Linear polynomials

p(3)=4×3+7=19 and p(1)=4×1+7=11. Therefore, p(3)-p(1)=19-11=8. Option 4 is only the coefficient, not the required difference. Exam tip: substitute each given value of x carefully before subtracting the polynomial values.

For a polynomial to be linear, the coefficient of \(x\) must be non-zero. Here, the coefficient of \(x\) is \(m+4\). When \(m=-4\), \(m+4=0\), so the expression becomes \(-9\), a constant polynomial rather than a linear polynomial. For example, when \(m=0\), the coefficient is \(4\), so the polynomial is still linear. Exam tip: In parameter-based linear polynomials, set the coefficient of the variable equal to zero to find when it stops being linear.

A zero of a polynomial is a value that makes the polynomial equal to 0. Here, \(6x-18=0\) gives \(6x=18\), so \(x=3\). Therefore, the correct answer is 3. Substituting \(-3\) gives \(6(-3)-18=-36\), so it is not a zero. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\).

On opening the bracket, \(5(x-2)=5x-10\). Therefore, \(5x-10-3x+4=2x-6\). Hence, the correct polynomial is \(2x-6\). The close distractor \(2x+6\) can result from incorrectly adding the constant terms \(-10\) and \(+4\). Exam tip: first use the distributive property to open brackets, then combine like terms.

A zero of 3 means that the polynomial becomes 0 when x=3. Thus, p(3)=3a+12=0. Hence, 3a=-12 and a=-4. If a=4, then 3(4)+12=24, not 0. Exam tip: When a zero is given, substitute that value for x and set p(x)=0.

In \(5x-3x+5\), \(5x\) and \(-3x\) are like terms. Combining them gives \(2x\), so the expression becomes \(2x+5\). Option B has the constant term \(-5\), while option C contains \(x^2\), so it is not linear. Exam tip: Combine only terms with the same variable and exponent.

First add the polynomials: \(p(x)+q(x)=(3x-5)+(x+7)=4x+2\). The highest power of \(x\) is 1, so the degree of the sum is 1. Degree 0 would apply only if the sum were a non-zero constant polynomial. Exam tip: after adding polynomials, combine like terms and identify the highest exponent in the result.

p(x)-q(x)=(5x+4)-(5x-6)=5x+4-5x+6=10. No term containing x remains, so the result is a non-zero constant polynomial. It is not a zero polynomial, which would require the result to be 0. Exam tip: while subtracting polynomials, change the sign of every term in the second polynomial.

If \((-2)\) is a zero of \(((k-7)x+14)\), substituting \(x=-2\) must make the polynomial equal to 0: \((k-7)(-2)+14=0\). Thus, \(-2k+14+14=0\), or \(-2k+28=0\), which gives \(k=14\). For \(k=7\), the coefficient of \(x\) becomes 0 and the expression is 14, so it has no zero. Exam tip: When a zero is given, substitute it for \(x\) and equate the polynomial to 0.

Given p(x)=2x-7, substitute -3 for x: p(-3)=2(-3)-7=-6-7=-13. Therefore, -13 is correct. The distractor -1 may result from incorrectly treating 2(-3) as 6. Exam tip: when substituting a negative value, write each sign step separately.

A zero of a polynomial is a value of x that makes the polynomial equal to 0. Substituting \(x=\frac{5}{4}\) in \(4x-5\) gives \(4\times\frac{5}{4}-5=5-5=0\). Hence, \(4x-5\) has zero \(\frac{5}{4}\). The closest distractor, \(4x+5\), has zero \(-\frac{5}{4}\). Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\).

Given \(p(x)=6x+c\). Substituting \(x=2\), we get \(p(2)=6(2)+c=12+c\). Since \(p(2)=17\), \(12+c=17\), so \(c=5\). If 7 were chosen, then \(p(2)=19\), not the given value 17. Exam tip: while evaluating a polynomial, substitute the given value of \(x\) carefully in every term.

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.