Quadratic expressions

\(7x^2\) is a quadratic monomial because it has only one term and the exponent of \(x\) is 2. \(x^2+7\) is a binomial, while \(x^2+7x+1\) is a trinomial, so neither is a monomial. Exam tip: first count the terms, then check the degree.

\(x^2-9\) is a binomial because it has two terms: \(x^2\) and \(-9\). Its highest power is 2, so it is a quadratic expression. \(x^2+3x+2\) is quadratic, but it has three terms and is therefore a trinomial. Exam tip: first count the terms, then check the highest power.

\(x^2+4x+4\) has three terms: \(x^2\), \(4x\), and \(4\). Its highest power is 2, so it is a quadratic trinomial. \(x^3+x+1\) also has three terms, but its degree is 3, making it a cubic trinomial. Exam tip: count the terms first, then check the highest power of the variable.

A quadratic expression must have highest power of \(x\) equal to 2, so the coefficient \(a\) of \(x^2\) cannot be zero. Values of \(b\) and \(c\) may be zero. If \(a=0\) and \(b\ne0\), the expression is linear. Exam tip: classify an expression by its highest non-zero power.

A quadratic expression must have degree 2, so the coefficient \(a\) of \(x^2\) cannot be zero. If \(a=0\), its degree becomes at most 1. Unlike \(a\), \(b\) and \(c\) may be zero. Exam tip: check the highest-power term first.

A quadratic polynomial must have highest power 2, so the coefficient of \(x^2\), namely \(a\), cannot be zero. \(c=0\) does not prevent it from being quadratic. Exam tip: check the highest non-zero power first.

A quadratic polynomial has the form \(ax^2+bx+c\), where \(a\ne0\). If \(a=0\), the highest possible power is 1: it is linear when \(b\ne0\) and constant when \(b=0\). Exam tip: check the leading non-zero coefficient before identifying the degree.

Since (16=4^2), the middle term is (2\cdot4\cdot x=8x). Exam tip: (b=2\sqrt{c}) is useful here.

Since (25=5^2), for the negative perfect square (b=-2\cdot5=-10). Exam tip: read the sign condition carefully.

Here (2\cdot2=4), and (1+4=5), so (x+4x) works. Exam tip: check the relation of (ac) and (b).

In \(x^2-6x+9\), the first and last terms are \(x^2\) and \(3^2\), while the middle term is \(-2\times x\times3=-6x\). Hence \(x^2-6x+9=(x-3)(x-3)=(x-3)^2\). Option A has unequal factors. Exam tip: check whether the middle term is \(\pm2ab\).

\(x^2-49=x^2-7^2\) is a difference of two squares, so it factorises as \((x+7)(x-7)\). Options C and D are perfect-square trinomials. Exam tip: first check for a missing middle \(x\)-term.

In \((x+a)^2=x^2+2ax+a^2\), we have \(p=2a\) and \(q=a^2\). Hence \(p^2=(2a)^2=4a^2=4q\). In exams, square the middle coefficient and compare it with \(4q\).

Use the identity


aaa(a+b)^2=a^2+2ab+b^2. Here, 2×x×4=8x, so the expansion is x^2+8x+16. Writing 4x misses the factor 2; always check the middle term.

Use \((a+b)^2=a^2+2ab+b^2\). With \(a=x\) and \(b=5\), the middle term is \(2\times x\times5=10x\), giving \(x^2+10x+25\). Exam tip: always check for the middle term.

(x^2+6x+9) is a perfect square, but (x^2+9) lacks the middle term. Exam tip: check whether the middle term is present.

(9x^2=(3x)^2) and (25=5^2), so the middle term will be (2\cdot3x\cdot5=30x). Exam tip: take the roots of both square terms and multiply twice.

A quadratic expression has the highest power (2). Exam tip: identify it by checking the highest power.

The number attached to (x^2) is (5). Exam tip: identify the term and its coefficient separately.

A quadratic expression must have highest power 2, so the coefficient \(a\) of \(x^2\) cannot be zero. If \(a=0\), it becomes linear or constant. Exam tip: check the highest non-zero power first.