Algebraic expressions

On opening the bracket, \(-3(4-x)=-12+3x\), because \(-3\) multiplies both 4 and \(-x\). Therefore, \(2x-3(4-x)=2x-12+3x=5x-12\). In \(5x+12\), the sign of the constant term is incorrect. Exam tip: when multiplying a bracket by a negative number, check the sign of every term.

In \(8rs-5r\), the term \(8rs\) can be written as \(8\times rs\). Hence, the coefficient of \(rs\) is 8. In \(rs+8\), the coefficient of \(rs\) is 1 because \(rs=1\times rs\). Exam tip: To find a coefficient, identify the numerical factor multiplying the term.

Here, \(2x^2\) and \(-4x^2\) are like terms, so \(2x^2-4x^2=-2x^2\). Similarly, \(3xy\) and \(5xy\) are like terms, giving \(3xy+5xy=8xy\). Therefore, the simplified expression is \(-2x^2+8xy\). The terms \(x^2\) and \(xy\) are not like terms, so they cannot be combined. Exam tip: add or subtract only terms with identical variables and powers.

Substituting the given values, (2x+y^2)=2(1)+5^2=2+25=27. Hence, the correct answer is 27. A result such as 12 may come from incorrectly treating y² as 2y. In exams, evaluate powers first, then multiplication and addition.

Use the distributive property: \(6(x+1)-x=6x+6-x\). Combining the like terms \(6x\) and \(-x\) gives \(5x\), so the expression equals \(5x+6\). The option \(7x+6\) results from incorrectly treating \(6x-x\) as \(7x\). Exam tip: after opening brackets, add or subtract coefficients only of like terms.

Simplifying a polynomial means combining only like terms. Like terms have the same variables with the same powers, so their coefficients can be added. In the expression \(x^2+x+x^2+x+1\), the two \(x^2\) terms combine to give \(2x^2\). The two x terms combine to give \(2x\). The number 1 is a constant and has no matching constant, so it remains 1. Therefore the simplified expression is \(2x^2+2x+1\), which is option A.

A useful check is to group the terms: \((x^2+x^2)+(x+x)+1\). Adding each group gives \(2x^2+2x+1\). We must not multiply the terms or change their powers; addition combines coefficients of identical variable parts. Thus option B incorrectly changes the coefficient and loses the x term, while C and D use incorrect powers. The grouping confirms option A.

When a variable is written without a visible exponent, its exponent is understood to be 1. In xy², x therefore has power 1 and y has power 2, exactly as required. Option B is correct. In x²y, the powers are 2 and 1; in x²y² both are 2; and in xy both powers are 1, so the other expressions do not satisfy both conditions.

Substituting \(n=-4\), we get \(n^2=(-4)^2=16\). Hence, \(\frac{n^2+n}{2}=\frac{16+(-4)}{2}=\frac{12}{2}=6\). Therefore, 6 is the correct answer. The distractor 8 results from wrongly ignoring the \(n=-4\) term in \(n^2+n\). Exam tip: always put a negative number in brackets before squaring it.

On expanding the brackets, \(3(a-b)+2(2a+b)=3a-3b+4a+2b\). Combining like terms gives \(3a+4a=7a\) and \(-3b+2b=-b\). Therefore, the simplified expression is \(7a-b\). The option \(7a+b\) results from combining the terms containing \(b\) incorrectly. Exam tip: multiply the number outside a bracket by every term inside it.

\(4x^2\) and \(3x^2\) are like terms, so they add to \(7x^2\). Similarly, \(-6xy\) and \(+xy\) are like terms, giving \(-6xy+xy=-5xy\). Hence, the simplified expression is \(7x^2-5xy\). \(12x^2-5xy\) is incorrect because \(4+3=7\), not 12. Exam tip: combine only terms with identical variable parts and exponents.

A constant term is a term that contains no variable. In \(2x^2+5x-1\), both \(2x^2\) and \(5x\) contain \(x\), whereas \(-1\) has no variable. Therefore, the constant term is \(-1\). The number \(5\) is the coefficient of \(5x\), not the constant term. Exam tip: identify the term with no letter or variable to find the constant term.

The expression is \(4(3x-2)-5(x+1)\). Expanding the brackets gives \(12x-8-5x-5\). Combining like terms, \(12x-5x=7x\) and \(-8-5=-13\). Therefore, the simplified form is \(7x-13\). The option \(7x+13\) results from an incorrect sign while expanding \(-5(x+1)\), which must be \(-5x-5\), not \(-5x+5\). Exam tip: When a minus sign precedes a bracket, apply it to every term inside the bracket.

Substituting x=-3 gives 2x^2+5x-4=2(-3)^2+5(-3)-4=2(9)-15-4=-1. Therefore, the correct value is -1. Option 1 may result from a sign error; remember that (-3)^2=9. Exam tip: evaluate powers first, then perform multiplication and addition or subtraction.

Combining like terms gives (6a^2-2a^2=4a^2), (-4a+7a=3a), and (9-5=4). Group terms by powers.

Subtracting \((2x+7)\) from \((8x-3)\) gives \((8x-3)-(2x+7)\). Since there is a minus sign before the second bracket, the signs of both its terms change: \(8x-3-2x-7=6x-10\). Hence, \(6x-10\) is correct. \(6x+4\) may result from incorrectly combining \(-3\) and \(+7\). Exam tip: When removing brackets after a minus sign, change the sign of every term inside the bracket.

Substituting the given values,
\(3x^2-2xy+y=3(2)^2-2(2)(-4)+(-4)=12+16-4=24\). Therefore, the correct answer is 24. Note that in
\(-2xy\), using
\(y=-4\) makes the product positive; treating it as
\(-16\) is a common error. In exams, substitute negative values in brackets to track signs correctly.

Using the distributive property, \(5(2m-3n)=10m-15n\) and \(-2(m+n)=-2m-2n\). Therefore, \(10m-15n-2m-2n=(10m-2m)+(-15n-2n)=8m-17n\). In \(8m-13n\), the \(-2n\) term has not been combined correctly. Exam tip: When a negative coefficient is outside brackets, apply it to every term inside the brackets.

First, the sum of p and 3 is \(p+3\). Four times this entire sum is \(4(p+3)\). Subtracting p from it gives \(4(p+3)-p\). In option A, only p is multiplied by 4, not the whole sum. Exam tip: when a multiple of a sum is stated, enclose the sum in brackets.

Substitute r=-2: r^3=(-2)^3=-8 and r^2=(-2)^2=4. Therefore, r^3-4r^2+6r=-8-4(4)+6(-2)=-8-16-12=-36. Hence, option C is correct. A common error is taking r^2 as -4, but the square of a negative number is positive. Exam tip: use brackets while evaluating powers after substitution.

First simplify the innermost bracket: \(4x-(2x-5)=4x-2x+5=2x+5\). Therefore, the complete expression becomes \(3x-(2x+5)=3x-2x-5=x-5\). Hence, \(x-5\) is correct. \(x+5\) results from incorrectly handling the minus sign before the outer bracket. Exam tip: When a bracket is preceded by a minus sign, change the signs of all terms inside it.