General chapter practice

The powers of the variable in the terms are 2, 1, and 0. The greatest of these is 2, so the degree of the polynomial is 2. Option 1 gives the power of only the linear term \(2a\), not of the entire polynomial. In an exam, identify the highest power of the variable to find the degree.

The common factor in both terms, 8k and −8, is 8. Taking 8 outside gives 8k ÷ 8 = k and −8 ÷ 8 = −1, so \(8k-8=8(k-1)\). Option A has the wrong sign and represents addition, while the other options are not equivalent to the original expression. Exam tip: After factoring, multiply the common factor back into the bracket to verify the original expression.

Expanding the bracket gives \(2(x+3)=2x+6\). Now combine like terms: \(2x+x=3x\). Therefore, the simplified form is \(3x+6\). Option B omits the final \(+x\). Exam tip: distribute the coefficient across every term inside the bracket before combining like terms.

In \(\frac{y}{4}=3\), multiply both sides by 4 to isolate \(y\): \(y=3\times4=12\). Therefore, 12 is correct. Option 4 is only the denominator; the value of \(y\) is found by multiplying 3 by 4. Exam tip: To undo division by a denominator, multiply both sides by that denominator.

A rational number can be written in the form \(p/q\), where \(p\) and \(q\) are integers and \(q\ne0\). For every integer \(n\), \(n=n/1\), and 1 is not zero; therefore, every integer is rational. Option B is wrong because integers are not irrational. Exam tip: when checking a rational form, make sure that the denominator is not zero.

Combine like terms separately: 4a - 2a = 2a and 5 + 1 = 6. Therefore, the expression simplifies to 2a + 6. Option B incorrectly adds the constant terms. Exam tip: combine variable terms and constant terms separately.

Using the distributive law, multiply 6 by each term inside the bracket: \(6(x+1)=6\times x+6 imes1=6x+6\). Therefore, option A is correct. Option B is wrong because the 1 has not been multiplied by 6. Exam tip: remember \(a(b+c)=ab+ac\).

In \(11c\), the variable \(c\) is multiplied by 11, so the coefficient of \(c\) is 11. Option A, 1, would be the coefficient if \(c\) appeared alone; here, 11 is the numerical multiplier. Exam tip: the numerical factor multiplying a variable is its coefficient.

The like terms \(a^2\) and \(-a^2\) cancel each other, so \(a^2-a^2=0\). Therefore, the expression becomes \(0+5=5\). Option B is incorrect because it retains one \(a^2\) term instead of cancelling the two opposite terms. Exam tip: combine like terms with their signs before simplifying the remaining expression.

Subtracting 3 from both sides gives \(2x=8\). Dividing both sides by 2 gives \(x=4\), so option C is correct. Option B is incorrect because substituting 5 gives \(2(5)+3=13\), not 11. As an exam tip, verify the solution by substituting it back into the original equation.

The expression \\(x+x^2\\) has two terms separated by the plus sign: \\(x\\) and \\(x^2\\). They are unlike terms because their powers are different, so they cannot be combined into a single term. Therefore, the correct answer is 2. Exam tip: Count algebraic terms by separating the expression at plus or minus signs.

Substitute (b=1) into (7b+8): (7\times1+8=7+8=15). Therefore, the correct answer is 15. The value 8 is an incomplete result because it ignores the term (7b). For such questions, substitute the value of the variable first and then perform the calculation.

All the terms are like terms because they contain the same variable, \(x\). Combine their coefficients: \(9+3-2=10\). Therefore, \(9x+3x-2x=10x\), so option B is correct. Option D does not correctly account for the subtraction. Exam tip: When adding or subtracting like terms, operate on the coefficients and keep the common variable unchanged.

When multiplying monomials, multiply the numerical coefficients first: \(2\times4=8\). Then add the exponents of the same variable: \(m\times m=m^2\). Thus, \(2m\cdot4m=8m^2\), so option C is correct. Option B misses the factor \(m^2\). Exam tip: when multiplying like variables, add their exponents.

In this equation, the order of the addends changes, but the sum remains the same: a + 2 = 2 + a. Therefore, it represents the commutative property of addition. The associative property changes the grouping with brackets, not the order, as in (a + 2) + 3 = a + (2 + 3). Exam tip: if changing the order does not change the result, identify the commutative property.

\(3q\) and \(3q\) are like terms. Subtracting a term from itself gives zero, so \(3q-3q=0\). Option B keeps only one term instead of performing the subtraction. In an exam, subtract the coefficients of like terms: \(3-3=0\).

Substitute 2 for \(s\) in the expression: \(5s+2=5\times2+2=10+2=12\). Therefore, the correct value is 12. The value 10 results from omitting the constant term 2. In an exam, substitute the given value first, then follow the order of operations.

The expression \(x^2-9\) can be written as \(x^2-3^2\), which is a difference of two squares. Therefore, using \(a^2-b^2=(a-b)(a+b)\), we get \(x^2-9=(x-3)(x+3)\). Option B is incorrect because the identity for the square of a difference contains the middle term \(-2ab\). Exam tip: when two perfect squares are separated by subtraction, check for the difference-of-squares identity first.

In the term \(-3y\), the coefficient of \(y\) is \(-3\), which is negative. The coefficients of \(5x\), \(z\), and \(\frac{1}{2}p\) are \(5\), \(1\), and \(\frac{1}{2}\), respectively, and all are positive. Exam tip: When no number is written before a variable, its coefficient is \(1\).

A constant term is a term that does not contain a variable. In \(4p-9\), \(4p\) contains the variable, whereas −9 does not; therefore, −9 is the constant term. In an exam, include the sign of the term when identifying it.