Linear polynomials

In \(9s+6\), the highest exponent of \(s\) is \(1\), so it is a linear polynomial. \(s^2-1\) is also a polynomial, but its degree is \(2\). \(s^{-1}+4\) has a negative exponent, while \(\sqrt{s}+3\) has \(s\) raised to \(\tfrac{1}{2}\); hence neither is a polynomial. Exam tip: exponents of variables in a polynomial must be non-negative integers such as \(0,1,2,\ldots\).

In \(C(x)=12x+50\), the highest exponent of the variable \(x\) is 1, and the coefficient of \(x\), 12, is non-zero. Therefore, it is a linear polynomial of degree 1. The number 50 is only the constant term; it does not make the entire expression constant. Exam tip: find a polynomial’s degree by identifying the highest exponent of its variable.

Open the brackets first: \(3(2x-5)=6x-15\) and \(-2(5x+1)=-10x-2\). Therefore, \(6x-15-10x-2=-4x-17\). Hence, the correct polynomial is \(-4x-17\). In \(-4x+17\), the sign of the constant term is incorrect. Exam tip: when a negative multiplier precedes a bracket, apply it to every term inside the bracket.

\(3x+6=3(x+2)\), so its zero is -2 and the highest power of x is 1; hence it is linear. \(x^2+4x+4\) also has zero -2, but it is quadratic. Exam tip: check the degree before selecting the expression.

Since \(a\ne0\), solving \(ax+b=0\) gives the single value \(x=-b/a\). Hence, a non-constant linear polynomial has exactly one zero. Exam tip: if \(a=0\), the expression is not linear.

\(3p(x)-4q(x)=3(5x-4)-4(2x+7)\). Expanding gives \(15x-12-8x-28=7x-40\), so \(7x-40\) is correct. \(7x+16\) results from failing to apply the negative sign to both terms of \(4(2x+7)\). Exam tip: when a minus sign precedes brackets, change the sign of every term inside them.

A zero equal to 3 means that the polynomial must have value 0 when \(x=3\). Thus, \(3(2a-3)-9=0\). Simplifying gives \(6a-18=0\), so \(a=3\). For example, if \(a=1\), the value is \(-12\), so it is not correct. Exam tip: Substitute the given zero into the polynomial and set its value equal to zero.

Since \(-4\) is a zero of the polynomial, \(p(-4)=0\). Thus, \(-4a+b=0\). Substituting \(a=3\) gives \(-12+b=0\), so \(b=12\). Taking \(-12\) would not satisfy the equation. Exam tip: if \(r\) is a zero, always use \(p(r)=0\).

We have \(p(6)=42+c\) and \(p(2)=14+c\). Therefore, \(p(6)-p(2)=(42+c)-(14+c)=28\). The constant \(c\) cancels on subtraction, so the given condition is true for every value of \(c\). Values such as \(c=0\) or \(c=28\) are merely particular cases, not necessary conditions. Exam tip: write both function values and subtract them; a common constant term often cancels.

Set each candidate polynomial equal to zero to identify its zero. For option C, 9x+4=0 leads to 9x=-4 and hence x=-4/9. Thus option C is correct. Option A has zero 4/9, while options B and D have zero -9/4. The coefficient of x becomes the denominator of the fractional zero after rearranging.

The direct answer is option B: \(x=12\). To find the zero of the polynomial, set its value equal to zero: \(\frac{5}{6}x-10=0\). Add 10 to both sides, giving \(\frac{5}{6}x=10\). Multiply both sides by the reciprocal \(\frac{6}{5}\): \(x=10\times\frac{6}{5}=12\). Substitution checks it: \(p(12)=\frac{5}{6}(12)-10=10-10=0\). Option A, \(x=10\), gives \(\frac{25}{3}-10\ne0\). Option B works exactly. Option C, \(x=15\), gives \(12.5-10=2.5\), not zero. Option D, \(x=20\), gives \(\frac{50}{3}-10\ne0\). The coefficient \(\frac56\) is undone by multiplying by \(\frac65\), not by simply ignoring the fraction. Memory cue: for a linear polynomial, set it to zero and isolate the variable.

Both p(x) and q(x) are linear polynomials, so each has degree 1. Their product is (x-7)(x+4)=x^2-3x-28. The highest power of x is 2, so the degree of the product is 2. Option 1 is the degree of each individual polynomial, not of their product. Exam tip: For non-zero polynomials, the degree of a product equals the sum of their degrees.

Substitute the entire expression \(x+2\) for \(x\): \(p(x+2)=6(x+2)-11=6x+12-11=6x+1\). Hence, \(6x+1\) is correct. The option \(6x-9\) results from not multiplying \(2\) correctly by \(6\). Exam tip: always put the substituted expression in brackets.

Given \(p(x)=15-5x\), substitute the entire input \(2x\) for \(x\): \(p(2x)=15-5(2x)=15-10x\). Hence, option B is correct. In option C, \(x\) has not been replaced at all. Exam tip: in \(p(2x)\), replace only the input \(x\) by \(2x\); the constant term 15 remains unchanged.

For a linear polynomial, setting \(ax+b=0\) gives \(x=-\frac{b}{a}\). Since \(a\ne0\), this is one definite real value, so there is exactly one zero. Exam tip: a polynomial with two distinct zeroes cannot be linear.

Given \(p(5)=p(-1)\), we get \(5k-8=-k-8\). Hence \(6k=0\), so \(k=0\). Then \(p(x)=-8\) is a constant polynomial of degree 0, not a linear polynomial. In option B, \(k\ne0\) would make the polynomial linear, but it would not satisfy \(p(5)=p(-1)\). Exam tip: \(ax+b\) is linear only when \(a\ne0\).

In \(7-\frac{x}{2}=-\frac12x+7\), the highest power of \(x\) is 1, so it is linear. Fractional coefficients are allowed; exponents must be non-negative integers. Exam tip: use the highest exponent, not the number of terms.

First simplify the expression: \(q(x)-2p(x)=(11x-2)-2(5x+3)=11x-2-10x-6=x-8\). The highest power of \(x\) in \(x-8\) is 1, so its degree is 1. Degree 0 would apply only to a non-zero constant polynomial; here the \(x\)-term does not cancel. Exam tip: When adding or subtracting linear polynomials, simplify first and check whether the variable terms cancel.

To find the zero of \(7x-19\), set \(7x-19=0\). This gives \(7x=19\), so \(x=\frac{19}{7}\). Hence, option A is correct. Option B, \(19x-7\), has zero \(\frac{7}{19}\), not the given value. Exam tip: The zero of \(ax+b\) is \(-\frac{b}{a}\).

Given \(p(x)=8x-13\), we get \(p(x+1)=8(x+1)-13=8x-5\). Therefore, \(p(x+1)-p(x)=(8x-5)-(8x-13)=8\). The option \(8x\) 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{15}{\lambda}). From (-\frac{15}{\lambda}>-5) and (\lambda>0), we get (\lambda>3).

In \(p(x)+q(x)=(x+3r)+(x-5r)\), combine like terms: \(x+x=2x\) and \(3r-5r=-2r\). Hence, the sum is \(2x-2r\). The expression \(x-2r\) incorrectly leaves out one \(x\) term. Exam tip: while adding polynomials, combine only terms with the same variable and exponent.

\(p(x)-q(x)=(x+3r)-(x-5r)=x+3r-x+5r=8r\). The terms containing \(x\) cancel, so no power of \(x\) remains. Hence, treating \(r\) as a constant, \(8r\) is a constant polynomial in \(x\). A linear polynomial must contain an \(x\)-term, so option A is not correct. Exam tip: identify the variable with respect to which the polynomial is being classified.

First evaluate the inner function: \(p(3)=6\times3-17=1\). Then substitute this result into \(p\): \(p(p(3))=p(1)=6\times1-17=-11\). Therefore, the correct answer is \(-11\). Option 1 is only the value of \(p(3)\), not of \(p(p(3))\). Exam tip: In a composite function, always evaluate the expression inside first.

First evaluate the inner function: \(p(0)=4(0)+3=3\). Then substitute this result into \(p\): \(p(p(0))=p(3)=4(3)+3=15\). Option 12 is only \(4\times3\), which misses the constant term 3. Exam tip: in a composite function, always evaluate the innermost function first.