Algebraic expressions

In the expression 2x - 3, the term containing x is 2x. It can be written as 2 × x, so the coefficient of x is 2. The number -3 is the constant term, not the coefficient of x. Exam tip: the number multiplying a variable is its coefficient.

In \(3x^2-5x+1\), the only variable is x and its exponents are 2, 1, and 0, all non-negative integers. Hence it is a polynomial in one variable. In \(1/x\), x has exponent −1. Exam tip: fractional or negative exponents are not allowed in polynomials.

In \(x^2+2x-1\), the terms separated by plus or minus signs are \(x^2\), \(2x\), and \(-1\). Therefore, the expression has 3 terms. \(x^2\) and \(2x\) cannot be treated as one term because they are separated by a plus sign. Exam tip: Count a negative constant as a separate term as well.

\(-9p^2\) has only one term, so it is a monomial. \(p+2\) and \(p^2+p\) each have two terms, while \(p+q+1\) has three terms; hence they are not monomials. Exam tip: Count terms by separating an expression at addition or subtraction signs.

Here, 5r, 2r, and -r are like terms, so their coefficients are combined: 5+2-1=6. Therefore, the simplified expression is 6r. The option 6r^2 is incorrect because adding like terms does not change the exponent of r. Exam tip: Add or subtract only terms with the same variable and the same exponent.

8s and -3s are like terms, so their coefficients are combined: 8s-3s=5s. The constant term +4 remains unchanged. Therefore, the simplified expression is 5s+4. In 5s-4, the sign of the constant term has been changed incorrectly. Exam tip: Combine or subtract only terms with the same variable and exponent.

Given x=5, substitute 5 for x in 3x-2: 3(5)-2=15-2=13. Therefore, the correct answer is 13. The close distractor 15 results from calculating 3×5 but not subtracting 2. Exam tip: after substitution, perform multiplication or division before addition or subtraction.

Given y=4, substitute it into y²−1: 4²−1=16−1=15. Therefore, 15 is the correct answer. 16 is only the value of y²; 1 still has to be subtracted. Exam tip: after substitution, evaluate powers before addition or subtraction.

The word ‘sum’ means that the two quantities must be added. Therefore, the sum of \(n\) and \(12\) is \(n+12\). \(12n\) represents multiplication, while \(12-n\) and \(n-12\) represent subtraction. Exam tip: Match words such as sum, difference, and product with \(+\), \(-\), and \(\times\), respectively.

‘5 less than 2p’ means subtract 5 from 2p, giving 2p-5. Option A adds 5, so it represents ‘5 more than 2p’. Exam tip: In phrases such as ‘less than’, subtract the number from the expression that follows it.

Product means multiplication of two quantities. Therefore, \(7 \times k = 7k\). The expression \(7+k\) represents addition, not a product. Exam tip: when a number and a variable are written together, such as \(7k\), they indicate multiplication.

The square of \(t\) is \(t^2\). “Subtracting 6 from the square of \(t\)” means \(t^2-6\), so option B is correct. In \(6-t^2\), the order is reversed; it means subtracting \(t^2\) from 6. Exam tip: In phrases using “from,” subtract the later-mentioned quantity from the earlier quantity.

A polynomial is named by its number of terms, not by its number of variables. \(7a\), \(-3b\), and \(5\) are three terms, so it is a trinomial. Exam tip: count terms separated by + or − signs.

In \(4x+5y-2z\), \(x\), \(y\), and \(z\) are letters whose values can vary, so they are variables. Hence, there are 3 variables. The numbers 4, 5, and −2 are coefficients, not variables. Exam tip: Count the distinct letters that can take values in an expression.

A constant expression contains no variable such as \(x\). Option A has only numbers and its value is \(1\), so it is a constant expression. Options B, C and D contain \(x\); although they may equal \(1\) for some values, they are variable expressions and are undefined when their denominators are zero. Exam tip: First check whether a variable is present in the expression.

The numerical coefficient is the number multiplying the variables in a term. Since 11xy = 11 × x × y, its numerical coefficient is 11. Here, xy is the variable part; 1 would be the coefficient only if no number were written, as in xy. Exam tip: write a term as a product of its number part and variable part to identify the coefficient.

An algebraic term can be separated into its numerical coefficient and its variable part. In 3a²b, the number 3 is the coefficient, while the variables and their powers form a²b. Thus option D is correct. Options B and C incorrectly retain the coefficient 3, and option A names only the numerical coefficient rather than the variable portion.

To find one-fourth of a quantity, divide that quantity by 4. Hence, one-fourth of \(x\) is \(\frac{x}{4}\). The expression \(4x\) means four times \(x\), while \(x+4\) and \(x-4\) mean adding and subtracting 4 respectively. Exam tip: Interpret “one-fourth of” as multiplying by \(\frac{1}{4}\), or dividing by 4.

Combine like terms: \(x+x=2x\) and \(2y+3y=5y\). Therefore, the simplified expression is \(2x+5y\). \(5xy\) is incorrect because adding terms containing \(x\) and \(y\) does not multiply them. Exam tip: Add only terms with the same variable and the same power.

Using the distributive property, multiply 2 by every term inside the bracket: \(2(a+3)=2\times a+2\times 3=2a+6\). Hence, \(2a+6\) is correct. In \(2a+3\), the constant 3 has not been multiplied by 2. Exam tip: When expanding brackets, multiply the outside factor by each term inside.