Linear polynomials

First, \(p(0)=10\times 0+a=a\). Hence, \(p(p(0))=p(a)=10a+a=11a\). Given \(11a=110\), we get \(a=10\). Option 11 may result from incorrectly taking the coefficient in \(p(a)=11a\) as the value of \(a\). Exam tip: In a composite function, evaluate the inner function first.

Given \(p(x)=mx+n\), we get \(p(-1)=-m+n=3\) and \(p(4)=4m+n=23\). Subtracting the first equation from the second gives \(5m=20\), so \(m=4\). Then \(-4+n=3\) gives \(n=7\). Therefore, \(m+n=4+7=11\). Option 7 is only the value of \(n\), not of \(m+n\). Exam tip: Convert each given value of a linear polynomial into an equation before solving.

Given \(p(x)=11x-6\), we get \(p(x+1)=11(x+1)-6=11x+5\) and \(p(x-2)=11(x-2)-6=11x-28\). Hence, \(p(x+1)+p(x-2)=(11x+5)+(11x-28)=22x-23\). The option \(22x-17\) results from an incorrect addition of the constant terms. Exam tip: simplify each substituted expression separately before adding them.

A polynomial of the form ax+b is linear when the coefficient a of x is nonzero. In the given expression, that coefficient is r²−5r+6. Therefore, the expression will stop being linear precisely when r²−5r+6=0. Factorising gives r²−5r+6=(r−2)(r−3). Hence (r−2)(r−3)=0, so r=2 or r=3. For either value, the x-term disappears and the expression becomes the constant polynomial 4, whose degree is 0 rather than 1. Thus option A is correct. Option B does not satisfy the coefficient equation, option C gives only one valid value, and option D is false because two values exist.

On opening the brackets, \(4(3x-2)=12x-8\) and \(-5(2x+7)=-10x-35\). Therefore, \(12x-8-10x-35=2x-43\). Hence, the required linear polynomial is \(2x-43\). Option \(2x-27\) results from incorrectly combining the constant terms \(-8\) and \(-35\). Exam tip: When a negative coefficient multiplies a bracket, apply it to every term and check the signs carefully.

In \(7x-3\), the highest power of x is 1, so its degree is 1 and it is linear. \(x^2+7\) is quadratic, while \(5/x\) and \(\sqrt{x}\) are not polynomials. Exam tip: in a linear polynomial, variable powers can only be 0 or 1.

Since 4 is a zero of the polynomial, \(p(4)=0\). Thus, \(4(k+2)-20=0\), so \(4k+8=20\). Hence, \(4k=12\) and \(k=3\). If \(k=2\), then \(p(4)=4(2+2)-20=-4\), so it cannot be the required value. Exam tip: When a zero is given, substitute it for \(x\) and equate the polynomial to 0.

\(2p(x)-3q(x)=2(7x-3)-3(4x+5)\). On expanding, \(14x-6-12x-15=2x-21\). Hence, \(2x-21\) is the correct answer. \(2x+21\) can result from failing to distribute the negative sign in \(-3(4x+5)\) to both terms. Exam tip: when a negative coefficient multiplies a bracket, check the sign of every term carefully.

If \(-2\) is a zero of the polynomial \((3a-4)x+16\), its value must be 0 when \(x=-2\). Thus, \((3a-4)(-2)+16=0\). This gives \(-6a+8+16=0\), or \(-6a+24=0\), so \(a=4\). For example, if \(a=2\), the polynomial evaluates to \(12\), not 0. Exam tip: Substitute the given zero into the polynomial and set the result equal to 0.

Since 6 is a zero of the polynomial, \(p(6)=0\). Therefore, \(6a+b=0\). Substituting \(a=-3\) gives \(6(-3)+b=0\), so \(b=18\). \(-18\) is a common distractor caused by missing the sign change. In exams, substitute the given zero in the polynomial and set the result equal to 0.

\(p(5)=9\times5+c=45+c\), while \(p(1)=9\times1+c=9+c\). Hence, \(p(1)+36=9+c+36=45+c=p(5)\). Since \(c\) occurs equally on both sides, its value cancels out; therefore, the condition holds for every \(c\). Values such as \(c=0\) and \(c=36\) are only particular cases, not requirements. Exam tip: substitute the given input values and check whether the constant term cancels.

The zero of a linear polynomial is the value of x that makes the polynomial equal to zero. For an expression of the form ax+b, solve ax+b=0, giving x=-b/a. To obtain the required zero \(-\frac{8}{11}\), the polynomial must have a positive constant 8 and coefficient 11, so that solving the equation produces a negative fraction with denominator 11.

For option D, set 11x+8=0. Subtracting 8 gives 11x=-8, and dividing by 11 gives x=-\frac{8}{11}. Thus its zero is exactly the required value. Option A would give \(\frac{8}{11}\), while the other options produce different values, so option D is correct.

From (\frac{7}{10}x-14=0), (\frac{7}{10}x=14) and (x=20). With a fractional coefficient, multiply by the reciprocal.

Since \(0x^2=0\), that term contributes nothing to the polynomial. The expression becomes \(7-3x\), whose highest power is 1, so it is linear. Exam tip: first remove all terms with zero coefficients before finding degree.

Given \(p(x)=8x-15\), replace every \(x\) in the expression by the complete quantity \((x-2)\): \(p(x-2)=8(x-2)-15=8x-16-15=8x-31\). Hence, \(8x-31\) is correct. \(8x-17\) results from an incorrect simplification of the constant terms. Exam tip: always use brackets when substituting an expression into a function.

Replace x in p with the complete expression \(3x\). Thus, \(p(3x)=17-6(3x)=17-18x\). In option B, the constant term 17 is incorrectly multiplied by 3; only x is replaced by \(3x\). Exam tip: In \(p(kx)\), replace every x by \(kx\), not the constant term.

Writing \(\frac{2}{x}=2x^{-1}\) gives an exponent of \(-1\). In a polynomial, variable exponents must be 0, 1, 2, …, so this is not linear. Exam tip: check for variables in denominators first.

From the condition \(p(6)=p(-2)\), we get \(6k+11=-2k+11\). Hence, \(8k=0\), so \(k=0\). Then \(p(x)=11\), which is a constant polynomial of degree 0, not a linear polynomial. Option A satisfies the given equality, but it does not keep the polynomial linear. Exam tip: \(ax+b\) is linear only when \(a\ne0\).

First simplify the expression: \(q(x)-2p(x)=(13x-1)-2(6x+4)=13x-1-12x-8=x-9\). The highest power of \(x\) in \(x-9\) 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: simplify like terms before finding the degree of a sum or difference of polynomials.

To find the zero of the linear polynomial \(9x-23\), set it equal to zero: \(9x-23=0\). Thus, \(9x=23\), so \(x=\frac{23}{9}\). Therefore, \(9x-23\) is the correct option. The close distractor \(23x-9\) has zero \(\frac{9}{23}\), not the given value. Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\).