Linear polynomials

\(x^2+1\) is a polynomial because the powers of \(x\) are non-negative integers. Its highest power is 2, so its degree is 2 and it is not a linear polynomial. \(4x-9\) is linear because its degree is 1. \(\frac{1}{x}+2\) and \(\sqrt{x}+1\) are not polynomials because they contain powers \(-1\) and \(\frac{1}{2}\), respectively. Exam tip: in a polynomial, the exponent of a variable must be a non-negative integer.

A non-zero linear polynomial has the form \(ax+b\), where \(a\ne0\). From \(ax+b=0\), we get \(x=-b/a\), so there is exactly one zero. Exam tip: first check that the coefficient of \(x\) is non-zero.

The point where the graph meets the x-axis represents a zero of the polynomial. Putting \(p(x)=0\) gives \(ax+b=0\), so \(x=-\frac{b}{a}\). Since \(a\ne0\), this is one definite real value; hence the line intersects the x-axis at exactly one point. Option B can occur for a quadratic polynomial, not for a linear polynomial. Exam tip: because \(a\ne0\), a linear polynomial always has exactly one zero.

In \(\sqrt{2}x-\frac{3}{5}\), the highest power of the variable \(x\) is 1, and \(\sqrt{2}\) is a real coefficient. Hence, it is a linear polynomial in \(x\). \(x^2-1\) is quadratic, while \(\frac{1}{x}+2\) has power \(-1\) of \(x\) and \(\sqrt{x}+1\) has power \(\frac{1}{2}\); therefore, these are not polynomials. Exam tip: powers of a variable in a polynomial must be non-negative integers such as 0, 1, 2, ... .

On opening the brackets, \(5(2x-3)=10x-15\) and \(-3(3x+1)=-9x-3\). Therefore, \(10x-15-9x-3=x-18\). Hence, the correct polynomial is \(x-18\). In \(10x-18\), the \(-9x\) term has not been combined. Exam tip: When opening a bracket preceded by a negative sign, change the sign of every term inside it.

For a polynomial to be linear in x, the coefficient of x must never be zero. In option A, \(p^2+1>0\) for every real p, so it always has degree 1. In option B, the coefficient becomes zero at \(p=0\) or \(1\). Exam tip: always test whether a parameter-dependent coefficient can vanish.

A zero of a polynomial is a value for which the polynomial becomes 0. So, put \(p(-3)=0\): \((k-2)(-3)+9=0\). This gives \(-3k+6+9=0\), or \(-3k+15=0\), hence \(k=5\). If \(k=2\), the polynomial reduces to the constant 9, so it cannot have \(-3\) as a zero. Exam tip: When a zero is given, substitute it for \(x\) and equate the polynomial to 0.

\(2p(x)-3q(x)=2(3x-2)-3(x+5)\). On expanding, \(6x-4-3x-15=3x-19\). Hence, the correct expression is \(3x-19\). The option \(x-19\) results from incorrectly simplifying \(6x-3x\) as \(x\). Exam tip: When a minus sign occurs before brackets, apply it to every term inside the bracket.

If \(x=4\) is a zero, substituting \(x=4\) in \((2m-5)x+10\) must give zero. Thus, \(4(2m-5)+10=0\). On simplifying, \(8m-20+10=0\), so \(8m=10\) and hence \(m=\frac{5}{4}\). For \(m=\frac{5}{2}\), the coefficient of \(x\) becomes zero and the expression is \(10\), which has no zero. Exam tip: Substitute the given zero in the polynomial and equate the result to zero.

A zero of 2 means \(p(2)=0\). Therefore, \(2a+b=0\). Substituting \(a=5\) gives \(2\times5+b=0\), so \(b=-10\). If \(b=10\), then \(p(2)=20\), not zero. Exam tip: for a zero \(r\) of a linear polynomial \(ax+b\), write \(ar+b=0\).

A linear polynomial must have highest power 1, so the coefficient \(a\) of \(x\) must be non-zero. \(b\) may be zero; \(b\ne0\) alone does not ensure linearity. Exam tip: check the coefficient of \(x\) first.

For \(p(x)=4x+c\), we have \(p(3)=12+c\) and \(p(1)=4+c\). Hence, \(p(3)-p(1)=(12+c)-(4+c)=8\), but the given condition requires this difference to be 10. The constant \(c\) cancels out, so no value of \(c\) can satisfy the condition. Option B is incorrect because the difference remains 8 for every \(c\). Exam tip: When comparing two values of a linear polynomial, the constant term often cancels out.

Both given polynomials are linear, so each has degree 1. On multiplying,
\((x+2)(x-5)=x^2-3x-10\). The highest power of the variable is 2, so the degree of the product is 2. Option 1 is the degree of each individual linear polynomial, not of their product. Exam tip: for non-zero polynomials, the degree of a product is generally the sum of their degrees.

Replace the entire variable \(x\) in the polynomial with \(x+3\). Thus, \(p(x+3)=2(x+3)-1=2x+6-1=2x+5\). Therefore, \(2x+5\) is correct. The option \(x+5\) incorrectly drops the coefficient \(2\). Exam tip: always use brackets when substituting an expression for a variable.

Replace the variable \(x\) in the polynomial by the complete expression \(2x\): \(p(2x)=5-3(2x)=5-6x\). Therefore, \(5-6x\) is correct. \(10-6x\) would result from incorrectly multiplying the constant term 5 by 2 as well. Exam tip: To find \(p(kx)\), substitute \(kx\) only for every \(x\).

\(\frac{3}{x}\) can be written as \(3x^{-1}\). A polynomial permits only non-negative integer powers such as 0, 1, and 2, so this expression is not a polynomial. Exam tip: check every exponent before finding the degree.

Given \(p(2)=p(-2)\), we get \(2k+4=-2k+4\). Thus \(4k=0\), so \(k=0\). Then \(p(x)=4\), which is a constant polynomial, not a linear polynomial. Hence, under the given condition, there is no value of \(k\) for which \(p(x)\) remains linear. Exam tip: \(ax+b\) is linear only when \(a\ne0\).

For \(a\ne0\), \(ax+b=0\) gives one value, \(x=-b/a\). Thus a non-constant linear polynomial has exactly one zero. A constant polynomial is excluded. Exam tip: first verify \(a\ne0\).

First simplify the expression: \(q(x)-2p(x)=(5x-1)-2(2x+3)=5x-1-4x-6=x-7\). The highest power of \(x\) in \(x-7\) is 1, with a non-zero coefficient, so its degree is 1. Option 0 would apply only to a constant polynomial; here the \(x\)-term does not cancel. Exam tip: simplify and combine like terms before deciding the degree of a polynomial.

A zero of a polynomial is a value of x that makes the polynomial equal to 0. For \(5x-11=0\), we get \(5x=11\), so \(x=\frac{11}{5}\). Hence, \(5x-11\) has zero \(\frac{11}{5}\). The close distractor \(11x-5\) has zero \(\frac{5}{11}\), which is different. Exam tip: To find the zero of \(ax+b\), solve \(ax+b=0\).

First find p(x+1): p(x+1)=3(x+1)-8=3x-5. Therefore, p(x+1)-p(x)=(3x-5)-(3x-8)=3. Hence, the correct answer is 3. The option 3x is incorrect because the x-terms cancel on subtraction. Exam tip: for a linear polynomial ax+b, p(x+1)-p(x) is always a.

The zero is (-\frac{6}{\lambda}). From (-\frac{6}{\lambda}>-2) and (\lambda>0), we get (\lambda>3).

Since \(a\ne0\), the graph is a non-horizontal straight line with zero \(x=-\frac{b}{a}\). Hence it meets the x-axis exactly once. Exam tip: put \(p(x)=0\) to locate the x-intercept.

Given \(p(x)=x+r\) and \(q(x)=x-r\), \(p(x)+q(x)=(x+r)+(x-r)=x+x+r-r=2x\). The terms \(+r\) and \(-r\) cancel each other. \(2r\) would result only if the \(x\)-terms cancelled, which they do not. Exam tip: While adding polynomials, group like terms together first.

On subtraction, (p(x)-q(x))=(x+r)-(x-r)=2r. Since r is a constant, 2r has no term involving x; therefore, it is a constant polynomial in x. A linear polynomial must contain an x-term, so option A is not correct. Exam tip: while subtracting polynomials, change the sign of every term inside the second bracket. If r=0 specifically, the result is the zero polynomial; normally, r is taken as a non-zero constant here.