Linear polynomials

Given \(p(x)=6x-1\), we get \(p(x+3)=6(x+3)-1=6x+17\) and \(p(x-2)=6(x-2)-1=6x-13\). Therefore, \(p(x+3)-p(x-2)=(6x+17)-(6x-13)=30\). Option 18 is only the change for an input increase of 3; here the inputs differ by \((x+3)-(x-2)=5\), so the change is \(6\times5=30\). Exam tip: for a linear polynomial \(ax+b\), use \(p(u)-p(v)=a(u-v)\).

First evaluate the inner function: \(p(3)=4\times3-9=3\). Therefore, \(p(p(3))=p(3)=3\). Hence, the correct answer is 3. The value 9 is only the constant term and is not the value of the function here. Exam tip: In a composite function, always evaluate the inner function first.

Given \(p(x)=5x+2\), \(p(x+1)=5(x+1)+2=5x+7\) and \(p(x-1)=5(x-1)+2=5x-3\). Therefore, \(p(x+1)+p(x-1)=(5x+7)+(5x-3)=10x+4\). The option \(10x\) would result only if the constant terms added to zero, but here \(7+(-3)=4\). Exam tip: substitute \(x+1\) and \(x-1\) carefully in the complete polynomial before adding.

For it not to be linear, the coefficient of (x) must be (0). (r^2-2r=r(r-2)=0) gives (r=0) or (r=2).

Given \(p(x)=ax+3\), first find \(p(0)=a(0)+3=3\). Therefore, \(p(p(0))=p(3)=3a+3\). Using \(3a+3=18\), we get \(3a=15\), so \(a=5\). If \(a=6\), then \(p(p(0))=21\), not 18. Exam tip: In a composite expression such as \(p(p(0))\), evaluate the inner function first.

From the two equations, (4m=12), so (m=3). From (2m+n=13), (n=7) and (m-n=-4), which is not in the options.

On expanding the brackets, \(7(2x-5)=14x-35\) and \(-5(3x+4)=-15x-20\). Therefore, \(14x-35-15x-20=-x-55\). Hence, the correct polynomial is \(-x-55\). In \(-x+55\), the sign of the constant term is incorrect. Exam tip: when a minus sign is multiplied with a bracket, change the sign of every term inside it.

In \(P(x)=\sqrt{2}x-5\), the highest power of \(x\) is 1 and \(\sqrt{2}\) is a real coefficient, so it is linear. Exam tip: determine degree from the exponent, not from the number of terms.

For a linear polynomial, the coefficient of \(x^2\) must be zero and the coefficient of \(x\) must be non-zero. From \(a^2-9=0\), \(a=\pm3\). At \(a=-3\), the \(x\)-coefficient is also 0, so only \(a=3\) works. Exam tip: check both conditions.

\(4p(x)-3q(x)=4(2x+5)-3(3x-7)\). Thus, \(=8x+20-(9x-21)=8x+20-9x+21=-x+41\). Therefore, \(-x+41\) is correct. The option \(x+41\) has the wrong sign of the \(x\)-term, since \(8x-9x=-x\). Exam tip: When a minus sign precedes a polynomial, change the sign of every term inside it.

A zero of a polynomial is a value of the variable for which the polynomial becomes 0. Substituting \(x=2\), we get \(2(2m+3)-14=0\). Thus, \(4m+6-14=0\), so \(4m-8=0\), giving \(m=2\). For \(m=\frac{7}{2}\), the polynomial does not evaluate to 0 at \(x=2\). Exam tip: When a zero is given, substitute it for \(x\) and equate the polynomial to 0.

A zero equal to 5 means that \(p(5)=0\). Hence, \(5a+b=0\). Substituting \(a=-2\) gives \(5(-2)+b=0\), so \(b=10\). If \(b=-10\), then \(p(5)=-20\), not zero. Exam tip: for a zero \(r\) of the linear polynomial \(ax+b\), write \(ar+b=0\).

Here, \(p(5)=6\times5+c=30+c\) and \(p(2)=6\times2+c=12+c\). Therefore, \(p(2)+18=12+c+18=30+c=p(5)\). Since \(c\) occurs equally on both sides, the condition holds for every value of \(c\). Values such as \(c=0\) or \(c=18\) are only particular cases, not necessary conditions. Exam tip: when subtracting two values of a linear polynomial, the constant term cancels out.

A polynomial’s zero is found by setting the polynomial equal to zero and solving for x. For option D, 9x + 7 = 0. Subtracting 7 from both sides gives 9x = −7, and dividing by 9 gives x = −7/9. Hence option D is correct. Checking the distractors clarifies the possible errors: 9x − 7 = 0 gives x = 7/9, so its sign is positive. Both 7x + 9 = 0 and x + 9/7 = 0 give x = −9/7, which reverses the required fraction. Therefore only 9x + 7 has the stated zero. The coefficient and constant must be used in the equation exactly as written; changing their order changes the zero.

The governing idea is the zero of a polynomial: it is the input value that makes the polynomial equal to zero. Set p(x) equal to zero and solve: (4/9)x − 8 = 0. Add 8 to both sides, obtaining (4/9)x = 8. Multiply both sides by the reciprocal of 4/9, namely 9/4: x = 8 × 9/4 = 2 × 9 = 18. Verification is important: p(18) = (4/9)(18) − 8 = 8 − 8 = 0. Therefore option A is correct. Values such as 36 or 72 can result from mishandling the fractional coefficient, while 27 does not satisfy the original equation. The substitution check confirms the unique zero because the coefficient of x is nonzero.

Both p(x) and q(x) are linear polynomials, so each has degree 1. Their product is (x+6)(x-2)=x^2+4x-12. 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: the degree of the product of non-zero polynomials equals the sum of their degrees.

Given \(p(x)=5x-8\). To find \(p(x-3)\), replace every \(x\) with the complete expression \(x-3\): \(p(x-3)=5(x-3)-8=5x-15-8=5x-23\). Therefore, \(5x-23\) is correct. \(5x-11\) results from an incorrect simplification of the constant terms. Exam tip: while substituting in a function, replace the variable with the entire expression in brackets.

Given \(p(x)=13-4x\), substitute the entire expression \(2x\) for \(x\): \(p(2x)=13-4(2x)=13-8x\). Therefore, option B is correct. \(26-8x\) would result from multiplying the whole polynomial by 2, which is not required here. Exam tip: in \(p(kx)\), replace only the variable \(x\) by \(kx\).

Putting \(p(x)=0\) gives \(ax+b=0\), so \(x=-b/a\). Since \(a\ne0\), this is one definite real value; hence a linear polynomial has exactly one zero. Exam tip: first check that the coefficient of \(x\) is non-zero.

Using \(p(4)=p(-2)\), we get \(4k+7=-2k+7\). Hence \(6k=0\), so \(k=0\). Then \(p(x)=7\), which is a constant polynomial, not a linear polynomial. For a polynomial to be linear, the coefficient of \(x\) must be non-zero. Therefore, under the given condition, \(p(x)\) can never remain linear. Exam tip: a linear polynomial has the form \(ax+b\), where \(a\ne0\).