Algebraic expressions

\(4a^2b\) and \(3a^2b\) are like terms, so their sum is \(7a^2b\). Similarly, \(-7ab^2\) and \(2ab^2\) add to \(-5ab^2\). Therefore, the simplified expression is \(7a^2b-5ab^2\). In option C, the coefficients of \(a^2b\) have not been added correctly. Exam tip: combine only terms that have exactly the same variables with the same powers.

In \(x^2-7\), the \(x^2\) term is present, but there is no first-degree \(x\) term. Therefore, the coefficient of \(x\) is \(0\). In \(2x^2+x\), the coefficient of \(x\) is \(1\), so it is not correct. Exam tip: If a term of a particular degree is absent, its coefficient is \(0\).

\(2(x-y)-3(y-x)=2x-2y-3y+3x\), because \(-3(y-x)=-3y+3x\). Combining like terms gives \(5x-5y\). \(5y-5x\) is the negative of the required expression, so it is not correct. Exam tip: When a bracket is preceded by a minus sign, change the sign of every term inside it.

Given u=-3 and v=2, we have u²=9, uv=(-3)(2)=-6, and v²=4. Hence, u²-2uv+v²=9-2(-6)+4=9+12+4=25. The value 13 can result from handling the sign of the negative product uv incorrectly. Exam tip: determine the sign of uv first, then evaluate the complete term -2uv.

The perimeter of a rectangle is \(2(\text{length}+\text{breadth})\). Thus, \(2[(2x+3)+(x-5)] = 2(3x-2) = 6x-4\). Therefore, \(6x-4\) is correct. \(3x-2\) is only the sum of the length and breadth, not the perimeter. Exam tip: Always multiply the sum of length and breadth by 2 for a rectangle’s perimeter.

Using distribution, \((x+3)(x-3)=x^2-3x+3x-9=x^2-9\). Rima used the wrong sign for \(3\times(-3)\). Exam tip: apply \((a+b)(a-b)=a^2-b^2\) to avoid expansion errors.

On expanding, \(4(2a-3b)=8a-12b\) and \(-3(a-2b)=-3a+6b\). Therefore, \(8a-12b-3a+6b=5a-6b\). In option C, \(-3a\) has incorrectly been added instead of subtracted. Exam tip: When a negative multiplier is outside brackets, apply it to every term inside the brackets.

For x=1, we have x^3=1 and x^2=1. Therefore, 5x^3-4x^2+3x-2 = 5(1)-4(1)+3(1)-2 = 5-4+3-2 = 2. Hence, the correct answer is 2. Getting 4 may result from an error in addition or subtraction. Exam tip: After substitution, write each term with its sign carefully, especially the negative signs.

First the square of the sum (x+y) is formed and then (xy) is subtracted. Brackets are necessary for the square of the whole sum.

First simplify the innermost bracket: 2x-(x-5)=2x-x+5=x+5. Then, 3x-(x+5)=3x-x-5=2x-5. Hence, the correct answer is 2x-5. In 2x+5, the negative sign before the outer bracket has not been applied to every term inside it. Exam tip: When a bracket is preceded by a minus sign, change the sign of each term inside while removing the bracket.

Substituting t=-1 gives 2(-1)^3-3(-1)^2+4(-1)+6. Thus, -2-3-4+6=-3, so the correct answer is -3. The option 3 can result from incorrectly handling the negative signs in (-1)^3 or 4(-1). Exam tip: an odd power of a negative number is negative, while an even power is positive.

The governing concept is simplifying an algebraic expression by applying the distributive property and combining like terms. First simplify the innermost product: 4(1 − x) = 4 − 4x. Thus the square bracket becomes 3x + 4 − 4x = 4 − x. The original expression is now 9x − 2(4 − x). Distribute −2 across the parentheses: −2(4 − x) = −8 + 2x. Finally combine like terms: 9x + 2x − 8 = 11x − 8. Therefore, option A is correct. A common error is to lose the negative sign before 2, which can produce 7x − 8 or a positive constant. The brackets must be simplified from the inside outward.

Given a+b=7 and ab=10, substitute these directly into 2(a+b)-ab: 2(7)-10=14-10=4. Therefore, the correct option is 4. The value 14 is only the value of 2(a+b); ab must still be subtracted. In exams, treat a grouped expression such as (a+b) as one unit before substituting.

\(3x^4-2x^2+7\) has three terms, and the highest exponent of x is 4, so it is a trinomial polynomial of degree 4. Option B has four terms, while D has a negative exponent and is not a polynomial. Exam tip: check term count and highest exponent separately.

Substituting the given values, \,\((x+y)^2-xy=(2+3)^2-(2\times3)=5^2-6=25-6=19\). Therefore, the correct answer is 19. The value 25 is only \,\((x+y)^2\); \,\(xy=6\) must also be subtracted. Exam tip: Evaluate the squared term first, then perform multiplication and subtraction in order.

In
-6x^2+5x-1, the term
-6x^2 can be written as
(-6)x^2. Therefore, the coefficient of
x^2 is
-6. In option A, the coefficient is 6, while in option D the coefficient of
x^2 is 1;
-6 is the coefficient of x there. Exam tip: To identify a coefficient, include the sign and consider the complete numerical factor multiplying the term.

Write (\frac{x}{2}) as (\frac{2x}{4}), then the sum is (\frac{5x}{4}). Make common denominators for fraction terms.

The number 3 more than k is \(k+3\). Squaring this entire quantity gives \((k+3)^2\), so option D is correct. In \(k^2+3\), 3 is added only to the square of k; it is not the square of \(k+3\). Exam tip: For “square of” a quantity, write the complete quantity in brackets before applying the exponent 2.

In \(3x^2-5x+7\), the powers of x are 2, 1 and 0, all non-negative integers. It has three terms, so it is a trinomial polynomial. \(4x-9\) is only a binomial. Exam tip: reject fractional or negative powers first.

Given \(x+y=6\). Taking 3 as a common factor, \(3x+3y-4=3(x+y)-4\). Substituting the given value gives \(3(6)-4=18-4=14\). Therefore, the correct answer is 14. Option 18 is a close distractor because it results from forgetting to subtract 4. Exam tip: First rewrite terms with a common coefficient as a multiple of the given sum.