Quadratic expressions

In \((2x-3)(x+4)\), the highest-degree term is \(2x\times x=2x^2\), so its degree is 2 and it is quadratic. Option B has degree 3. Exam tip: check for variables in denominators or radicals before calling an expression a polynomial.

Since (16=4^2), the middle term is (2\cdot4\cdot x=8x). Exam tip: take the positive middle term for positive (p).

The parts are (2x) and (5), so the middle term is (-2\cdot2x\cdot5=-20x). Exam tip: choose the negative value using (p<0).

Combining like terms gives (-15x+2x=-13x). Exam tip: expand first and then combine like terms.

\(x^2+6x+9=x^2+2\cdot x\cdot3+3^2=(x+3)^2\), so option A is a perfect square. Option B is the closest distractor, but its constant term is \(8\); it must be \(9\) to form \((x+3)^2\). Exam tip: for \(x^2+bx+c\) to be a perfect square, check whether \(c=(b/2)^2\).

\((x+3)^2-6x=x^2+6x+9-6x=x^2+9\), whose highest power is 2, so it is quadratic. Options A and D simplify to linear expressions. Exam tip: combine like terms before identifying the degree.

In \(5x^2-9\), the highest power of x is 2, so it is quadratic. A linear x-term need not be present. \(5x-9\) is linear, while \(5x^3-9\) is cubic. Exam tip: identify the degree from the highest exponent.

On subtracting, (5x^2+2x-7-(2x^2-3x+4)=3x^2+5x-11). Exam tip: change signs when removing brackets.

Since (9=3^2), the negative middle term is (-2\cdot3\cdot x=-6x). Exam tip: match the sign of the middle term with the condition.

Here (x^2-4=x^2-2^2) and (x^2+4x+4=(x+2)^2). Exam tip: identify the type of identity separately.

In \(x^2+10x+25\), the last term is \(5^2\) and the middle term is \(2\times x\times5=10x\). Hence it equals \((x+5)^2\). In exams, verify the middle term using \(2ab\).

From (a^2=36), (a=6), and the middle term is (-2ax=-12x). Exam tip: the middle term is negative in a subtraction square.

Squaring gives (x^2-2+\frac{1}{x^2}=16), so (x^2+\frac{1}{x^2}=18). Exam tip: add (2) in the subtraction identity.

Here (x^2+4x+7=x^2+4x+4+3=(x+2)^2+3). Exam tip: add (4) to make a square and balance it.

A perfect square has the form \((x+p)^2=x^2+2px+p^2\). Thus \(b=2p\) and \(c=p^2\), giving \(b^2=4c\). Option \(b^2=c\) misses the factor 4. Exam tip: square the middle coefficient and compare it with \(4c\).

Since ((x+3)^2=x^2+6x+9), multiply by (2) and subtract (5). Exam tip: apply the outside multiplier to the whole bracket.

For \(x^2-5x+6\), \(a=1, b=-5, c=6\), so \(b^2-4ac=25-24=1>0\). Hence it has two distinct real zeroes. In option B, the discriminant is 0, giving equal zeroes. Exam tip: check \(D>0\) for distinct real zeroes.

\(x^2-6x+9=(x-3)^2\). The square of every real number is zero or positive, so this trinomial can never be negative. Option B equals \((x-3)^2-1\), which can be negative. Exam tip: look for a perfect-square identity.

Since (x^2+x-12=(x+4)(x-3)), the values are (-4) and (3). Exam tip: do not get confused by the order in options.

On subtracting, (x^2) and (1) cancel, leaving (4x). Exam tip: subtract like terms separately.