Quadratic expressions

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.

The expression has a positive first term, a negative middle term, and a positive constant. This suggests the pattern \(a^2-2ab+b^2=(a-b)^2\). The first term \(4x^2\) is \((2x)^2\), and the last term 9 is \(3^2\). The middle term from the identity is \(-2\cdot2x\cdot3=-12x\), exactly as required.

Therefore \(4x^2-12x+9=(2x)^2-2(2x)(3)+3^2=(2x-3)^2\). Option C is correct. Option A would give a positive middle term, producing \(4x^2+12x+9\), not the given expression. Options B and D have incorrect leading coefficients or constant terms. Including the coefficient 2 in the square root of \(4x^2\) is essential.

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

The expression has the standard perfect-square pattern \(a^2+2ab+b^2=(a+b)^2\). Here the first square is \(x^2\), the last square is \(y^2\), and the middle term is \(2xy\). Thus the two quantities being added are \(x\) and \(y\), giving the square \((x+y)^2\). This identity works for all values of the variables.

Expanding confirms the result: \((x+y)^2=(x+y)(x+y)=x^2+xy+xy+y^2=x^2+2xy+y^2\). Hence option A is correct. The expression \((x-y)^2\) would have a negative middle term, namely \(-2xy\). The product \((x+y)(x-y)\) and the expression \(x^2-y^2\) both represent a difference of squares, not the given sum.

\(x^2+6x+9=x^2+2\cdot x\cdot3+3^2=(x+3)^2\), 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.

Direct answer: option C, \(x^3+x+1\). A quadratic expression has highest power exactly 2, and its general form is \(ax^2+bx+c\) with \(a\) not zero. Option A, \(x^2+2x+1\), has highest power 2, so it is quadratic. Option B, \(3x^2-5\), also has highest power 2; the missing x term is allowed because its coefficient may be zero. Option C has \(x^3\), so its degree is 3 and it is cubic, not quadratic. Option D, \(5x^2\), has degree 2, so it is quadratic even though the x and constant terms are absent. Check the greatest exponent, not the number of terms. Memory cue: degree 2 means quadratic; degree 3 means cubic.

\(7x^2\) 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.

The direct answer is option B: \(b=8\). A quadratic is a perfect square when it can be written as \((x+a)^2=x^2+2ax+a^2\). Here the constant term is \(16=4^2\), so the possible square is \((x+4)^2=x^2+8x+16\). Comparing this with \(x^2+bx+16\), the coefficient of \(x\) is \(b=8\). Option A, 4, is only the number whose square gives 16; it is not the middle coefficient. Option B, 8, is correct because the middle term is \(2\cdot4\cdot x=8x\). Option C, 16, is the constant term, not the coefficient of \(x\). Option D, 32, has no basis in the identity. A useful exam cue is: when the first coefficient is 1 and the constant is a square, take the positive square root and double it. Thus \(x^2+bx+16=(x+4)^2\), so the positive value is 8.

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

To split the middle term of \(2x^2+5x+2\), multiply the leading coefficient and constant term: \(2\cdot2=4\). We need two numbers whose product is 4 and whose sum is 5. Those numbers are 1 and 4. Therefore the middle term \(5x\) is written as \(x+4x\), giving \(2x^2+x+4x+2\).

This split is useful because it allows grouping: \(2x^2+x+4x+2=x(2x+1)+2(2x+1)=(x+2)(2x+1)\). Hence option B correctly describes the split. Option A gives \(2x+3x=5x\), but the associated numbers 2 and 3 have product 6, not the required product 4, so that split would not support correct grouping.

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, \(2\times x\times4=8x\), 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.