Algebraic expressions

While subtracting the second polynomial, the sign of every term in it changes: \((12x^2-5x+4)-(7x^2-9x+10)=12x^2-5x+4-7x^2+9x-10\). Combining like terms gives \(5x^2+4x-6\). In option C, the \(-9x\) term has not been correctly changed to \(+9x\) during subtraction. Exam tip: when a minus sign precedes brackets, reverse every sign inside the brackets.

Like terms must have the same variables raised to the same powers. In \(r^3s^2\), r has exponent 3 and s has exponent 2. In \(r^2s^3\), r has exponent 2 and s has exponent 3. The powers of both variables are therefore different in corresponding positions. Even though the same letters occur, the variable parts are not identical, so the terms are unlike. Option B is correct.

To see why option D is not suitable, multiplying the terms would produce a different expression, \(r^5s^5\), but that is not their common variable part and is not the test for like terms. Constants contain no variables, so C is false. Option A is false because matching letters alone do not make terms like. Comparing the exponent pair \((3,2)\) with \((2,3)\) clearly shows the difference and confirms B.

Substituting the given values, \(a^3b=(-2)^3\times3=-24\), \(-2ab^2=-2\times(-2)\times3^2=36\), and \(b^3=3^3=27\). Therefore, \(a^3b-2ab^2+b^3=-24+36+27=39\). Hence, 39 is correct. The option \(-39\) can result from handling the negative signs incorrectly. Exam tip: evaluate powers first and check the sign of every term separately.

On expanding, \(3(5x-2y)=15x-6y\) and \(-4(2x+y)=-8x-4y\). Thus the expression becomes \(15x-6y-8x-4y+6y\). Combining like terms gives \(15x-8x=7x\) and \(-6y-4y+6y=-4y\), so the simplified expression is \(7x-4y\). Exam tip: when multiplying a bracket by a negative number, apply the negative sign to every term inside the bracket.

The term containing x is kx, so the coefficient of x is k. Given that the coefficient of x is -11, we get k=-11. Therefore, k+4=-11+4=-7. Option -11 is the value of k only, not of k+4. Exam tip: To identify a coefficient, look at the number or variable multiplying the given variable in its term.

The square of \(y\) is \(y^2\), so five times its square is \(5y^2\). Three times \(x\) is \(3x\). Taking the difference in the stated order gives \(5y^2-3x\), so option A is correct. Option C reverses the order and therefore changes the sign of the difference. Exam tip: “the difference between A and B” is usually written as \(A-B\).

Combine like terms: \(6x^2-4x^2=2x^2\), \(-7xy+2xy=-5xy\), and \(3y^2-5y^2=-2y^2\). Therefore, the simplified expression is \(2x^2-5xy-2y^2\). Option D has the correct \(xy\) coefficient but the wrong sign for the \(y^2\) term. Exam tip: add or subtract only terms with the same variables raised to the same powers.

First simplify (3x(x-2)-2x^2=x^2-6x). Substituting (x=4) gives (16-24=-8), so check the options carefully.

Like terms must have the same variables raised to the same respective powers. Here, the powers of \(a\) are 2 and 1, while those of \(b\) are 1 and 2. Their total degree is 3, but they are still unlike terms. Exam tip: compare each variable’s exponent separately.

Using the distributive property, \(x(6x-13)=x\cdot6x-x\cdot13=6x^2-13x\). Hence, option A is correct. Option B gives \(+13x\), while option C gives \(-78x\) as the second term. Exam tip: To check a factored expression, multiply the outside factor by every term inside the bracket.

Adding the given equations gives \(2p=14\), so \(p=7\). Substituting \(p=7\) in \(p+q=10\) gives \(q=3\). Hence, \(2p+q=2(7)+3=17\). Option 14 is only the value of \(2p\); it does not include \(q\). Exam tip: Use addition or subtraction of the two linear equations to eliminate one variable quickly.

While subtracting the second polynomial, the sign of every term in it changes: \((7a^2-4ab+6b^2)-(3a^2-9ab+2b^2)=7a^2-4ab+6b^2-3a^2+9ab-2b^2\). Combining like terms gives \((7-3)a^2+(-4+9)ab+(6-2)b^2=4a^2+5ab+4b^2\). Option C results from incorrectly retaining the negative sign of \(-9ab\). Exam tip: when a minus sign precedes brackets, reverse the signs of all terms inside the brackets.

In \(4x^2+7x-9\), \(-9\) is the constant term because it has no \(x\). Removing \(-9\) leaves \(4x^2+7x\), which has exactly two terms and no constant term. If \(7x\) were removed, the constant term \(-9\) would still remain. Exam tip: A term without a variable is called a constant term.

\(4x^3-2x+7\) has three terms: \(4x^3\), \(-2x\), and \(7\). Its highest power of \(x\) is 3, so it is a trinomial polynomial of degree 3. Option B has four terms, C has degree 2, and D is not a polynomial because \(\frac{2}{x}=2x^{-1}\). Exam tip: count the terms first, then check the highest non-negative integer exponent of the variable.

Expand the brackets: \(3(2x+y)=6x+3y\) and \(-5(x-2y)=-5x+10y\). Therefore, \(6x+3y-5x+10y+2x=3x+13y\). Hence, option A is correct. In \(13x+13y\), the \(x\)-terms have been added incorrectly. Exam tip: When a bracket has a negative coefficient before it, apply the sign to every term inside the bracket.

The total degree of the monomial \(x^4y^3\) is the sum of the exponents: \(4+3=7\). The power of \(x\) is \(4\), so the difference is \(7-4=3\). Note that \(4\) is only the exponent of \(x\), not the total degree. Exam tip: To find the total degree of a monomial, add the exponents of all its variables.

Given \,\(x^2-2x=15\), rewrite the expression as \,\(2x^2-4x+7=2(x^2-2x)+7\). Therefore, its value is \,\(2\times15+7=37\). Option 30 would result from forgetting to add the constant term 7. Exam tip: first rewrite the expression to identify the given group \,\(x^2-2x\).

Combining like terms gives \(6x^2y+9x^2y=15x^2y\) and \(-4xy+7xy=3xy\). Therefore, the simplified expression is \(15x^2y+3xy-12\), so the term \(11xy\) does not remain. \(3xy\) is a close distractor because it is actually obtained by adding the \(xy\) terms. Exam tip: combine only terms having the same variables with the same powers.

Factoring out the common factor \,\(xy\) from \,\(x^2y-xy^2\) gives \,\(xy(x-y)\), which is exactly the second expression. For the given values, the first expression is \,\(3^2\times2-3\times2^2=18-12=6\), while the second is \,\(3\times2\times(3-2)=6\). Hence, both values are equal. The zero option is incorrect because \,\(x-y=1\), not zero. Exam tip: Factor out the common term to compare algebraic expressions quickly.

Combine like terms: \(9x^2-2x^2=7x^2\), \(-5xy+8xy=3xy\), and \(4y^2-7y^2=-3y^2\). Therefore, the simplified expression is \(7x^2+3xy-3y^2\). 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.