General chapter practice

Like terms have exactly the same variables with the same exponents; their coefficients may differ. In 4x and -7x, the variable part is x, with exponent 1 in both terms, so they are like terms. In 5a and 5a², the exponents of a are 1 and 2, so they are unlike terms. Exam tip: Ignore the coefficients and compare only the variables and their exponents.

Repeated addition of identical terms can be written as multiplication. Since x is added four times, x+x+x+x=4x. Option A, x^4, represents multiplying x by itself four times, while options C and D are not equivalent to four copies of x. Exam tip: when the same term is added n times, write it as n times that term.

Substitute the given value x = 6 into x + 8: 6 + 8 = 14. Therefore, 14 is correct. The value 12 comes from 6 − 8, while 48 comes from 6 × 8, so neither matches the given expression. In such questions, substitute the variable’s value before performing the operation.

Substitute \(a=5\) into \(3a\): \(3\times 5=15\). Therefore, the correct answer is 15. \(35\) is incorrect because \(3a\) means 3 multiplied by \(a\), not the digits written together. Exam tip: replace the variable with its given value before performing the operation.

Terms in an expression are separated by addition (+) or subtraction (−) signs. In \(2m+7\), \(2m\) and \(7\) are the two terms. \(2m\) is not counted as two separate terms because 2 and \(m\) are multiplied. Exam tip: Count only the parts separated by + or − signs.

In \((p+q)\), the plus sign separates two terms: \(p\) and \(q\). Therefore, it is a binomial. It is not a monomial because it has two terms rather than one. In an exam, identify the type of an expression by counting its terms.

The degree of a polynomial is the highest exponent of the variable with a non-zero coefficient. In \(2x^2+3x+1\), the highest exponent of \(x\) is \(2\), so its degree is 2. The number 3 represents the number of terms, not the degree. In an exam, identify the highest exponent rather than the coefficient.

Expanding the bracket gives 5(x−1) = 5x−5. Adding 2x results in 5x−5+2x = 7x−5, so option A is correct. In option C, the sign of the constant term is incorrect. Exam tip: expand brackets first, then combine like terms.

In the equation \(x+5=12\), subtract 5 from both sides to isolate \(x\): \(x=12-5=7\). Therefore, 7 is correct. The value 12 is the right-hand side of the equation, not the value of \(x\). Exam tip: use the inverse operation to isolate the unknown.

To isolate \(m\), add 4 to both sides of the equation: \(m-4+4=9+4\). Therefore, \(m=13\). Option A results from subtracting 4 instead of using the required inverse operation. Exam tip: Apply the same inverse operation to both sides when solving a linear equation.

In the equation \(2t=18\), divide both sides by 2 to isolate \(t\): \(t=18\div2=9\). Therefore, option B is correct. Option 8 is incorrect because \(2\times8=16\), not 18. Exam tip: to undo multiplication by a coefficient, divide both sides by that same coefficient.

In \(\frac{n}{3}=6\), multiply both sides by the denominator 3 to isolate \(n\): \(n=6\times3=18\). Therefore, the correct answer is C, 18. The option 9 results from not carrying out the required multiplication correctly. Exam tip: apply the same operation to both sides of a linear equation when isolating the variable.

In the equation \(3p-2=10\), add 2 to both sides to get \(3p=12\). Then divide both sides by 3, giving \(p=4\). Therefore, option C is correct. Exam tip: Apply the same operation to both sides while isolating the variable.

Using the identity \((a+b)^2=a^2+2ab+b^2\) with \(a=x\) and \(b=1\), we get \((x+1)^2=x^2+2x+1\). Therefore, option A is correct. Option B has the middle term \(-2x\), while option C gives \(x^2-1\). In exams, match the sign and coefficient of the middle term first.

Using the identity \((a+b)^2=a^2+2ab+b^2\), the expanded form is \(a^2+2ab+b^2\), so option B is correct. Option C has a negative middle term and represents \((a-b)^2\), not \((a+b)^2\). Exam tip: the square of a sum has the middle term \(+2ab\), while the square of a difference has \(-2ab\).

Using the identity \((x-y)^2=x^2-2xy+y^2\), with \(x=a\) and \(y=b\), we get \((a-b)^2=a^2-2ab+b^2\). Therefore, option B is correct. Option A omits the middle term, option C has the wrong sign for \(2ab\), and option D represents the difference of two squares, not the square of a difference. Exam tip: the middle term in \((a-b)^2\) is always \(-2ab\).

This is the identity for the product of the sum and difference of two terms: (a+b)(a-b)=a^2-b^2. Therefore, option B is correct. Option A adds the two squares, while options C and D represent the expansions of (a+b)^2 and (a-b)^2, respectively. Exam tip: In (x+y)(x-y)=x^2-y^2, the middle terms cancel each other.

Since \(9x^2=(3x)^2\), its principal square root is \(3x\) when \(x\ge0\). Without this condition, the general answer would be \(3|x|\), because a principal square root is always non-negative. Exam tip: rewrite a perfect square as \((\text{quantity})^2\) before taking its square root.

Here, x816 = x848. Applying the difference of squares formula a8b8 = (ab)(a+b), we get x848 = (x4)(x+4), so option A is correct. Option C, (x4)8, expands to x88x+16 and is therefore not equivalent. Exam tip: whenever an expression has the form a8b8, use the difference of squares formula.

The terms of \(2x+6\) are \(2x\) and \(6\). Both are exactly divisible by 2, so their greatest common factor is 2. The number 3 is not a factor of \(2x\), and \(x\) is not a factor of 6. In an exam, factor each term and select the greatest factor common to all terms.