Algebraic expressions

Add the like terms in the two expressions: 2x^2+x^2=3x^2, -5x+4x=-x, and 3+(-8)=-5. Therefore, the sum is (3x^2-x-5). Option B incorrectly adds the x-terms. Exam tip: while adding polynomials, combine only terms with the same variable and exponent.

While subtracting the second polynomial, change the sign of each of its terms: \((9x^2-4x+1)-(3x^2+2x-5)=9x^2-4x+1-3x^2-2x+5\). Combining like terms gives \(6x^2-6x+6\). Option A has an incorrect linear term because \(-4x-2x=-6x\). Exam tip: change every sign inside the polynomial being subtracted before removing the brackets.

Two algebraic terms are like terms only when the exponent of every corresponding variable is the same. In \(x^2y\), x has exponent 2 and y has exponent 1. In \(xy^2\), x has exponent 1 and y has exponent 2. Although both terms contain the same two variables, their powers do not match. Therefore they are unlike terms, so option B is correct.

The order of writing the variables is not the real issue; the exponents attached to each variable must be compared. For instance, \(4x^2y\) and \(-7x^2y\) would be like terms because both have the pattern \(x^2y\). But \(x^2y\) and \(xy^2\) have different variable parts and cannot be combined by ordinary addition. They are not constants, and multiplying their variable parts is not what the question asks. Hence B is the correct statement.

Given \(a=1\) and \(b=-2\), we have \(a^2=1\), \(b^2=4\), and \(b^3=-8\). Thus, \(4a^2b-3ab^2+b^3=4(1)(-2)-3(1)(4)+(-8)=-8-12-8=-28\). Therefore, \(-28\) is correct. The option \(-12\) is only the value of the second term, not of the complete expression. Exam tip: an odd power of a negative number is negative, while an even power is positive.

On expanding the brackets, \(2(4x-3y)=8x-6y\) and \(-5(x-y)=-5x+5y\). Thus, the expression becomes \(8x-6y-5x+5y+3y\). Combining like terms gives \((8x-5x)+(-6y+5y+3y)=3x+2y\). The distractor \(3x-8y\) results from incorrectly handling the sign of the \(y\)-term while expanding \(-5(x-y)\). Exam tip: When a negative coefficient multiplies a bracket, apply it to every term inside the bracket.

In \(x^2+kx+16\), the term containing \(x\) is \(kx\). The coefficient of \(x\) in this term is \(k\). Since the coefficient of \(x\) is given as \(12\), \(k=12\). Therefore, \(k-5=12-5=7\). Option \(12\) is the value of \(k\), not of \(k-5\). Exam tip: identify the factor multiplying the variable to find its coefficient.

The square of \(x\) is \(x^2\), so three times its square is \(3x^2\). Twice \(y\) is \(2y\). The phrase “the difference between A and B” means \(A-B\), so the required expression is \(3x^2-2y\). Option C reverses the order of subtraction and represents \(2y-3x^2\). Exam tip: Write the square first, then apply the numerical coefficient.

Combine like terms: \(4x^2-x^2=3x^2\), \(-3xy+5xy=2xy\), and \(2y^2-7y^2=-5y^2\). Hence, the simplified expression is \(3x^2+2xy-5y^2\). Option B uses an incorrect coefficient for \(x^2\). Exam tip: add or subtract only terms having the same variables with the same powers.

Substitute the value directly into the expression:

\(2x(x-5)+4x^2=2\times3\times(3-5)+4\times3^2\)

\(=6\times(-2)+4\times9=-12+36=24\). Therefore, the correct answer is 24. A value such as 30 may result from an error while calculating the negative term \(-12\). In exams, evaluate powers such as \(x^2\) first and check signs inside brackets carefully.

Only like terms can be added or subtracted. The \(x^2y\) terms are \(3x^2y\) and \(-5x^2y\), so their sum is \((3-5)x^2y=-2x^2y\). Therefore, the coefficient of \(x^2y\) is \(-2\). The terms \(2xy^2\) and \(8xy^2\) are not like terms of \(x^2y\), so they do not affect this coefficient. Exam tip: Before combining terms, check that both the variables and their powers are identical.

Using the distributive law, \(x(4x-9)=x\cdot4x-x\cdot9=4x^2-9x\). Therefore, option A is correct. Option B expands to \(4x^2-36x\), so it is not equivalent even though it looks similar. Exam tip: To check a factored expression, multiply each term outside the bracket by every term inside it.

Adding \(p+q=6\) and \(p-q=2\) gives \(2p=8\), so \(p=4\). Substituting \(p=4\) into \(p-q=2\) gives \(q=2\). Hence, \(3p-q=3(4)-2=10\). The value \(12\) is only \(3p\); \(q\) must also be subtracted. Exam tip: Add or subtract a pair of linear equations first to eliminate one variable.

To subtract the second polynomial, change the sign of each of its terms: \((5a^2-2ab+3b^2)-(2a^2+5ab-b^2)=5a^2-2ab+3b^2-2a^2-5ab+b^2\). Combining like terms gives \(3a^2-7ab+4b^2\). In option B, the sign of the \(ab\) term is incorrect. Exam tip: when a minus sign comes before brackets, reverse the signs of every term inside the brackets.

The terms of 2x² + 5x + 3 are 2x², 5x, and 3. Here, 3 is the constant term because it contains no x. Removing 3 leaves 2x² + 5x, which has exactly two terms and no constant term. If 5x were removed, the constant term 3 would still remain. Exam tip: A term with no variable is called a constant term.

Substituting the given values, \(x^2y^2=(-1)^2\times2^2=1\times4=4\), while \(2xy=2\times(-1)\times2=-4\). Hence, \(x^2y^2-2xy=4-(-4)=8\). The option 4 results from missing the rule for subtracting a negative number. Exam tip: when a negative term is subtracted, change it to addition.

On expanding, \(2(3x-y)=6x-2y\) and \(4(x+2y)=4x+8y\). Therefore, \(6x-2y+4x+8y-5x=(6+4-5)x+(-2+8)y=5x+6y\). In option B, the sign of the \(y\)-term is incorrect because \(-2y+8y=+6y\). Exam tip: after removing brackets, combine terms with the same variable separately.

The total degree of the monomial \(x^3y^2\) is the sum of the exponents of all its variables: \(3+2=5\). The power of \(y\) is \(2\). Therefore, the required difference is \(5-2=3\). Note that \(5\) is the total degree, not the difference. Exam tip: To find the total degree of a monomial, add the exponents of every variable in it.

Given \(x^2+x=20\). Taking 3 common from the first two terms, \(3x^2+3x-4=3(x^2+x)-4\). Substituting the given value gives \(3(20)-4=60-4=56\). Therefore, 56 is correct. Option 60 is a close distractor because it results from forgetting to subtract 4. Exam tip: Rewrite the expression to contain the exact group given in the question before substituting.

Combining like terms gives \(7x^2y+4x^2y=11x^2y\) and \(-3xy+2xy=-xy\). Therefore, the simplified expression is \(11x^2y-xy-6\), which does not contain the term \(5xy\). The close distractor \(-xy\) does remain because the coefficients of the \(xy\) terms add to \(-3+2=-1\), not \(5\). Exam tip: Add or subtract only those terms that have exactly the same variables raised to the same powers.

Taking the common factor \(xy\) from the first expression gives \(x^2y+xy^2=xy(x+y)\). Therefore, the two expressions are equal for every value of \(x\) and \(y\). Substituting the given values, \(2^2\times1+2\times1^2=6\) and \(2\times1\times(2+1)=6\). Hence, neither expression is greater. Exam tip: factor out the common term \(xy\) to check such equivalence quickly.

Combine like terms: \(8x^2-3x^2=5x^2\), \(-6xy+9xy=3xy\), and \(y^2-5y^2=-4y^2\). Hence, the simplified expression is \(5x^2+3xy-4y^2\). In option B, the coefficients of the \(x^2\) and \(y^2\) terms have been combined incorrectly. Exam tip: add or subtract only terms with the same variables raised to the same powers.

When a whole polynomial is subtracted, the minus sign must be distributed to every term inside its brackets. Starting with \(4x^2-7x+6-(2x^2-5x-3)\), change the signs in the second bracket: \(-(2x^2-5x-3)=-2x^2+5x+3\). The expression becomes \(4x^2-7x+6-2x^2+5x+3\).

Now combine like terms. The quadratic terms give \(4x^2-2x^2=2x^2\), the linear terms give \(-7x+5x=-2x\), and the constants give \(6+3=9\). Hence the simplified expression is \(2x^2-2x+9\), which is option A. Options B and C result from failing to combine or change signs correctly, while option D has the wrong sign for the linear term. The supplied answer and explanation are correct.

Direct answer: Option B, \\(-7\\). Substitute \\(x=-2\\) carefully into every occurrence of x: \\(x^4+2x^3-3x^2+5=(-2)^4+2(-2)^3-3(-2)^2+5\\). Now calculate the powers: \\((-2)^4=16\\) because an even power is positive; \\((-2)^3=-8\\) because an odd power keeps the negative sign; and \\((-2)^2=4\\). Therefore the expression becomes \\(16+2(-8)-3(4)+5=16-16-12+5\\). Next, \\(16-16=0\\), \\(0-12=-12\\), and \\(-12+5=-7\\). Option A, 7, has the wrong final sign. Option B, -7, matches the calculation. Option C, 23, may result from mishandling the negative odd-power term or subtraction. Option D, 9, also does not follow the correct substitution and sign rules. Always use brackets around a negative substituted value. Memory cue: even powers become positive, odd powers retain the negative sign.

Only like terms can be added or subtracted. Here, 9a^2b and -6a^2b are like terms, so they give 3a^2b. Similarly, -4ab^2 and 11ab^2 give 7ab^2. Hence, the simplified expression is 3a^2b+7ab^2. Option A results from adding the coefficients of a^2b incorrectly. Exam tip: Before combining terms, check that both the variables and their exponents are identical.

displaystyle (2k+1)x-4x=[(2k+1)-4]x=(2k-3)x. Since its simplified form is 9x, the coefficients of x must be equal: 2k-3=9. Thus, 2k=12 and k=6. For example, choosing 5 gives coefficient 7, not 9. Exam tip: while combining like terms, add or subtract their coefficients only.