Quadratic expressions

Using \((a+b)^2=a^2+2ab+b^2\), put \(a=x\) and \(b=5\). The middle term is \(2\times x\times5=10x\), so A is correct. \(5x\) misses the factor 2. Exam tip: always check the \(2ab\) term.

For a quadratic expression, the coefficient of \(x^2\) must be non-zero; hence \(a\ne0\). The values of \(b\) or \(c\) may be zero, as in \(x^2+5\). Exam tip: check the highest non-zero power of the variable.

A quadratic polynomial must have 2 as the highest power of \(x\), so the coefficient \(a\) of \(x^2\) cannot be zero. If \(a=0\), the expression becomes at most linear. Exam tip: first check the coefficient of the highest-power term.

The expression is \(x^2+6x+9\). Here \(9=3^2\) and the middle term is \(6x=2\cdot x\cdot3\), so it equals \((x+3)^2\). Hence it is a perfect-square trinomial. Exam tip: check the middle-term pattern \(2ab\).

Area of a rectangle = length × breadth. Thus, \((x+4)(x-2)=x^2-2x+4x-8=x^2+2x-8\). Option B has an incorrect middle term, \(6x\). Exam tip: multiply each term before combining like terms.

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect-square trinomial. In option B, the constant should be \(5^2=25\), not 20. Exam tip: match the middle term with \(2ab\).

A quadratic expression must have \(x^2\) as its highest power, so its coefficient \(a\) must be non-zero. If \(a=0\), the \(x^2\) term disappears and the expression can be linear. Exam tip: check the leading coefficient first.

For an expression to be quadratic, the coefficient of \(x^2\) must be non-zero; hence \(a\neq0\). The values of \(b\) and \(c\) may be zero. If \(a=0\), the highest power becomes at most 1. Exam tip: identify the expression by its highest non-zero power.

\(x^2-25=x^2-5^2\), so using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-5)(x+5)\). \(x^2+25\) is a sum of squares. Exam tip: look for a subtraction sign between perfect squares.

\(9p^2-16q^2=(3p)^2-(4q)^2\). Hence, it is the difference of the squares of two monomials and factors as \((3p+4q)(3p-4q)\). Option C is a perfect-square trinomial, \((3p-4q)^2\), not a difference of squares. Exam tip: check for two perfect-square terms separated by a minus sign.

This expression is a difference of squares: \(25x^2-1=(5x)^2-1^2\). Applying \(a^2-b^2=(a-b)(a+b)\) gives \((5x-1)(5x+1)\). Options A and D are perfect squares; on expansion, they contain middle terms \(-10x\) and \(+10x\), respectively, which are not present in the expression. Exam tip: before factorising, check whether both terms are perfect squares separated by subtraction.

\(x^2-49=x^2-7^2\), so it matches \(a^2-b^2=(a-b)(a+b)\) and factorises as \((x-7)(x+7)\). \(x^2+49\) is a sum of squares, not a difference. Exam tip: check whether both terms are perfect squares with a minus sign.

\(x^2-25=x^2-5^2\), so it matches \(a^2-b^2=(a-b)(a+b)\) and factors as \((x-5)(x+5)\). \(x^2+6x+9\) is a perfect square instead. Exam tip: check for two squares separated by a minus sign.

A quadratic expression must have highest power 2, so the coefficient \(a\) of \(x^2\) cannot be zero. The coefficients \(b\) and \(c\) may be zero. Exam tip: first check the coefficient of the highest-power term.

Area = length × width = \((x+5)(x-2)\). Expanding gives \(x^2-2x+5x-10=x^2+3x-10\). Option B adds the linear terms incorrectly. Exam tip: multiply each term in one bracket by each term in the other.

In \(7x^2-3x+1\), the highest power of \(x\) is 2, so it is quadratic. Option B has highest power 3 and is cubic. Exam tip: identify the degree from the greatest exponent.

Using \((a+b)^2=a^2+2ab+b^2\), put \(a=2x\) and \(b=1\). The middle term is \(2\times2x\times1=4x\). \(2x\) misses the factor 2 in 2ab. Exam tip: calculate the middle term separately.

(bx) is the linear term because the power of (x) is (1). Exam tip: identify the term type from its power.

In a quadratic expression, the coefficient of the \(x^2\) term must not be zero. If \(a=0\), the term \(ax^2\) disappears and the expression becomes \(bx+c\), which may be linear (or constant when \(b=0\)), not quadratic. Option C refers to a linear expression with highest power 1. Exam tip: The degree of a polynomial is determined by the highest power with a non-zero coefficient.

A quadratic expression must have 2 as the highest power of \(x\), so the coefficient \(a\) of \(x^2\) must be non-zero. The values of \(b\) and \(c\) may be zero. Exam tip: check the leading coefficient first.

A polynomial whose greatest exponent of x is 2 has degree 2, so it is called a quadratic polynomial. A linear polynomial has highest power 1. Exam tip: identify a polynomial’s type by its greatest exponent.

Using \((a+b)^2=a^2+2ab+b^2\), put \(a=x\) and \(b=5\) to get \(x^2+10x+25\). Thus, the \(10x\) middle term is missing. Exam tip: always check the \(2ab\) term.

Using the distributive property, \((2x+3)(x-2)=2x\cdot x+2x\cdot(-2)+3\cdot x+3\cdot(-2)\). Thus, \(2x^2-4x+3x-6=2x^2-x-6\). Therefore, option D is correct. In option C, the constant term is incorrectly written as \(+6\) instead of \(-6\). Exam tip: multiply each term of one binomial by both terms of the other, then combine like terms.

Using the distributive property, \((3x-1)(x+4)=3x\cdot x+3x\cdot4-1\cdot x-1\cdot4=3x^2+12x-x-4=3x^2+11x-4\). Therefore, option D is correct. In option A, \(12x-x\) has incorrectly been simplified to \(x\). Exam tip: after expanding, check the sign of every product before combining like terms.

\(9p^2-25q^2=(3p)^2-(5q)^2\), so it is a difference of two squares. Its factorisation is \((3p-5q)(3p+5q)\). Option C is \((3p-5q)^2\), which is a perfect square, not a difference of squares. Exam tip: when two squared terms have a minus sign and no middle term, check the form \(a^2-b^2\).