Linear polynomials

Given \(p(x)=x-4\) and \(q(x)=2x+1\), \(2p(x)+q(x)=2(x-4)+(2x+1)=2x-8+2x+1=4x-7\). Hence, \(4x-7\) is correct. The option \(3x-7\) may result from incorrectly combining the \(x\)-terms in \(2p(x)\). Exam tip: Multiply each polynomial by its coefficient first, then combine like terms.

A zero of a polynomial is a value of \(x\) for which the polynomial equals 0. Setting \(p(x)=3x-6=0\) gives \(3x=6\), so \(x=2\). For example, substituting \(x=1\) gives \(-3\), so it is not a zero. Exam tip: To find a zero, always solve \(p(x)=0\).

To find the difference, subtract the second expression from the first: \((2x+5)-(x-1)\). On removing the bracket, \(2x+5-x+1=x+6\). In option B, the order of subtraction is reversed, so it gives \(-x-6\). Exam tip: when removing brackets preceded by a minus sign, change every sign inside the bracket.

Riya is incorrect. The degree depends on the highest exponent of the variable: in \(5-2x=-2x+5\), the exponent of \(x\) is 1, so it is linear. A negative coefficient does not change the degree. Exam tip: check the highest power, not the number of terms.

Substituting \(x=8\), \(p(8)=\frac{3}{4}\times 8-6=6-6=0\). Hence, the correct answer is 0. Option 6 is only the value of \(\frac{3}{4}\times8\); subtracting 6 is also necessary. Exam tip: While evaluating a polynomial, substitute the given value of \(x\) carefully in every term.

Here, \(p(x)^2=(x+3)^2=x^2+6x+9\). The highest power of \(x\) is 2, so its degree is 2 and it is a quadratic polynomial. A linear polynomial has degree 1, so that option is incorrect. Exam tip: Simplify the expression first, then identify the highest power of the variable.

In \(5x+0=5x\), the coefficient of \(x\) is \(5\), which is non-zero. Therefore, its degree is 1 and it is a linear polynomial. \(0x+9\) and \(0x-3\) simplify to \(9\) and \(-3\), so they are constant polynomials; \(7\) is also constant. Exam tip: for a linear polynomial, the coefficient of the variable must be non-zero.

Here, p(3)=2(3)+b=6+b and p(1)=2(1)+b=2+b. Thus, p(3)-p(1)=(6+b)-(2+b)=4, whereas the given condition requires the difference to be 8. The value of b cancels out, so no value of b can satisfy the condition. Exam tip: In such questions, calculate both function values first and then compare their difference.

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.