Algebraic expressions

Use the distributive property: \(2x^2\times 3x=6x^3\) and \(2x^2\times(-4)=-8x^2\). Therefore, the expanded expression is \(6x^3-8x^2\). In option D, the second term is positive, but multiplication by \(-4\) gives a negative term. Exam tip: while multiplying powers of \(x\), add their exponents; for example, \(x^2\times x=x^3\).

\(5p^2q\) and \(2p^2q\) are like terms, so adding their coefficients gives \((5+2)p^2q=7p^2q\). The term \(-3pq\) has variable part \(pq\), which is different from \(p^2q\), so it cannot be combined with \(7p^2q\). Therefore, the simplified expression is \(7p^2q-3pq\). Exam tip: combine only terms with exactly the same variables raised to the same powers.

Combining the like variable terms gives \(4x-9x=-5x\), and the constant terms give \(7+2=9\). Therefore, the simplified expression is \(-5x+9\). In \(5x+9\), the sign of \(4x-9x\) has been handled incorrectly. Exam tip: add or subtract only like terms.

Using the distributive property, \(7(x-2)=7\times x-7\times 2=7x-14\), so option C is correct. Option A expands to \(7x+14\), while option B expands to \(14x-14\); neither matches the given expression. Exam tip: multiply the number outside the bracket by every term inside it.

In the term \(3x^2y^2\), \(x\) and \(y\) are letters whose values can vary, so they are variables. The number \(3\) is the numerical coefficient, while \(2\) is an exponent; neither is a variable. Exam tip: identify the letters in a term, but do not count their exponents as variables.

Substituting k=3 gives 2(3+4)-3^2. The bracket equals 7, so 2×7=14, and 3^2=9. Therefore, 14-9=5, so 5 is correct. The value 14 is only the result of the multiplication part; 9 still has to be subtracted. Exam tip: after substitution, evaluate brackets, powers, multiplication, and then subtraction in order.

Combine like terms: \(11a^2-5a^2=6a^2\) and \(-4a+6a=2a\). Therefore, the simplified expression is \(6a^2+2a\). In option A, the coefficients of the squared terms have been added incorrectly. Exam tip: Add or subtract only terms that have the same variable with the same exponent.

First, square \(x\) and add 5 to get \(x^2+5\). Twice this entire sum is \(2(x^2+5)\), so option C is correct. In option A, only \(x^2\) is doubled, not 5. Exam tip: When you read “twice the sum,” put the complete sum inside brackets.

Use the distributive property: \(7(2r-3)=14r-21\) and \(4(r+1)=4r+4\). Thus, \(14r-21+4r+4=18r-17\). Therefore, the correct expression is \(18r-17\). The option \(18r+25\) may result from incorrectly adding the constant terms with their signs. Exam tip: while expanding brackets, multiply the outside number by every term inside the bracket.

\(5a^2b\) and \(3a^2b\) are like terms, so they add to \(8a^2b\). Similarly, \(2ab-7ab=-5ab\). Therefore, the simplified expression is \(8a^2b-5ab\). \(10a^3b\) is incorrect because exponents are not added when terms are added or subtracted. Exam tip: combine only terms with identical variable parts and powers.

Given x=2 and y=-1, x^2+xy+y^2=2^2+(2)(-1)+(-1)^2=4-2+1=3. Therefore, the correct answer is 3. The value 1 is only y^2; the terms x^2 and xy must also be included. Exam tip: the square of a negative number is positive, so (-1)^2=1.

A linear term has a variable with exponent 1. In the given expression, 5x^2 has exponent 2, 2 is a constant term, and -3x has x raised to the power 1. Therefore, -3x is the linear term. In exams, identify the type of each term by checking the exponent of its variable.

On expanding the bracket, \(2(a+b)-a=2a+2b-a\). Combining like terms gives \(2a-a=a\), so the expression equals \(a+2b\). The option \(2a+b\) does not simplify the \(a\)-terms correctly. Exam tip: while expanding brackets, multiply the outside coefficient by every term inside the bracket.

The powers of the variable s in the expression are 3, 2, and 1, while the constant term −8 has power 0. Therefore, the highest power is 3, found in the term 4s^3. The number 8 is only part of the constant term, not a power. Exam tip: To find the degree of a polynomial, identify the greatest exponent of its variable.

In an algebraic term, the number that multiplies the variable part is called the numerical coefficient. In 6x^2y, 6 multiplies x^2y, so the numerical coefficient is 6. Here, x^2y is the variable part, while 2 is only the exponent of x. Exam tip: To identify a coefficient, separate the variables and their exponents from the term; the remaining number is the coefficient.

Using the distributive property, \(3(4x-5)=12x-15\) and \(-2(2x+1)=-4x-2\). Therefore, \(12x-15-4x-2=8x-17\). Hence, \(8x-17\) is correct. In \(8x-13\), the constant terms have been combined incorrectly. Exam tip: when a negative coefficient is outside a bracket, apply it to every term inside the bracket.

The difference between m and n is \(m-n\). Taking five times this entire difference gives \(5(m-n)\), so option C is correct. In \(5m-n\), only m is multiplied by 5, not the whole difference. Exam tip: When a phrase says “times the difference,” put the difference in brackets before multiplying.

Combine like terms: \(10z^2-4z^2=6z^2\) and \(z-6z=-5z\). There is no other constant term to combine with \(3\), so it remains \(3\). Therefore, the simplified expression is \(6z^2-5z+3\). Option B incorrectly adds the \(z^2\) terms. Exam tip: add or subtract only terms having the same variable and exponent.

Given \(c=-2\), \(2c^3+c^2=2(-2)^3+(-2)^2=2(-8)+4=-16+4=-12\). Therefore, \(-12\) is correct. Since \(c^3\) has an odd exponent, it remains negative, whereas \(c^2\) has an even exponent and is positive. Exam tip: When substituting a negative value, check whether each exponent is odd or even.

Like terms are algebraic terms that have exactly the same variables with exactly the same exponents. Their coefficients may be different, but the variable pattern must match. In \(a^2b\), the exponent of a is 2 and the exponent of b is 1. In \(ab^2\), the exponent of a is 1 and the exponent of b is 2. The exponents are therefore interchanged, so the terms are not like terms. Option C states this idea.

The terms may look similar because both contain a and b, but merely having the same letters is not enough. For example, \(3a^2b\) and \(-5a^2b\) are like terms, while \(a^2b\) and \(ab^2\) are not. Options A, B and D do not describe the actual difference: the coefficients need not be the issue, variables are present, and these are not constants. Hence option C is correct.