Linear polynomials

Because \(x\) is in the denominator, the original expression is undefined at \(x=0\), so it is not a polynomial. It equals \(2x-6\) only for \(x\ne0\). Exam tip: always check for variables in the denominator before cancelling.

Since 4 is a zero of the polynomial, \(p(4)=0\). Thus, \((k+3)\times4-12=0\), so \(4k+12-12=0\). Hence \(4k=0\) and \(k=0\). If \(k=3\), then \(p(4)=12\), not zero. Exam tip: If \(a\) is given as a zero, substitute it and use \(p(a)=0\).

We have \(3p(x)-5q(x)=3(4x-7)-5(2x+3)\). Thus, \(=12x-21-10x-15=2x-36\). Hence, \(2x-36\) is correct. The distractor \(22x-36\) can result from incorrectly handling the negative sign while expanding \(-5(2x+3)\). Exam tip: distribute a negative multiplier to every term inside the bracket.

A zero is a value of \(x\) that makes the polynomial equal to 0. Substituting \(x=-2\) in \((3m+1)x+8\) gives \((3m+1)(-2)+8=0\). So, \(-6m-2+8=0\), or \(-6m+6=0\), hence \(m=1\). For instance, if \(m=-1\), the polynomial evaluates to 12, not 0. Exam tip: substitute the given zero for \(x\) first, then solve the resulting equation.

Putting \(p(x)=0\) gives \(ax+b=0\), so \(x=-\frac{b}{a}\). Since \(a\ne0\), this is one definite real number; hence there is exactly one zero. Two distinct zeros may occur for a quadratic polynomial. Exam tip: a linear polynomial has degree 1.

\(p(4)=5\times4+c=20+c\), while \(p(1)=5\times1+c=5+c\). Hence, \(p(1)+15=(5+c)+15=20+c=p(4)\). The \(c\)-term occurs equally on both sides, so the condition is true for every real value of \(c\). \(c=0\) and \(c=15\) are only particular values, not necessary ones. Exam tip: substitute the given \(x\)-values carefully and simplify the constant terms before deciding the parameter's value.

To test a proposed zero, set the polynomial equal to zero and solve for x. For option C, 8x+5=0 gives 8x=-5 and therefore x=-5/8. Thus option C is correct. Option A gives x=5/8, while options B and D both give x=-8/5, so their signs or numerator and denominator are not in the required arrangement.

Both given polynomials are linear. On multiplying,
\((x-3)(x+8)=x^2+5x-24\). The highest-power term with a non-zero coefficient is \(x^2\), 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 two non-zero polynomials equals the sum of their degrees.

Given \(p(x)=4x-9\), substitute the entire expression \((x+2)\) for \(x\): \(p(x+2)=4(x+2)-9=4x+8-9=4x-1\). Hence, \(4x-1\) is correct. \(4x-7\) results from incorrectly handling the constant term \(-9\). Exam tip: always enclose the substituted expression in brackets.

Given \(p(x)=11-2x\), replace the entire input \(x\) by \(3x\): \(p(3x)=11-2(3x)=11-6x\). Hence, option B is correct. \(33-6x\) would result from incorrectly multiplying the constant term 11 by 3. Exam tip: In \(p(3x)\), substitute \(3x\) only for the variable \(x\).

The student is wrong. In \(7-\sqrt{5}x\), the highest exponent of \(x\) is 1, so it is linear. \(\sqrt{5}\) is a real coefficient; in exams, determine degree from the variable’s exponent.

Using \(p(3)=p(-1)\), we get \(3k-5=-k-5\). Hence \(4k=0\), so \(k=0\). Then \(p(x)=-5\), which is a constant polynomial of degree 0, not a linear polynomial. Taking \(k=5\) does not satisfy the given condition. Exam tip: for a polynomial to be linear, the coefficient of \(x\) must be non-zero.

In \(p(x)=ax+b\), the highest power of \(x\) is 1, so its graph is a straight line. A parabola comes from a quadratic polynomial. Exam tip: degree 1 indicates a linear graph.

First simplify the expression: \(q(x)-2p(x)=(7x-4)-2(3x+2)=7x-4-6x-4=x-8\). The highest power of \(x\) is 1, so the degree of the polynomial is 1. Degree 0 would apply only to a non-zero constant polynomial; here the \(x\)-term remains. Exam tip: simplify fully before identifying the highest exponent.

A zero of a polynomial is a value of \(x\) for which the polynomial becomes 0. Setting \(6x-13=0\) gives \(6x=13\), so \(x=\frac{13}{6}\). Therefore, \(6x-13\) is the correct option. The zero of \(13x-6\) is \(\frac{6}{13}\), so it is a close but incorrect distractor. Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\).

Given \(p(x)=5x-6\), we get \(p(x+1)=5(x+1)-6=5x-1\). Hence, \(p(x+1)-p(x)=(5x-1)-(5x-6)=5\). Therefore, 5 is the correct answer. Choosing \(5x\) ignores the cancellation of the variable terms and the difference between the constant terms. Exam tip: for a linear polynomial \(ax+b\), \(p(x+1)-p(x)=a\), the coefficient of \(x\).

The zero is (-\frac{10}{\lambda}). From (-\frac{10}{\lambda}<-2) and (\lambda>0), we get (0<\lambda<5).

x+5 is linear because the highest power of x is 1. Setting x+5=0 gives x=-5, which is negative. x^2+5 is not linear. Exam tip: first check that the highest power of x is 1.

\(p(x)+q(x)=(x+s)+(x-2s)\). Combining like terms gives \(x+x=2x\) and \(s-2s=-s\). Therefore, the sum is \(2x-s\). \(2x+s\) is incorrect because \(s-2s=-s\), not \(s\). Exam tip: Remove brackets first, then combine terms with the same variable or parameter.

On subtraction, \(p(x)-q(x)=(x+s)-(x-2s)=3s\). There is no term containing \(x\), so with respect to \(x\), it is a constant polynomial. A linear polynomial must have a non-zero coefficient of \(x\), but the \(x\)-terms cancel here. Exam tip: simplify like terms first, then identify the highest power of \(x\). If \(s=0\), the result is the zero polynomial; the intended answer generally treats \(s\) as non-zero and is therefore a constant polynomial.

First evaluate the inner function: p(3)=3×3−7=2. Now substitute this value into p again: p(2)=3×2−7=−1. Therefore, p(p(3))=−1. Option 2 is only the value of p(3), not the final value. Exam tip: In a composite function, always evaluate the innermost expression first.

First evaluate the inner function: \(p(1)=4(1)+1=5\). Then \(p(5)=4(5)+1=21\). Therefore, \(p(p(1))=21\). Option 20 results from taking only \(4\times5\) and missing the constant term \(+1\). Exam tip: In a composite function, always evaluate the innermost input first.

A zero equal to 1 means that the polynomial must have value 0 when \(x=1\). Thus, \((n+2)(1)+3n-1=0\), which gives \(4n+1=0\). Hence, \(n=-\frac{1}{4}\). For instance, with \(n=-1\), the value of the polynomial is \(-4\), so it is not correct. Exam tip: When a zero is given, substitute that value for \(x\) and set the expression equal to 0.

Given \(p(x)=\frac{x}{4}+a\). Substituting \(x=8\) gives \(p(8)=\frac{8}{4}+a=2+a\). Since \(p(8)=7\), we get \(2+a=7\), so \(a=5\). Option 7 is the given value of \(p(8)\), not the value of \(a\). Exam tip: First substitute the given value of \(x\) into the polynomial, then solve the resulting equation.

\(p(x)+q(x)=(3x+a)+(2x-5)=5x+a-5\). Since its zero is \(1\), put \(x=1\): \(5(1)+a-5=0\). Hence, \(a=0\). If \(a=1\), the polynomial has value \(1\), not zero, at \(x=1\). Exam tip: if \(r\) is a zero of a polynomial, substituting \(x=r\) must make the polynomial equal to \(0\).