Algebraic expressions

Adding the two equations gives (2x=16). In such questions, try adding the given expressions.

On expanding the brackets, \(4(2x-y)=8x-4y\) and \(-3(x+2y)=-3x-6y\). Thus, the expression becomes \(8x-4y-3x-6y+5y\). Combining like terms gives \((8x-3x)+(-4y-6y+5y)=5x-5y\). The distractor \(11x-5y\) can result from incorrectly treating \(-3x\) as positive. Exam tip: when a negative sign is outside a bracket, change the sign of every term inside it.

The term containing x is px. In px, the coefficient of x is p. Since the coefficient of x is given as -7, p=-7. Option 7 is incorrect because it ignores the negative sign. Exam tip: The number or letter multiplying a variable in a term is its coefficient.

To add the polynomials, combine like terms: \(2x^2+x^2=3x^2\), \(-3x+x=-2x\), and \(1-4=-3\). Therefore, the resulting expression is \(3x^2-2x-3\). Option B has an error in adding the \(x\)-terms. Exam tip: add the \(x^2\)-terms, \(x\)-terms, and constants separately.

Substituting the given values: (a^3+3a^2b+2b^3)=2^3+3(2^2)(-1)+2(-1)^3=8-12-2=-6. Therefore, the correct answer is -6. The value -2 may result from omitting the term 3a²b. Exam tip: An odd power of a negative number remains negative; for example, (-1)^3=-1.

First simplify the innermost bracket: \(2x-(3x-4)=2x-3x+4=-x+4\). Then \(5x-(-x+4)=5x+x-4=6x-4\). Therefore, the correct answer is \(6x-4\). In \(6x+4\), the sign of the final \(-4\) has been changed incorrectly. Exam tip: when a bracket is preceded by a minus sign, change the sign of every term inside it.

Like terms have exactly the same variable part, including the exponent of every variable; only their numerical coefficients may differ. The variable part of 6x^2yz is x^2yz, so 9x^2yz is a like term. In 9xy^2z, the exponents of x and y are different, so it is not a like term. Exam tip: Ignore the coefficient and compare the exponent of each variable.

In a polynomial, variable exponents must be non-negative integers such as \(0,1,2,\dots\). Here \(x^{-1}=1/x\), so the expression is not a polynomial. Exam tip: check for negative exponents first.

On substituting \(x=3\), the numerator is \(3^2+2(3)-3=9+6-3=12\). Therefore, the complete expression is \(\frac{12}{3}=4\), so option A is correct. Option 12 is only the value of the numerator; it still has to be divided by the denominator, \(x=3\). Exam tip: In a fractional expression, evaluate the numerator and denominator separately before dividing.

In any polynomial in x, every term containing x has at least one factor of x. On substituting x=0, all such terms become 0, leaving only the constant term. Hence, the value of every polynomial at x=0 equals its constant term. “Only quadratic” or “only linear” is incorrect because this rule applies to polynomials of every degree. Exam tip: To find a polynomial’s constant term quickly, substitute x=0.

On expanding, \(2(3a-2b)+5(a+b)-4a=6a-4b+5a+5b-4a\). Combining like terms gives \((6a+5a-4a)+(-4b+5b)=7a+b\). Therefore, the coefficient of \(b\) is \(1\). The value \(-1\) would result from incorrectly adding \(-4b\) and \(5b\). Exam tip: to find a variable’s coefficient, first combine only the terms containing that variable.

Given x² = 9 and x = 3, substitute each value into its corresponding term: 4x² - 5x + 2 = 4(9) - 5(3) + 2 = 36 - 15 + 2 = 23. Therefore, 23 is correct. The value 17 can result from a sign error, such as subtracting 2 instead of adding it. Exam tip: Substitute the given values for x² and x separately in their respective terms.

Add the like terms in the two polynomials: \(2x^2-x^2=x^2\), \(-3xy+5xy=2xy\), and \(y^2-4y^2=-3y^2\). Therefore, the resulting expression is \(x^2+2xy-3y^2\). Option D incorrectly combines the \(y^2\) terms. Exam tip: while adding polynomials, combine only terms with the same variables raised to the same powers.

Given (3x-2=10), adding 2 to both sides gives 3x=12, so x=4. Hence, x^2+x=4^2+4=16+4=20. The value 16 is only x^2; the +x term must also be included. Exam tip: First find x, then substitute it into the complete expression.

The minus sign before the bracket changes the sign of every term inside it: \(10x^2-3x+1-4x^2+8x-6\). Combining like terms gives \((10-4)x^2+(-3+8)x+(1-6)=6x^2+5x-5\). In option B, the sign of \(-8x\) has not been changed correctly. Exam tip: When removing brackets preceded by a minus sign, reverse the sign of every term inside the bracket.

Given \(a+b=7\) and \(ab=10\), substitute these values directly into \(3(a+b)-2ab\): \(3\times7-2\times10=21-20=1\). Therefore, the correct answer is 1. Option 31 may result from incorrectly adding instead of subtracting. Exam tip: When the values of grouped expressions are given, substitute the complete groups first.

In option A, the powers of \(x\) are 4, 1 and 0, so the highest power is 4. Hence it is a fourth-degree polynomial, and \(-\sqrt{3}\) is an irrational coefficient. Option B contains \(x^{-1}\), so it is not a polynomial. Exam tip: variable exponents in a polynomial must be non-negative integers.

A constant term is a term with no variable. In 2x^2+3x+5, only 5 has no x, so it is the constant term. Removing 5 leaves 2x^2+3x, which has no constant term. Removing 3x would still leave 5 as the constant term. Exam tip: Identify the term without a variable to find the constant term.

Taking the common factor \(xy\) from the first expression gives \(x^2y+xy^2=xy(x+y)\). Hence, the two expressions are algebraically identical. For \(x=2\) and \(y=3\), the first value is \(2^2\cdot3+2\cdot3^2=12+18=30\), while the second is \(2\cdot3\cdot(2+3)=30\). Therefore, both values are equal. Exam tip: Factor out the common term to compare algebraic expressions quickly.

Combine like terms: \(7x^2-4x^2=3x^2\), \(-2xy+6xy=4xy\), and \(3y^2-8y^2=-5y^2\). Therefore, the simplified expression is \(3x^2+4xy-5y^2\). Option B results from incorrectly adding the coefficients of the \(x^2\) and \(y^2\) terms. Exam tip: combine only terms having the same variables with the same powers.