Linear polynomials

A zero of a polynomial is a value that makes the polynomial equal to 0. On substituting \(x=-4\), \(x+4=-4+4=0\); hence, \(x+4\) is the correct linear polynomial. In contrast, \(x-4\) gives \(-8\) at \(x=-4\), so it is not correct. Exam tip: the zero of \(x+a\) is \(-a\).

Opening the brackets gives \(3(x+1)+2(x-4)=3x+3+2x-8\). Combining like terms, \(3x+2x=5x\) and \(3-8=-5\), so the polynomial is \(5x-5\). In \(5x+5\), the sign of the constant term is incorrect. Exam tip: multiply the number outside each bracket by every term inside it.

To find the zero, set the polynomial equal to zero: \(9-3x=0\). Thus, \(3x=9\), so \(x=3\). Therefore, 3 is the correct answer. Substituting \(-3\) gives \(9-3(-3)=18\), not zero. In an exam, verify a zero by substituting the value back into the polynomial and checking whether the result is 0.

A linear polynomial has degree 1 and is of the form \(ax+b\), where \(a\ne0\). For \(p(x)=x-6\), \(p(0)=0-6=-6\), so option B is correct. Although \(p(x)=-6\) gives \(p(0)=-6\), it is a constant polynomial, not a linear one. Exam tip: To find \(p(0)\), substitute 0 for \(x\).

Given
p(t)=21
and
p(x)=4x+1
, we get
p(t)=4t+1
. Hence,
4t+1=21
, so
4t=20
and
t=5
. If
t=4
, then
p(4)=17
, not 21. Exam tip: substitute the given input into the polynomial and equate it to the stated value before solving.

In option C, \(x+5\) is a linear polynomial because its degree is 1. Setting \(x+5=0\) gives \(x=-5\), so its zero is negative. In contrast, the zero of \(x-6\) is \(6\), which is positive. Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\).

A zero of a polynomial is a value for which the polynomial equals 0. Thus, putting \(p(-2)=0\), we get \(3(-2)+a=0\). Hence, \(-6+a=0\), so \(a=6\). If \(a=-6\), then \(p(-2)=-12\), not 0. Exam tip: Substitute the given zero for \(x\) and equate the polynomial to 0.

q(x)-p(x)=(x+5)-(x+2)=x+5-x-2=3. Since the result has no x-term and 3 is a non-zero constant, it is a constant polynomial. A linear polynomial must contain a term with x to the power 1, but those terms cancel here. Exam tip: While subtracting polynomials, change the sign of every term in the second bracket.

Since \(4\) is a zero, the polynomial must have value \(0\) when \(x=4\). Thus, \(2(4)+b=0\), or \(8+b=0\). Therefore, \(b=-8\). If \(b=-4\), then \(2(4)-4=4\), not zero. Exam tip: Substitute the given zero into the polynomial and equate its value to \(0\).

Substituting \(x=6\), we get \(p(6)=\frac{2}{3}\times 6+5=4+5=9\). Hence, 9 is the correct option. Option 8 may result from incorrectly evaluating \(\frac{2}{3}\times6\) as 3; its value is 4. Exam tip: while evaluating a polynomial, substitute the given value for every \(x\), then perform multiplication before addition or subtraction.

The expression \((3a-6)x+1\) is linear only when the coefficient of \(x\) is non-zero. For it to stop being linear, \(3a-6=0\). Thus, \(3a=6\), so \(a=2\). At this value, the expression becomes \(1\), a constant polynomial rather than a linear polynomial. Exam tip: In parameter-based polynomials, set the coefficient of the highest-degree term equal to zero to find when the degree changes.

Given \(p(x)=8x-12\) and \(p(x)=4\), we get \(8x-12=4\). Adding 12 to both sides gives \(8x=16\), so \(x=2\). For example, \(x=1\) gives \(p(1)=-4\), so it is not correct. Exam tip: when a value of a polynomial is given, substitute that value for \(p(x)\) and solve the resulting equation.

For a linear polynomial \(ax+b\), substituting \(x=0\) gives \(b\), the constant term. For \(x+9\), we get \(0+9=9\), so it is correct. In contrast, \(x-9\) gives \(-9\), while \(9x\) gives \(0\). Exam tip: At \(x=0\), the value of a polynomial is its constant term.

Given p(x)=7x+c, substituting x=1 gives p(1)=7(1)+c=7+c. Since p(1)=2, we have 7+c=2, so c=-5. If c were 5, then p(1) would be 12, not 2. Exam tip: In polynomial-value questions, substitute the given value of x first and then solve the resulting equation.

Given p(6)=0, substitute x=6 in p(x)=x+a. This gives p(6)=6+a, so 6+a=0 and hence a=-6. If a were 6, then p(6) would be 12, not 0. Exam tip: on substituting a zero of a polynomial, the polynomial value must be 0.

Given p(x)=3x-2, p(2)=3(2)-2=4 and p(-2)=3(-2)-2=-8. Hence, p(2)+p(-2)=4+(-8)=-4. The option 4 is only the value of p(2), not the required sum. Exam tip: when substituting a negative value, take care of the sign of the term 3x.

In the polynomial \(5x+5\), the coefficient of \(x\) is \(5\). To find its zero, set \(5x+5=0\). This gives \(5x=-5\), so \(x=-1\). Although \(5x-5\) also has coefficient \(5\), its zero is \(1\), not \(-1\). Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\).

A linear polynomial has the form \(ax+b\), where \(a\ne0\). Here \(a=-2\) and \(b=7\), so the degree is 1. A negative coefficient is valid; only \(a=0\) makes it constant. Exam tip: check the highest power, not the sign of the coefficient.

Given \(p(x)=5x+8\), we get \(p(x)-8=(5x+8)-8=5x\). The highest power of \(x\) in \(5x\) is \(1\), so its degree is \(1\). Option \(0\) would apply only if a non-zero constant term alone remained. Exam tip: simplify the expression first, then identify the highest exponent of the variable.

Given p(x)=3x-4, p(x)-3x=(3x-4)-3x=-4. Since -4 is a non-zero constant and contains no term in x, it is a constant polynomial. It is not a zero polynomial, because a zero polynomial must equal 0. Exam tip: simplify first, then identify the polynomial type from the degree of the remaining expression.