Linear polynomials

For a linear polynomial ax+b, the zero is found by setting the expression equal to zero and solving for x. The required value is \(-\frac{9}{13}\), so the equation should lead to 13x=-9. A positive constant term 9 together with coefficient 13 produces exactly this result. The sign is important: changing the sign of the constant changes the sign of the zero.

In option C, solve 13x+9=0. Subtracting 9 gives 13x=-9, and division by 13 gives x=-\frac{9}{13}. Therefore option C has the required zero. Option A gives a positive zero \(\frac{9}{13}\), while options B and D lead to different fractions, so they do not satisfy the condition.

From (\frac{8}{15}x-16=0), (\frac{8}{15}x=16) and (x=30). With a fractional coefficient, multiply by the reciprocal.

Both p(x) and q(x) are linear polynomials, so each has degree 1. Their product is (x-10)(x+3)=x²-7x-30. The highest power of x is 2, so the degree of the product is 2. Option 1 is incorrect because it is the degree of each individual polynomial, not of their product. Exam tip: For non-zero polynomials, the degree of a product is usually the sum of their degrees.

Given \(p(x)=9x-17\), replace the entire \(x\) with \(x+3\): \(p(x+3)=9(x+3)-17=9x+27-17=9x+10\). Hence, \(9x+10\) is correct. \(9x+27\) results from forgetting to subtract 17. Exam tip: always put the substituted expression in brackets when evaluating a function.

Given \(p(x)=19-7x\), replace the input \(x\) by the entire expression \(2x\): \(p(2x)=19-7(2x)=19-14x\). Hence, option B is correct. \(38-14x\) would result from incorrectly doubling the constant term 19 as well; substitution changes only the input. Exam tip: In \(p(2x)\), replace every \(x\) in the rule by \(2x\).

A polynomial may have irrational real coefficients. In option A, the highest power of \(x\) is 1, so its degree is 1. B and D are not polynomials, while C has degree 2. Exam tip: check exponents, not coefficients.

Given p(7)=p(-1), we get 7k-14=-k-14. Thus, 8k=0 and k=0. Then p(x)=-14 is a constant polynomial, not a linear polynomial, because the coefficient of x must be non-zero for a linear polynomial. Therefore, under the given condition, p(x) can never remain linear. Exam tip: In a linear polynomial ax+b, a must satisfy a≠0.

First simplify the expression: (q(x)-2p(x))=(15x-6)-2(7x+2)=15x-6-14x-4=x-10. The highest power of x is 1, so its degree is 1. A constant polynomial would have degree 0, but here the coefficient of x is 1, not zero. Exam tip: When adding or subtracting linear polynomials, combine like terms first because leading terms may cancel.

The zero of a polynomial is the value of x that makes the polynomial equal to zero. To test option A, set 11x−29=0. Then 11x=29, giving x=29/11, exactly the required zero. The alternatives give different values: 29x−11=0 gives x=11/29; 11x+29=0 gives x=−29/11; and x−11/29=0 gives x=11/29. Thus only option A has zero 29/11. Every displayed expression is linear because its coefficient of x is nonzero, so the deciding step is finding the root, not checking whether the expression is linear. Therefore option A is unambiguously correct.

Given \(p(x)=12x-25\), \(p(x+1)=12(x+1)-25=12x-13\). Therefore, \(p(x+1)-p(x)=(12x-13)-(12x-25)=12\). The option \(12x\) is the linear term, not the difference. Exam tip: for a linear polynomial \(ax+b\), \(p(x+1)-p(x)=a\).

The zero is (-\frac{21}{\lambda}). From (-\frac{21}{\lambda}>-7) and (\lambda>0), we get (\lambda>3).

Since \(a\ne0\), the graph of \(p(x)=ax+b\) is a non-horizontal straight line. It meets the \(x\)-axis once, at \(x=-\frac{b}{a}\). Exam tip: a zero is the graph’s \(x\)-intercept.

\(p(x)+q(x)=(x+5r)+(x-8r)\). Combining like terms gives \(x+x=2x\) and \(5r-8r=-3r\). Therefore, the sum is \(2x-3r\). The option \(2x+13r\) results from incorrectly ignoring the negative sign before \(8r\). Exam tip: When adding polynomials, remove brackets carefully and check the sign of every term.

We get \(p(x)-q(x)=(x+5r)-(x-8r)=x+5r-x+8r=13r\). The \(x\)-terms cancel, so no power of \(x\) remains. Hence, with respect to \(x\), \(13r\) is a constant polynomial. A linear polynomial must have a non-zero coefficient of \(x\). Exam tip: simplify like terms first, then identify the highest power of the variable.

First evaluate the inner function: \(p(3)=10\times3-29=1\). Then \(p(p(3))=p(1)=10\times1-29=-19\). Therefore, the correct answer is \(-19\). Option 1 is only the value of \(p(3)\); the question requires applying \(p\) once again to that result. Exam tip: In \(p(p(a))\), always evaluate the inner \(p(a)\) first.

First evaluate the inner function: p(0)=6(0)+5=5. Then p(5)=6(5)+5=35. Therefore, p(p(0))=35. Option 30 results from calculating only 6×5 and forgetting to add the constant term 5. Exam tip: In a composite function, always evaluate the innermost input first.

If \(x=1\) is a zero of \((n+6)x+4n-3\), substituting \(x=1\) must make the polynomial equal to 0. Thus, \((n+6)(1)+4n-3=0\), or \(5n+3=0\). Hence, \(n=-\frac{3}{5}\). For example, \(n=-1\) gives the value \(-2\), not 0, so it is incorrect. Exam tip: substitute the given zero into the polynomial and equate the result to 0.

Given \(p(x)=\frac{x}{8}+a\), substitute \(x=24\): \(p(24)=\frac{24}{8}+a=3+a\). Since \(p(24)=17\), we get \(3+a=17\), so \(a=14\). Option 17 is the given value of \(p(24)\), not the value of \(a\). Exam tip: When a polynomial value is given, first substitute the specified value of \(x\) into the expression.

First add the polynomials: \(p(x)+q(x)=7x+a+4x-12=11x+a-12\). Since 2 is a zero, put \(x=2\): \(22+a-12=0\). Thus \(a+10=0\), so \(a=-10\). If \(a=0\), the value of the polynomial at \(x=2\) would be 10, not zero. Exam tip: when a zero is given, substitute that value in the polynomial and set the result equal to 0.

If \(-4\) is a zero of \(p(x)=(c-7)x+8c\), then \(p(-4)=0\). Thus, \(-4(c-7)+8c=0\). On simplifying, \(-4c+28+8c=0\), so \(4c+28=0\), which gives \(c=-7\). For \(c=7\), the constant term is \(56\), so \(-4\) cannot be a zero. Exam tip: for a given zero \(a\), always substitute it and use \(p(a)=0\).