Linear polynomials

\(4p(x)-3q(x)=4(2x+5)-3(3x-7)\). 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.

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 \(a\ne0\).

A linear polynomial has the form \(ax+b\), where \(a\neq0\) and powers of \(x\) are non-negative integers. Here \(\frac{4}{x}=4x^{-1}\), so it is not a polynomial. Exam tip: if a variable is in the denominator, rewrite it using a negative exponent first.

First simplify the expression: \(q(x)-2p(x)=(9x-3)-2(4x+1)=9x-3-8x-2=x-5\). The highest power of \(x\) is 1, so its degree is 1. Option 0 is incorrect because \(x-5\) is not a constant polynomial. Exam tip: After adding or subtracting polynomials, determine the degree from the highest non-zero power in the simplified result.

A zero of a polynomial is a value that makes the polynomial equal to 0. For \(8x-17=0\), we get \(8x=17\), so \(x=\frac{17}{8}\). Hence, \(8x-17\) is the correct polynomial. Both \(17x-8\) and \(x-\frac{8}{17}\) have zero \(\frac{8}{17}\), not \(\frac{17}{8}\). Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\).

Given \(p(x)=7x-10\), we get \(p(x+1)=7(x+1)-10=7x-3\). Therefore, \(p(x+1)-p(x)=(7x-3)-(7x-10)=7\). The option \(7x\) 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 coefficient of \(x\).

The zero is (-\frac{12}{\lambda}). From (-\frac{12}{\lambda}>-3) and (\lambda>0), we get (\lambda>4).

A linear polynomial has zero \(x=-b/a\). Since \(a\ne0\), this is one definite real number, so its graph meets the x-axis exactly once. Exam tip: a non-constant linear polynomial has exactly one zero.

\(p(x)+q(x)=(x+2s)+(x-3s)\). Combining like terms gives \(x+x=2x\) and \(2s-3s=-s\). Therefore, the sum is \(2x-s\). In \(2x+s\), the coefficients of \(s\) have been added with the wrong sign. Exam tip: while adding polynomials, combine coefficients of like terms carefully, including their signs.

On subtracting, \(p(x)-q(x)=(x+2s)-(x-3s)=x+2s-x+3s=5s\). There is no term containing \(x\), so it is a constant polynomial with respect to \(x\). It becomes the zero polynomial only in the special case \(s=0\); normally, \(s\) is treated as a constant parameter. Exam tip: simplify and cancel like \(x\)-terms before identifying the degree of a polynomial.

First evaluate the inner function: \(p(3)=4\times3-11=1\). Now substitute this value into \(p\) again: \(p(1)=4\times1-11=-7\). Therefore, \(p(p(3))=-7\). The value \(1\) is only \(p(3)\), not the final answer. Exam tip: In a composite function, always evaluate the innermost function first.

First evaluate the inner function: p(0)=5(0)+2=2. Then p(p(0))=p(2)=5(2)+2=12. Therefore, the correct answer is 12. The value 10 results from taking only 5×2 and missing the constant term +2. Exam tip: In a composite function, always evaluate the inner function first.

If \(x=1\) is a zero of the polynomial, substituting \(x=1\) must make its value 0. Thus, \((n-3)(1)+2n+1=0\), so \(3n-2=0\). Hence, \(n=\frac{2}{3}\). For example, when \(n=1\), the polynomial has value 1 at \(x=1\), not 0. Exam tip: To find a parameter for a given zero, substitute that zero and equate the polynomial to 0.

Given \(p(x)=\frac{x}{5}+a\). Substituting \(x=15\), we get \(p(15)=\frac{15}{5}+a=3+a\). Since \(p(15)=11\), \(3+a=11\), so \(a=8\). If \(a=6\), then \(p(15)=9\), not 11. Exam tip: To use a given polynomial value, substitute the specified value of \(x\) first and then simplify.

Adding the polynomials gives (p(x)+q(x)=4x+a+x-6=5x+a-6). Since 2 is a zero, substitute x=2: (5×2+a-6=0), so (a+4=0). Hence, (a=-4). Option 4 results from a sign error. Exam tip: when a zero is given, substitute it into the polynomial and set the result equal to 0.

If \(-2\) is a zero of \(p(x)\), then \(p(-2)=0\). Thus, \((c+4)(-2)+2c=0\). On simplifying, \(-2c-8+2c=-8=0\), which is impossible. Hence, there is no value of \(c\). In particular, \(c=-4\) makes the polynomial equal to the constant \(-8\), which has no zero. Exam tip: to test whether \(a\) is a zero, always substitute it and use \(p(a)=0\).

If \(-1\) is a zero of the polynomial, then \(p(-1)=0\). Thus, \((c+4)(-1)+2c=0\), or \(-c-4+2c=0\). Hence, \(c-4=0\), giving \(c=4\). If \(c=-4\), the polynomial becomes \(-8\), which has no zero. Exam tip: When a zero is given, substitute it into the polynomial and set the result equal to zero.

Given \(p(x)=rx+3\) and \(p(4)=23\), we get \(4r+3=23\). Hence \(4r=20\) and \(r=5\). Now \(p(-2)=5(-2)+3=-10+3=-7\). Therefore, \(-7\) is correct. \(-10\) is only the product \(5\times(-2)\); the constant term \(+3\) must also be added. Exam tip: first use the given function value to find the unknown coefficient, then substitute the required input.