Quadratic expressions

\(x^2+6x+9=x^2+2\cdot x\cdot3+3^2=(x+3)^2\), so it is the product of two identical binomials. \(x^2+6x+8=(x+2)(x+4)\), whose factors differ. Exam tip: in a perfect square, the middle term is \(2ab\).

((x+3)^2=x^2+6x+9), so the given expression is not (1) less but (4) less. The correct comparison should be (4) less.

To make (x^2+6x) a perfect square, (9) is needed, and (5) is already present, so add (4). In exams, square half the middle coefficient.

For (x^2-8x), the perfect square constant should be (16), so add (16-10=6). In exams, subtract the current constant.

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

The direct answer is option A. For the first expression, use the difference-of-squares identity \(a^2-b^2=(a-b)(a+b)\): \(x^2-16=x^2-4^2=(x-4)(x+4)\). For the second, use \((x-4)^2=x^2-8x+16\), so \(x^2-8x+16=(x-4)^2\). Thus option A gives both forms correctly. Option B reverses the two forms: \((x-4)^2\) is not equal to \(x^2-16\), while \((x-4)(x+4)\) is not equal to the second expression. Option C uses \((x+4)^2=x^2+8x+16\) for the first, which has the wrong middle structure, even though its second form is correct. Option D is wrong because \((x-16)(x+1)=x^2-15x-16\), not \(x^2-16\), and \((x-8)^2=x^2-16x+64\), not the second expression. Memory cue: no middle term often signals difference of squares; a negative middle term and positive last term can signal \((x-4)^2\).

For an expression to be quadratic, the coefficient of \(x^2\) must not be zero. In A, \(a^2+1>0\) for every real a, so it is always quadratic. In B, taking \(a=0\) makes it linear. Exam tip: check the \(x^2\) coefficient first.

(16=4^2), and the negative middle term is (-2\cdot x\cdot4=-8x). In exams, choose the coefficient according to the given sign.

A perfect-square trinomial has zero discriminant, so \(b^2=4ac\). Substituting \(c=\frac{b^2}{4a}\) gives \(a\left(x+\frac{b}{2a}\right)^2\). For this to be a square of a real linear expression, \(a\) must be positive. Exam tip: check the sign of \(a\) along with the discriminant.

For \((px+q)^2\), we have \(a=p^2,\ b=2pq\), and \(c=q^2\). Thus \(b^2-4ac=(2pq)^2-4p^2q^2=0\). A positive discriminant gives distinct real roots, not a repeated root. Exam tip: test a perfect-square quadratic using its discriminant.

Using \(a^2-b^2=(a-b)(a+b)\), put \(a=p\) and \(b=q\) to get \(p^2-q^2=(p-q)(p+q)\). In \((p-q)^2\), the middle term is \(-2pq\), so it is different. Exam tip: look for a subtraction between two squares.

The expression must be \((x\pm4)^2\). Expanding gives \((x+4)^2=x^2+8x+16\) and \((x-4)^2=x^2-8x+16\), so \(p=\pm8\). Exam tip: take the square root of the constant term and check the middle term.

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

In \((x-3)(2x+1)\), the highest-degree product is \(x\times2x=2x^2\), so its degree is 2 and it is quadratic. B and C have degree 3, while D contains \(x^{-1}\), so it is not a polynomial. Exam tip: check the highest power first.

\(x^2+4x+5=(x+2)^2+1\). Since \((x+2)^2\geq0\), its minimum value is 1, so it is positive for every real \(x\). In the closest distractor, option B, substituting \(x=2\) gives \(-1\), so it is not always positive. Exam tip: complete the square to find the minimum or maximum value of a quadratic.

A perfect square trinomial has the pattern \((a-b)^2=a^2-2ab+b^2\) or \((a+b)^2=a^2+2ab+b^2\). To recognize the pattern, compare the first and last terms with squares and then check whether the middle term equals twice their product, with the correct sign. This prevents confusing a subtraction square with an addition square.

In the expression, \(x^2\) is the square of x and \(81=9^2\). The middle term is \(-18x\), which equals \(-2\cdot x\cdot9\). Hence the expression matches \(x^2-2\cdot x\cdot9+9^2\), so it can be written as \((x-9)^2\). Therefore option A is correct. Expanding it confirms the result: \((x-9)^2=x^2-18x+81\).

(16x^2=(4x)^2), (25=5^2), and (40x=2\cdot4x\cdot5). In exams, check the square roots of first and last terms.

(49=7^2), so the middle term is (2\cdot x\cdot7=14x). In exams, take the positive middle coefficient for a positive perfect square.

For \(2x^2+3x+5\), the discriminant is \(b^2-4ac=3^2-4(2)(5)=-31\). Since it is negative, the expression has no real linear factors, hence no integer-coefficient factorisation. Exam tip: check the discriminant first.

The direct answer is A: \((x-6)^2-16\). Completing the square means creating a perfect square from the first two terms and adjusting the constant term. Half of the coefficient of x in \(x^2-12x\) is \(-6\), whose square is 36. Add and subtract 36: \(x^2-12x+20=x^2-12x+36-36+20=(x-6)^2-16\). Option A is correct. Option B, \((x+6)^2-16\), expands to \(x^2+12x+20\), so the middle sign is wrong. Option C, \((x-12)^2+20\), uses 12 instead of half of 12; it gives the wrong middle term and constant. Option D, \((x-6)^2+16\), expands to \(x^2-12x+52\), so the final constant is too large. The key is to add and subtract the same number, so the value does not change. Take half the middle coefficient, square it, and compensate outside the square.

\((x-5)^2=x^2-2\cdot5x+5^2=x^2-10x+25\), so the correct factorisation is \((x-5)^2\). In contrast, \((x-5)(x+5)\) gives \(x^2-25\). Exam tip: check whether the middle term is \(\pm2ab\).

\(p^2-16q^2=p^2-(4q)^2=(p-4q)(p+4q)\), so its factors are conjugate binomials. C is a repeated-binomial square. Exam tip: spot two squares separated by a minus sign.

Since \(a\ne0\), the \(x^2\) term is present, so the highest power is 2. \(b\) or \(c\) may be zero, and real roots are not guaranteed. Exam tip: check the highest power first.

(-4-8=-12) and ((-4)(-8)=32). In exams, use both negative signs when the constant is positive and middle term is negative.

(25x^2=(5x)^2) and (36=6^2), so the middle term is (2\cdot5x\cdot6=60x). In exams, multiply twice the two square roots.