Quadratic expressions

\(a^2+10a+25=(a+5)^2\) because the middle term is \(2\times a\times5=10a\). In option B, the constant 20 is not \(5^2=25\). Exam tip: use the square roots of the first and last terms to check the middle term.

A quadratic of the form \(x^2+bx+k\) is a perfect square when it matches \((x-r)^2=x^2-2rx+r^2\). Comparing \(x^2-10x+k\) with this identity gives \(-2r=-10\), so \(r=5\). The constant term must then be \(r^2=25\). Thus the expression is \((x-5)^2\), and option C is correct.

The same result follows from completing the square: half of the coefficient of \(x\) is \(-10/2=-5\), and its square is \((-5)^2=25\). Therefore \(x^2-10x+25=(x-5)^2\). The sign disappears after squaring, but the middle term must remain negative. Options A, B, and D do not produce the required middle coefficient and constant term together.

The direct answer is option B: 2. A quadratic expression is defined by the highest power of its variable being 2. It may look like ax²+bx+c, where a is not zero; the other terms can have powers 1 or 0, but none can have a power greater than 2. For example, 3x²−5x+7 is quadratic because the largest exponent is 2. Option A, 1, describes a linear expression, whose highest power is 1. Option B is correct because the word quadratic is connected with square, and x² is the highest power. Option C, 3, describes a cubic expression, not a quadratic one. Option D, 0, describes a constant expression, which has no variable term. Always inspect the highest exponent, not the number of terms or the coefficient. Memory cue: linear 1, quadratic 2, cubic 3.

A quadratic expression must have 2 as the highest power of \(x\), so the coefficient \(a\) of \(x^2\) cannot be zero. For example, \(3x^2-5x+1\) is quadratic. Both \(b\) and \(c\) may be zero. Exam tip: check the non-zero coefficient of the highest power first.

A quadratic expression has highest power 2, so the coefficient of \(x^2\) must be non-zero. The \(x\)-term and constant may be zero. Exam tip: identify the highest non-zero power first.

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

The conjugate binomials \((p+q)\) and \((p-q)\) give \((p+q)(p-q)=p^2-q^2\), as the middle terms cancel. Option B is a binomial square. Exam tip: opposite signs with matching terms indicate a difference of squares.

Using \((a-b)^2=a^2-2ab+b^2\), with \(a=x\) and \(b=5\), gives \((x-5)^2=x^2-2(x)(5)+25=x^2-10x+25\). Hence, option B is correct. Option C has an incorrect positive middle term, while option D is the expansion of \((x-5)(x+5)\), not \((x-5)^2\). Exam tip: in \((a-b)^2\), the middle term is always negative, \(-2ab\).

A perfect square trinomial has the pattern \(a^2+2ab+b^2=(a+b)^2\). In the given expression, the first term is \(x^2\), the last term is 36, and \(36=6^2\). The middle term should then be \(2\cdot x\cdot6=12x\), which is exactly the term present. This confirms that the expression is the square of \(x+6\).

Therefore \(x^2+12x+36=(x+6)^2\), so option C is correct. Expanding it gives \((x+6)(x+6)=x^2+6x+6x+36=x^2+12x+36\). Option B would produce a negative middle term, while A and D give middle coefficients 24 and 6 respectively. Checking the last term and the middle term together prevents such errors.

The direct answer is option D: \((x-7)^2\). Use \((a-b)^2=a^2-2ab+b^2\) with \(a=x\) and \(b=7\). This gives \((x-7)^2=x^2-2\cdot x\cdot7+7^2=x^2-14x+49\), exactly the given expression. Option A, \((x+7)^2\), would have a positive middle term \(+14x\). Option B, \((x-14)^2\), would expand to \(x^2-28x+196\), so both its middle and constant terms are wrong. Option C, \((x+14)^2\), would give \(x^2+28x+196\). Option D has the correct square term, negative middle term, and positive constant. Memory cue: a negative middle term in a perfect square means the bracket uses subtraction.

A quadratic expression must have a non-zero coefficient of \(x^2\), so \(a\ne0\) is necessary. If \(a=0\), the highest possible power becomes 1. Exam tip: first check the coefficient of the squared term.

A quadratic expression has highest variable power 2, with non-negative integer powers. In \(7y^2-3y+4\), the highest power is 2. Option B is cubic. Exam tip: check the greatest exponent first.

The direct answer is D: 0. A quadratic expression has the form \(ax^2+bx+c\). The linear term is the term containing exactly one power of x, namely \(bx\). In \(4x^2+0x-9\), the linear term is explicitly \(0x\), so its coefficient is 0. Option D is therefore correct. Option A, 4, is the coefficient of the quadratic term \(x^2\), not the linear term. Option B, -9, is the constant term, which contains no x. Option C, 9, is not even the constant coefficient because the expression contains -9. The zero is written to make the missing linear term visible; even if \(0x\) were omitted, its coefficient would still be zero. Memory cue: match the coefficient with the power: x² uses a, x uses b, and the number alone is c.

\(x^2-49=x^2-7^2=(x+7)(x-7)\), 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.

The direct answer is A: x. A common factor is a quantity that divides every term. The expression is \(2x^2+5x\). Rewrite the terms as \(2x\cdot x+5\cdot x\). Both terms contain x, so x is common and the expression can be factored as \(x(2x+5)\). Option A is therefore correct. Option B, 2x, is not common to both terms because 5x is not divisible by 2. Option C, 5x, is not common because 2x² is not divisible by 5. Option D, 10x, is larger than the numerical factors available and cannot divide either term in the required integer factorisation. Notice that 1 is also technically a common factor, but x is the intended greatest useful variable factor and is the only listed correct choice. Memory cue: compare numbers and variables separately; take only what appears in every term.

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

Direct answer: C, 16. A perfect square should have the form \((x-m)^2=x^2-2mx+m^2\). Compare this with \(x^2-8x+k\). We need \(-2m=-8\), so \(m=4\), and therefore \(k=m^2=16\). Indeed, \(x^2-8x+16=(x-4)^2\). Option A, 4, is only half the needed coefficient and gives \(x^2-8x+4\), not a square. Option B, 8, is the magnitude of the middle coefficient, not its half squared. Option C is correct. Option D, 64, squares 8 directly, but the rule requires squaring half of the coefficient of x. Memory cue: for \(x^2+bx+c\), the completing-square value is \(c=(b/2)^2\); the sign disappears after squaring.

\(x^2-49=x^2-7^2\) 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 \((a+b)^2\).