Quadratic expressions

\(x^2-49=x^2-7^2=(x+7)(x-7)\), so it follows the difference-of-squares identity. \(x^2+49\) is a sum of squares, not this form. Exam tip: first check whether both terms are perfect squares.

The standard form of a quadratic expression is \(ax^2+bx+c\). Here \(a=-2\), \(b=5\), and \(c=7\), so it is \(-2x^2+5x+7\). In option B, the coefficients of x² and x are interchanged. Exam tip: arrange terms as x², x, then constant.

A quadratic expression must have highest power of \(x\) equal to 2, so the coefficient \(a\) of \(x^2\) cannot be zero. Values of \(b\) or \(c\) may be zero. Exam tip: check the leading coefficient first.

For an expression to be quadratic, the coefficient of \(x^2\) must be non-zero; hence \(a\ne0\). The coefficients \(b\) and \(c\) may be zero, as in \(3x^2+5\). Exam tip: check the highest non-zero power of the variable.

The common factor is (3), and (12\div3=4). Exam tip: check inner terms after taking out the factor.

When \(a\ne0\), the \(x^2\) term remains, so the highest power of x is 2. If \(a=0\), the expression becomes linear or constant. Exam tip: check the coefficient of the highest power first.

Only (x) is common in both terms. Exam tip: check common numerical and variable parts separately.

Half of (6) is (3), so the constant term should be (3^2=9). Exam tip: square half the coefficient of (x).

Half of (-8) is (-4), and ((-4)^2=16). Exam tip: the square of half the coefficient is positive.

\(x^2-49=x^2-7^2\) is a difference of two perfect squares, so it factorises as \((x-7)(x+7)\). \(x^2+49\) is a sum of squares. Exam tip: first check whether both terms are perfect squares.

Using \((a+b)^2=a^2+2ab+b^2\), take \(a=3x\) and \(b=2\). The missing middle term is \(2\times3x\times2=12x\), so the expression is \(9x^2+12x+4\). Exam tip: always check the \(2ab\) term.

\((x+4)^2=x^2+2\cdot x\cdot4+4^2=x^2+8x+16\), so the missing term is \(8x\). \(4x\) misses the factor 2 in the middle term. Exam tip: always check the \(2ab\) term in \((a+b)^2\).

Here, \(x^2=(x)^2\) and \(25=5^2\). Since the middle term is \(2\cdot x\cdot5=10x\), it matches \(a^2+2ab+b^2\). Exam tip: always check the sign of the middle term too.

This matches \((a+b)(a-b)=a^2-b^2\). Hence, \((x+4)(x-4)=x^2-4^2=x^2-16\). Reena used the wrong sign for the constant term. Exam tip: opposite binomials give a difference of squares.

(4x^2=(2x)^2), (9=3^2), and the middle term is (-12x). Exam tip: take square roots with coefficients.

(9x^2=(3x)^2), (16=4^2), and (24x=2\cdot3x\cdot4). Exam tip: look for all three perfect square clues.

This is a perfect square trinomial in two variables. Exam tip: treat (2xy) as the middle double product.

\(x^2+6x+9=x^2+2\cdot x\cdot3+3^2=(x+3)^2\), so it is a perfect-square trinomial. Option B has \(8\), not the required \(9\). Exam tip: match the middle term with \(2ab\).

A polynomial is read according to the powers of its variable. In x^2+0x+16, the term without x is the constant term, so the constant is 16. The coefficient of the first-degree or linear term x is the number multiplying x; here the term is 0x, so its coefficient is 0. The coefficient of x^2 is 1, but that is the quadratic coefficient, not the requested linear coefficient. Therefore the requested pair, in the stated order, is constant term 16 followed by linear coefficient 0. Option B reverses the order, while options C and D confuse the coefficient of x^2 with another requested value. Thus A is correct.

The highest power in (x^3+x+1) is (3). Exam tip: for a quadratic, the highest power must be (2).