Algebraic expressions

In a polynomial, exponents of variables must be 0 or positive integers. Here \(a^{-1}=\frac{1}{a}\), so the negative exponent makes it non-polynomial. Exam tip: check every variable exponent.

Given x=-2, substitute it into the expression: x^2+3x+1=(-2)^2+3(-2)+1=4-6+1=-1. Therefore, the correct value is -1. The option 1 may result from incorrectly taking 3x as +6. Exam tip: always use brackets while squaring a negative number, for example, (-2)^2=4.

Terms in an expression are separated by addition or subtraction signs. Here the terms are \(7mn\), \(-3m\), and \(5\), so there are 3 terms. The term \(-3m\) must be counted separately, with its negative sign. Exam tip: Count parts separated by \(+\) or \(-\), not factors joined by multiplication.

Using the distributive property, \(2(x-5)=2x-10\). Adding \(3x\) to \(2x-10\), the like terms \(2x\) and \(3x\) combine to give \(5x\), so the simplified expression is \(5x-10\). \(5x+10\) is incorrect because \(2\times(-5)=-10\), not \(+10\). Exam tip: When opening brackets, multiply the outside number by every term inside the bracket.

The linear term containing x is -3x. Since -3x = (-3)×x, the coefficient of x is -3. Here, 4 is the coefficient of x², while 6 is the constant term. In exams, identify the number multiplying the variable and keep its sign.

Substituting the given values, \(3a-2b=3(2)-2(-1)=6+2=8\). Since \(b=-1\), \(2b=-2\), so subtracting \(2b\) means subtracting \(-2\), which gives \(+2\). Option 6 is only the value of \(3a\); it ignores the \(-2b\) term. Exam tip: When subtracting a negative number, use brackets first to avoid sign errors.

A constant term is a term that contains no variable. In \(5u^2+2u-8\), both \(5u^2\) and \(2u\) contain the variable \(u\), whereas \(-8\) does not. Therefore, the constant term is \(-8\). Also, \(8u\) is not a term in the given expression. Exam tip: always retain the sign while identifying a constant term.

Here, 4r^2 and -6r^2 are like terms, so their coefficients are combined: 4-6=-2. Thus, 4r^2-6r^2=-2r^2, while 5r remains unchanged because it has a different power of r. Therefore, the simplified expression is -2r^2+5r. In option D, the sign of 5r is incorrectly changed. Exam tip: Combine only terms with the same variable raised to the same power.

In 7ab and -2ab, the variable part ab is exactly the same: both a and b have exponent 1. Therefore, they are like terms even though their coefficients, 7 and -2, are different. In option C, the exponents of a and b are interchanged, so those terms are not like terms. Exam tip: compare only the variables and their exponents; coefficients need not be the same.

Substituting t=3 gives 2t^2-t=2(3^2)-3=2(9)-3=18-3=15. Therefore, the correct answer is 15. The nearby distractor 18 results from forgetting to subtract 3 in the final step. Exam tip: after substitution, evaluate powers first, then multiplication, and finally addition or subtraction.

On expanding the brackets, \(5(x+2)-2(x-1)=5x+10-2x+2\). Multiplying \(-2\) by both terms in \((x-1)\) gives \(-2x+2\). Combining like terms gives \(5x-2x+10+2=3x+12\). The option \(3x+8\) may result from using the wrong sign for the constant term in \(-2(x-1)\). Exam tip: when a negative coefficient is outside a bracket, distribute it carefully to every term inside.

“5 more than 2y” means add 5 to 2y. Therefore, the expression is 2y+5. The expression 2y-5 represents 5 less than 2y, so it is not correct. Exam tip: Use addition (+) for “more than.”

Given \(p=-3\), \(p^2+2p=(-3)^2+2(-3)=9-6=3\). Therefore, the correct value is \(3\). The option \(-3\) can result from not handling the square of the negative value correctly. Exam tip: Always use brackets when squaring a negative number, as in \((-3)^2\).

\(8a^2b\) and \(-3a^2b\) are like terms because both have \(a\) raised to 2 and \(b\) raised to 1. Adding their coefficients gives \(8-3=5\), so they combine to form \(5a^2b\). The term \(4ab\) is not like \(a^2b\), since the power of \(a\) is 1, so it remains separate. Hence, the simplified expression is \(5a^2b+4ab\). Exam tip: combine coefficients only when the variables and all their exponents are exactly the same.

The square of \(x\) is \(x^2\), and three times \(x\) is \(3x\). In the difference between these quantities, the second quantity is subtracted from the first, so the expression is \(x^2-3x\). \(3x-x^2\) reverses the order of subtraction. Exam tip: read “the difference between A and B” as \(A-B\).

A term in algebra can be separated into its numerical coefficient and its variable part. In \(12x^2y\), the number 12 tells how many times the variable product is taken, so 12 is the coefficient. The letters and their powers together form the variable part. Hence the variable part is \(x^2y\), making option B correct. The exponent 2 belongs to x, while y has an understood exponent of 1.

To verify this, rewrite the term as \(12\times x^2\times y\). The first factor is numerical, and the remaining factors contain variables, so they form the variable part. Option A names only the coefficient. Option C contains both a number and a variable, so it is not the complete variable part. Option D leaves out \(x^2\). Therefore, separating the term into coefficient and literal or variable factors leads to option B.

A minus sign before a bracket changes the sign of every term inside it: \(4m-(2m-7)=4m-2m+7=2m+7\). Therefore, \(2m+7\) is correct. \(2m-7\) would result if the sign of \(-7\) were not changed. Exam tip: Before removing brackets, check the sign immediately preceding them.

\(x\) and \(2x\) are like terms because both have \(x\) raised to the power 1. Adding their coefficients gives \(x+2x=3x\). The term \(x^2\) is not like \(x\), so it remains separate. Therefore, the expression becomes \(x^2+3x\). Exam tip: combine terms only when both the variable and its exponent are the same.

In \(-4x^2+7\), the \(x^2\)-term is \(-4x^2\), which can be written as \((-4)\times x^2\). Hence, the coefficient of \(x^2\) is \(-4\). In option C, \(4-x^2=4+(-1)x^2\), so its coefficient is \(-1\), not \(-4\). Exam tip: Always include the sign when identifying a coefficient.

The cost of n notebooks is \(12n\) rupees because each notebook costs ₹12. The ₹5 cover charge is paid only once, so the total cost is \(12n+5\). In \(17n\), the cover charge has incorrectly been added for every notebook. Exam tip: multiply the per-item cost by the variable, then add any one-time charge as a constant term.

Combine like terms in the expression: \(2a+5a=7a\) and \(-3b-b=-4b\). Therefore, \(2a-3b+5a-b=7a-4b\), so option B is correct. In option A, the \(b\)-terms have been combined incorrectly, while option D wrongly multiplies \(a\) and \(b\). Exam tip: Add or subtract only terms with the same variable and the same power.

In \(3x^2y+4xy^2-5\), the only distinct variables are \(x\) and \(y\), so there are 2 variables. The exponent 2 in \(x^2\) and \(y^2\) does not create an additional variable, and \(-5\) is a constant. Exam tip: count distinct letters representing variables, not their powers.

Using the distributive property, \(7(z-1)=7z-7\). Then \(7z-7+2=7z-5\), so the correct answer is \(7z-5\). The expression \(7z-9\) would result from incorrectly treating \(+2\) as \(-2\). Exam tip: after expanding brackets, combine only like terms.

In \(5s^2\) and \(5s\), the variable \(s\) is the same, but its powers are 2 and 1 respectively. Their literal parts are therefore different, so they are unlike terms. In contrast, \(4c\) and \(-9c\) are like terms because both contain \(c\) to the power 1. Exam tip: for like terms, every variable and its exponent must match exactly; only the coefficients may differ.

Given k = 4, substitute 4 for k in the expression: 4² − 2(4) + 3 = 16 − 8 + 3 = 11. Therefore, the correct answer is 11. The value 9 may result if the final +3 is mistakenly omitted. Exam tip: substitute the value first, then apply powers, multiplication, and addition/subtraction in order.