(p(0))=p(5)=5a+5\). Using \(5a+5=35\), we obtain \(5a=30\), so \(a=6\). Option 5 is the value of \(p(0)\), not the value of \(a\). Exam tip: In a composite function, evaluate the inner function first.

Given \(p(x)=x+a\), we get \(p(3)=3+a\). Therefore, \(p(p(3))=p(3+a)=(3+a)+a=3+2a\). Hence \(3+2a=13\), so \(2a=10\) and \(a=5\). Option 10 may result from stopping at \(2a=10\) without dividing by 2. Exam tip: In a composite function, evaluate the inner function first.

Using \(p(-3)=p(1)\), we get \(-3m+n=m+n\). Hence, \(-4m=0\), so \(m=0\). Then \(p(x)=n\), which is a constant polynomial, not a linear polynomial. Whether \(n=0\) or not does not change this conclusion. Exam tip: for \(mx+n\) to be linear, \(m\ne0\) is essential.

A linear polynomial has the form \(ax+b\), where \(a\ne0\), so its highest power of x is 1. For example, \(3x-5\) is linear, whereas a constant has degree 0. Exam tip: identify the degree by checking the highest exponent.

Given \(p(x)=9x-4\), \(p(x+2)=9(x+2)-4=9x+14\) and \(p(x-1)=9(x-1)-4=9x-13\). Hence, \(p(x+2)-p(x-1)=(9x+14)-(9x-13)=27\). Option 18 may result from considering only the shift of 2, but the difference between the inputs is \((x+2)-(x-1)=3\). Exam tip: for a linear polynomial \(ax+b\), an input difference of \(h\) gives an output difference of \(ah\).

First evaluate the inner function: \(p(3)=7\times3-20=1\). Now substitute this value into \(p\) again: \(p(1)=7\times1-20=-13\). Therefore, \(p(p(3))=-13\). 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.