Quadratic expressions

In \(x^2+10x+25\), we have \(x^2=x^2\), \(25=5^2\), and the middle term \(10x=2\times x\times5\). Therefore, using \(a^2+2ab+b^2=(a+b)^2\), it becomes \((x+5)^2\). In option B, the middle term is \(10x\), but the constant term is \(20\) instead of \(25\), so it is not a perfect-square trinomial. Exam tip: for a perfect square, compare the middle coefficient with twice the square root of the constant term.

A perfect-square trinomial has the form \(x^2+2ax+a^2=(x+a)^2\). In option A, \(36=6^2\) and the middle term is \(12x=2\times 6\times x\). Therefore, \(x^2+12x+36=(x+6)^2\). Option D has the constant term \(36\), but a square ending in \(6^2\) must have middle term \(12x\), not \(6x\). Exam tip: take the square root of the constant term and compare twice that value with the coefficient of the middle term.

The middle term is (3x+5x=8x) and the constant term is (15). Exam tip: multiply every term of both binomials.

Using the distributive property, \((p-2)(p+7)=p^2+7p-2p-14=p^2+5p-14\). Hence, option D is correct. In option A, the sign of the middle term is incorrect because \(7p-2p=5p\), not \(-5p\). Exam tip: multiply each term of one binomial by both terms of the other, then combine like terms.

In a perfect-square trinomial, the middle coefficient is \(2ab\) and the constant is \(b^2\). For \(x^2+10x+21\), \(2\times5=10\), but \(5^2=25\), not 21. Exam tip: check the constant term against the middle term.

\(x^2-49=x^2-7^2=(x+7)(x-7)\), so its factors are conjugates. Options C and D are perfect squares, while \(x^2+49\) is not a difference of squares. Exam tip: \(a^2-b^2\) has no middle term.

A quadratic expression in one variable contains only one variable, with highest exponent 2. In 3x² - 5x + 2, x is the only variable and its highest power is 2, so it is quadratic. x³ + x is cubic because its degree is 3, while x + y² has two variables. Exam tip: first check the number of variables, then identify the highest exponent.

The governing concept is the coefficient of a term. A coefficient is the numerical factor multiplying a variable term, while the exponent identifies the power of that variable. In 3x² + 2x − 1, the term containing x to the first power is 2x, so the coefficient of x is 2. Therefore, option B is correct. The number 3 is the coefficient of x², not of x, because it belongs to the quadratic term. The number −1 is the constant term and has no x factor. Although the coefficient of a separately written x is 1, this expression contains 2x, not x alone. Carefully matching the requested power prevents confusing the linear coefficient with the quadratic coefficient or the constant term.

A quadratic expression has degree 2, meaning that the highest power of its variable is 2 with a non-zero coefficient. In option B, 2x^2 - 3x + 1, the highest power of x is 2, so it is quadratic. Option A is linear, option C is cubic, and option D is a constant expression. Exam tip: determine an expression’s degree by identifying the highest exponent of the variable.

For area, multiply the dimensions: \((x+5)(x-2)=x^2-2x+5x-10=x^2+3x-10\). The student omitted the \(3x\) term. Option C also changes terms incorrectly. Exam tip: use distributive multiplication term by term.

In a quadratic expression, the coefficient of \(x^2\) must be non-zero, so \(a\ne0\) is necessary. If \(a=0\), the highest possible power becomes 1. Exam tip: first identify the highest power with a non-zero coefficient.

In \(x^2-5x+6\), the highest power of \(x\) is 2, and it is a polynomial in \(x\); therefore, it is a quadratic expression. \(4x-9\) is linear, \(2x^3+x\) is cubic, and \(\frac{3}{x}+2\) is not a polynomial because \(x\) occurs in the denominator. Exam tip: a quadratic polynomial has highest exponent 2.

In a quadratic expression, the coefficient of \(x^2\) must be non-zero; hence \(a\ne0\). If \(a=0\), the highest possible power is 1, making it linear. Exam tip: always check the coefficient of the highest power first.

Using \((a-b)^2=a^2-2ab+b^2\), take \(a=2x\) and \(b=5\). The middle term is \(-2\times2x\times5=-20x\). Writing only the squared terms is incorrect. Exam tip: always check the middle term and its sign.

(49=7^2), and a perfect square needs middle term (2\cdot7x=14x). Exam tip: take the square root of the last term.

None of the options is a difference of squares. Option B equals \((2a-3b)^2\), which is a square of a binomial, not \(A^2-B^2\). Hence the item is defective. Exam tip: a basic difference of squares has no middle term, as in \(u^2-v^2\).

\(x^2+6x+9=x^2+2\cdot x\cdot3+3^2=(x+3)^2\), so both linear factors are identical. In contrast, \(x^2+6x+8=(x+2)(x+4)\) has different factors. Exam tip: compare the middle term with \(2ab\).

\(x^2-4y^2=x^2-(2y)^2\), so it fits \(a^2-b^2=(a-b)(a+b)\) and factorises as \((x-2y)(x+2y)\). A and C are perfect-square trinomials, not differences. Exam tip: check for two squares separated by a minus sign.

Identical linear factors, such as \((x-3)(x-3)=(x-3)^2\), give a repeated root. Hence the parabola touches the x-axis at one point rather than crossing it. Exam tip: equal factors indicate a double root.

(x^2+6x+9=(x+3)^2), and the minimum value of a square is (0). Exam tip: recognise the perfect square form.