Linear polynomials

Given \(p(x)=10x-21\), we get \(p(x+1)=10(x+1)-21=10x-11\). Hence, \(p(x+1)-p(x)=(10x-11)-(10x-21)=10\). The option \(10x\) is incorrect because the terms containing \(x\) cancel on subtraction. Exam tip: for a linear polynomial \(ax+b\), \(p(x+1)-p(x)=a\).

The zero is (-\frac{18}{\lambda}). From (-\frac{18}{\lambda}>-6) and (\lambda>0), we get (\lambda>3).

For the zero, set \(D(n)=0\): \(600-75n=0\), so \(n=8\). Thus, the debt is cleared after 8 days. Option A describes half the debt, not zero. Exam tip: substitute 0 for the polynomial value to find its zero.

\(p(x)+q(x)=(x+4r)+(x-7r)\). Combining like terms gives \(x+x=2x\) and \(4r-7r=-3r\). Therefore, the sum is \(2x-3r\). In \(2x+11r\), the signs of \(4r\) and \(-7r\) have been combined incorrectly. Exam tip: while adding polynomials, add coefficients of like terms along with their signs.

On subtracting,
\(p(x)-q(x)=(x+4r)-(x-7r)=x+4r-x+7r=11r\). No term involving \(x\) remains, so it is a constant polynomial in \(x\). A zero polynomial would arise only if the constant itself were 0; the simplified expression is generally \(11r\). Exam tip: simplify first, then identify the highest power of the variable.

First evaluate the inner function: \(p(3)=8\times3-23=24-23=1\). Substituting this result into \(p\) again gives \(p(p(3))=p(1)=8\times1-23=-15\). Therefore, -15 is the correct option. Option 1 is only the value of \(p(3)\), not of \(p(p(3))\). Exam tip: In a composite function, always evaluate the innermost expression first.

First evaluate the inner function: \(p(0)=5\times0+4=4\). Then \(p(4)=5\times4+4=24\). Therefore, \(p(p(0))=24\). The value 20 comes from only \(5\times4\), missing the constant term 4. Exam tip: In composition questions, always evaluate the innermost function first.

If 1 is a zero of the polynomial, its value must be 0 when \(x=1\). Thus, \((n-5)(1)+3n+2=0\), which gives \(4n-3=0\). Therefore, \(n=\frac{3}{4}\). For instance, putting \(n=1\) gives the value \(-3\), so it is not correct. Exam tip: Substitute the given zero in the polynomial and equate the result to 0.

Given \(p(x)=\frac{x}{7}+a\). Substituting \(x=21\) gives \(p(21)=\frac{21}{7}+a=3+a\). Since \(p(21)=16\), we have \(3+a=16\), so \(a=13\). Option 16 is the given value of \(p(21)\), not the value of \(a\). Exam tip: when a polynomial value is given, substitute the stated value of \(x\) first.

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 \(x\).