Linear polynomials

At a zero, the value of the polynomial is zero. Substituting \(x=-3\) gives \(5(-3)+b=0\), so \(-15+b=0\). Hence, \(b=15\). If \(b=-15\), the polynomial value would be \(-30\), not zero. Exam tip: substitute the given zero into a linear polynomial and equate the result to \(0\).

Substituting \(x=10\), \(p(10)=\frac{3}{5}\times10+4=6+4=10\). Hence, 10 is the correct option. A value such as 9 can result from incorrectly evaluating \(\frac{3}{5}\times10\). Exam tip: To find the value of a polynomial, substitute the given number for \(x\) and follow the order of multiplication and addition carefully.

The expression \((4a+8)x-5\) is linear only when the coefficient of \(x\) is non-zero. For it to stop being linear, \(4a+8=0\), which gives \(a=-2\). At this value, the expression becomes \(-5\), a constant polynomial rather than a linear polynomial. For example, at \(a=0\), the coefficient of \(x\) is \(8\), so it is still linear. Exam tip: in a parameter-based linear polynomial, set the coefficient of the variable equal to zero to test when it is no longer linear.

Given \(p(x)=12x-20\) and \(p(x)=4\), we get \(12x-20=4\). Adding 20 to both sides gives \(12x=24\), so \(x=2\). If \(x=1\), then \(p(x)=-8\), not 4. Exam tip: when a value of a polynomial is given, equate the polynomial to that value and solve the resulting linear equation.

Given \(p(x)=8x+c\). Substituting \(x=2\), we get \(p(2)=8\times2+c=16+c\). Since \(p(2)=19\), \(16+c=19\), so \(c=3\). If \(c=4\), then \(p(2)=20\), not 19. Exam tip: To use a given value of a polynomial, first substitute the stated value of \(x\).

Given \(p(x)=x+a\) and \(p(-9)=0\), substitute \(-9\) for \(x\): \(-9+a=0\). Hence, \(a=9\). If \(a=-9\), then \(-9+(-9)=-18\), not zero. Exam tip: When a zero of a polynomial is given, substitute it for \(x\) and set the polynomial equal to zero.

Given p(x)=4x-3, p(2)=4(2)-3=5 and p(-2)=4(-2)-3=-11. Therefore, p(2)+p(-2)=5+(-11)=-6. Choosing 6 may result from an error with the sign of p(-2). Exam tip: when substituting a negative value, check the sign of every product carefully.

To find the zero of \(6x+12\), set \(6x+12=0\). This gives \(6x=-12\), so \(x=-2\). Its coefficient of \(x\) is also \(6\), so option B is correct. Although \(6x-12\) has coefficient 6, its zero is \(2\). Exam tip: the zero of a linear polynomial \(ax+b\) is \(-b/a\).

In \(7x-3\), the highest power of \(x\) is 1, so its degree is 1 and it is linear. \(9\) has degree 0, not 1. Exam tip: identify the highest exponent to find the degree.

Given \(p(x)=7x+9\), we get \(p(x)-9=(7x+9)-9=7x\). The highest exponent of \(x\) in \(7x\) is \(1\), so its degree is \(1\). Here, \(7\) is a coefficient, not the degree. Exam tip: Simplify the polynomial first, then identify the highest exponent of the variable.

Given p(x)=4x+11, p(x)-4x=(4x+11)-4x=11. Since 11 has no x-term, its degree is 0; hence it is a non-zero constant polynomial. A linear polynomial has degree 1, so option A is not correct. Exam tip: simplify by combining like terms first, then identify the polynomial from its degree.

A zero of a polynomial is a value of x that makes the polynomial equal to 0. Substituting \(x=-\frac{7}{2}\) in \(2x+7\) gives \(2\left(-\frac{7}{2}\right)+7=-7+7=0\). Hence, \(2x+7\) is the correct option. The close distractor \(2x-7\) has zero \(\frac{7}{2}\), not \(-\frac{7}{2}\). Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\).

Substitute 3 for x: \(p(3)=16-5\times3=16-15=1\). Therefore, the correct answer is 1. The value 31 would result from adding \(16+15\), but the expression contains subtraction. Exam tip: after substituting the given value, perform multiplication before subtraction.

\(2p(x)+q(x)=2(x-6)+(3x+2)\). Using the distributive property, \(2(x-6)=2x-12\). Combining like terms gives \(2x-12+3x+2=5x-10\). Hence, \(5x-10\) is correct. The option \(4x-10\) results from incorrectly adding \(2x\) and \(3x\). Exam tip: multiply each polynomial by its coefficient first, then combine like terms.

To find a zero, set \(p(x)=0\): \(5x-15=0\). Thus, \(5x=15\), so \(x=3\). On checking, \(p(3)=5(3)-15=0\); hence, \(x=3\) is the zero. Substituting \(x=5\) gives \(10\), so it is not a zero. Exam tip: verify a proposed zero by substituting it into the polynomial and checking whether the result is \(0\).

To find the difference, subtract the second expression from the first: \(4x+9-(x-3)=4x+9-x+3=3x+12\). Therefore, option A is correct. In option B, the order of subtraction is reversed, so it gives \(-3x-12\). Exam tip: When a bracket is preceded by a minus sign, change the signs of all terms inside it.

For a zero, set \(4x-9=0\). Then \(4x=9\), so \(x=\frac{9}{4}\). The value \(-\frac{9}{4}\) comes from a sign error. In exams, substitute the answer to verify it.

Substituting \(x=8\), we get \(p(8)=\frac{7}{8}\times 8-7=7-7=0\). Hence, the correct answer is 0. \(-7\) is only the constant term, not the value of \(p(8)\). Exam tip: To evaluate a polynomial, substitute the given value of \(x\) and simplify step by step.

Here, \(p(x)^2=(x+5)^2=x^2+10x+25\). The highest power of \(x\) is 2, so the polynomial has degree 2 and is a quadratic polynomial. A linear polynomial has degree 1, so option A is not correct. Exam tip: identify a polynomial’s type from its highest power of the variable.

Given p(x)=4x+b, we get p(5)=20+b and p(2)=8+b. Therefore, p(5)-p(2)=(20+b)-(8+b)=12. The term b cancels because it occurs equally in both values. Note that 8 is only the value of the variable part of p(2), not the required difference. Exam tip: Evaluate the polynomial at both inputs separately before subtracting.