Linear polynomials

For a polynomial to be linear, the coefficient of \(x\) must be non-zero. Here, the coefficient of \(x\) is \(k+9\). Thus, \(k+9\ne0\), or \(k\ne-9\), makes the expression a linear polynomial. If \(k=-9\), the expression becomes \(-1\), which is a constant polynomial, not a linear one. Exam tip: in \(ax+b\), always check that \(a\ne0\).

To find the zero, set the polynomial equal to 0: \(5x-20=0\). Thus, \(5x=20\), so \(x=4\). Therefore, 4 is the zero of the polynomial. Substituting \(-4\) gives \(5(-4)-20=-40\), not 0. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\).

A zero of a polynomial is a value that makes the polynomial equal to 0. Setting \(x+21=0\) gives \(x=-21\). Therefore, \(-21\) is the correct answer. If \(21\) is substituted, the value is \(21+21=42\), so it is not a zero. Exam tip: To find a zero, equate the polynomial to 0 and solve for \(x\).

Given \(p(x)=4x+7\). To find \(p(3)\), substitute 3 for \(x\): \(p(3)=4\times3+7=12+7=19\). Hence, the correct answer is 19. The value 21 would result from an incorrect calculation instead of adding 7 to 12. Exam tip: when evaluating a polynomial, carefully replace every \(x\) with the given number.

Given \(p(x)=8x-13\), substitute \(x=0\): \(p(0)=8\times0-13=-13\). Hence, option C is correct. Choosing \(13\) is a common sign error because the constant term is \(-13\), not \(+13\). Exam tip: at \(x=0\), only the constant term of a polynomial remains.

To find the zero, equate the polynomial to 0: \(18-6x=0\). Thus, \(6x=18\), so \(x=3\). Therefore, 3 is the correct answer. Substituting \(-3\) gives \(18-6(-3)=36\), not 0. Exam tip: Verify a zero by substituting its value in the polynomial; the result must be 0.

In the expression (9r-14), the only variable is r. The highest power of r is 1, so it is a linear polynomial in r. The variables x, y, and z do not occur in this expression. Exam tip: A polynomial is linear in a variable when its highest power is 1.

The polynomial \(ax+b\) has degree 1 only when the coefficient of \(x\), \(a\), is non-zero. If \(a=0\), the expression becomes the constant \(b\), so it is not a linear polynomial. Also, \(b\) need not be non-zero: \(ax\), where \(a\neq 0\), is still linear. Exam tip: determine the degree from the highest power of the variable with a non-zero coefficient.

When adding polynomials, add only like terms. Here, \(3x\) and \(2x\) are like terms, so \(3x+2x=5x\). The constant terms give \(6+5=11\). Therefore, \(p(x)+q(x)=5x+11\). Option \(6x+11\) incorrectly adds the coefficients of the \(x\)-terms. Exam tip: add variable terms and constant terms separately.

\(p(x)-q(x)=(9x+8)-(4x+3)\). On removing the brackets, we get \(9x+8-4x-3\). Combining like terms gives \((9x-4x)+(8-3)=5x+5\), so the correct answer is \(5x+5\). In \(5x+11\), the constant terms have been added instead of subtracted. Exam tip: While subtracting polynomials, change the sign of every term in the second polynomial.

Using the distributive property, multiply 4 by each term inside the bracket: \(4(3x-2)=4\times3x+4\times(-2)=12x-8\). Therefore, the correct linear polynomial is \(12x-8\). In \(12x+8\), the negative sign of 2 has been handled incorrectly. Exam tip: When removing brackets, multiply the outside number by every term and check the signs carefully.

Using the distributive property, multiply \(-4\) by each term inside the bracket: \(-4\times x=-4x\) and \(-4\times 3=-12\). Therefore, \(-4(x+3)=-4x-12\). In \(-4x+12\), the sign of the constant term is incorrect. Exam tip: When multiplying by a negative number, check the sign of every term carefully.

The polynomial 23x contains only a term with x; it has no term without the variable x. Hence, its constant term is 0. Here, 23 is the coefficient of x, not the constant term. Exam tip: identify the term that has no variable to find the constant term.

Given \(p(x)=-5x+25\). Substituting \(x=5\), \(p(5)=-5\times5+25=-25+25=0\). Therefore, the correct answer is 0. Option 5 is not the value of the polynomial; it is the \(x\)-value at which the polynomial becomes zero. Exam tip: To find a polynomial’s value, substitute the given \(x\)-value carefully and keep track of signs.

To find a zero, equate the polynomial to 0: \(20-5x=0\). Thus, \(5x=20\), so \(x=4\). Checking, \(20-5(4)=0\); hence 4 is the zero. If 5 is substituted, the value is \(-5\), not 0. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\).

In the polynomial \(30-11v\), \(v\) is the letter whose value can change, so it is the variable. \(30\) is the constant term, while \(-11\) is the coefficient of \(v\); neither is a variable. Exam tip: Identify the letter whose value can vary.

The governing concept is the degree of a polynomial. A polynomial p(x)=ax+b is linear only when its degree in x is exactly 1. That requires the coefficient a of x to be nonzero. If a=0, the x-term disappears and p(x)=b becomes a constant polynomial of degree 0, so it is no longer linear. The coefficient does not have to be 1: a=2, a=−3, and a=1/2 all produce linear polynomials, provided each is nonzero. Also, there is no rule requiring a=b, because the coefficient of x and the constant term are independent quantities. Therefore option B is the only correct choice.

In a linear polynomial, the term without a variable is called the constant term. In \(3x-8\), the variable-free term is \(-8\), so option A is correct. In \(-8x+3\), \(-8\) is the coefficient of \(x\), while the constant term is \(3\). Exam tip: identify the term that has no variable to find the constant term.

To find the zero of the linear polynomial \(x-8\), set \(x-8=0\). This gives \(x=8\), so its zero is 8. In contrast, \(x+8\) has zero \(-8\). Exam tip: Set a polynomial equal to 0 and solve for \(x\) to find its zero.

A zero of a polynomial is a value that makes the polynomial equal to 0. Substituting \(x=-10\) in \(x+10\) gives \(-10+10=0\), so its zero is \(-10\). The close distractor \(x-10\) has zero \(10\), not \(-10\). Exam tip: the zero of \(x+a\) is \(-a\), while the zero of \(x-a\) is \(a\).