Linear polynomials

Multiply 5 by each term inside the bracket: \(5(2x+3)=10x+15\). Then combine like terms: \(10x+15-4x=6x+15\). Therefore, \(6x+15\) is correct. \(14x+15\) would result from incorrectly adding \(4x\) instead of subtracting it. Exam tip: after opening brackets, combine only like terms.

Given \(p(x)=9-2x\). To find \(p(1)\), substitute 1 for \(x\): \(p(1)=9-2(1)=9-2=7\). Therefore, 7 is the correct answer. The number 9 is only the constant term, not the value of the polynomial at \(x=1\). Exam tip: When evaluating a polynomial, substitute the given value carefully for the variable.

In \(8x-17\), the highest power of \(x\) is \(1\), so its degree is \(1\). A polynomial of degree \(1\) is called a linear polynomial. A quadratic polynomial would have the highest power of the variable equal to \(2\), so it is not correct here. Exam tip: identify the type of a polynomial by checking the highest power of its variable.

The governing concept is the degree of a polynomial. The degree is the greatest exponent of the variable whose coefficient is not zero. A polynomial is called linear when its degree is exactly 1; it may have one term, two terms, or more terms, provided the greatest nonzero exponent is 1. For example, 3x+5 and -7x are linear polynomials, whereas x^2+1 has degree 2 and is quadratic. Thus option C correctly identifies a linear polynomial by its highest power being 1. Power 0 describes a nonzero constant polynomial, power 2 describes a quadratic polynomial, and the number of terms does not determine whether a polynomial is linear.

In \(13u+9\), the variable \(u\) can be written as \(u^1\), so its power is 1. The number 13 is the coefficient of \(u\), and 9 is the constant term; neither is the power. Hence, this is a linear polynomial in \(u\). Exam tip: When no exponent is written on a variable, its power is 1.

The general form of a linear polynomial is (ax+b) where (a\neq0). If (a=0), the degree will not remain (1).

The degree of a polynomial is the greatest exponent of the variable with a non-zero coefficient. In \(p(x)=12x+1\), the highest exponent of \(x\) is \(1\), so its degree is \(1\). The number \(12\) is only the coefficient of \(x\), not the degree. Exam tip: Look for the highest power of the variable, not its coefficient.

In the polynomial 25-9x, the term containing x is -9x. The number multiplying x is -9, so the coefficient of x is -9. The option 9 is incorrect because the negative sign is part of the coefficient. Exam tip: Always include the sign immediately before the variable when identifying a coefficient.

A constant term is a term that contains no variable. In \(x+18\), \(x\) is the variable term, while \(18\) has no variable. Therefore, the constant term is \(18\). Although 0 can be a constant, it is not the constant term in this polynomial. Exam tip: identify the term with no letter or variable to find the constant term.

The polynomial 7x+4 has an x-term and the constant term 4, but no x²-term. It can therefore be written as 0x²+7x+4, so the coefficient of x² is 0. Here, 7 is the coefficient of x, not of x². Exam tip: If a term of a given power is missing, its coefficient is 0.

In \(7x+5\), the highest power of x is 1, so it is a linear polynomial. \(4x^2-3\) has degree 2 and is quadratic. Exam tip: identify a polynomial’s degree by its highest exponent.

The given polynomial is \(p(x)=x-20\). It can be written as \(1x+(-20)\), so it is a linear polynomial of the form \(ax+b\), where \(a=1\) and \(b=-20\). The form \(ax^2+b\) is quadratic and must contain an \(x^2\) term. Exam tip: identify the type of a polynomial from its highest power; here, the highest power of \(x\) is 1.

In (0x+12), the coefficient of x is 0, so 0x=0 and the expression simplifies to 12. Since 12 is a non-zero constant, it is a constant polynomial with degree 0. A linear polynomial must have a non-zero coefficient of x. Exam tip: Remove terms with zero coefficients before finding the degree of a polynomial.

In a linear polynomial, the coefficient of \(x\) must be non-zero. Substituting \(m=8\) gives \(m-8=0\), so the expression becomes \(0x+6=6\). This is a constant polynomial, not a linear polynomial. For example, when \(m=7\), the coefficient of \(x\) is \(-1\), so it is still linear. Exam tip: a linear polynomial has highest power 1 with a non-zero coefficient of the variable.

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.