Quadratic expressions

Direct answer: option B, \(x^2-4x+7\). A quadratic expression in x must be a polynomial whose highest power of x is 2. Option A, \(3x+5\), has highest power 1, so it is linear. Option B contains \(x^2\) and no higher power, so it is quadratic. Option C, \(x^3+2x\), has highest power 3, so it is cubic. Option D, \(2/x+1\), contains x in the denominator; it can be written with \(x^{-1}\) and is not a quadratic polynomial. Thus B is the only correct choice. Do not decide by how complicated an expression looks; inspect the largest exponent after simplifying. Memory cue: the highest power must be exactly 2.

A coefficient is the numerical factor multiplied by a variable term. In the expression given, the terms are formed with different powers of x: the first term contains x squared, the second contains x, and the last is a constant. The coefficient must be read from the term containing the exact power asked for, not from another term or from the exponent itself.

The term containing x squared is 5x squared, written mathematically as \(5x^2\). Therefore, its numerical coefficient is 5, so option D is correct. The number 3 belongs to the x term and 9 is the constant term. The exponent 2 tells the power of x; it is not the coefficient. This distinction is useful when identifying coefficients in any polynomial.

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

In \(x^2+10x+25\), we have \(x^2=x^2\), \(25=5^2\), and the middle term \(10x=2\times x\times5\). Therefore, using \(a^2+2ab+b^2=(a+b)^2\), it becomes \((x+5)^2\). In option B, the middle term is \(10x\), but the constant term is \(20\) instead of \(25\), so it is not a perfect-square trinomial. Exam tip: for a perfect square, compare the middle coefficient with twice the square root of the constant term.

A perfect-square trinomial has the form \(x^2+2ax+a^2=(x+a)^2\). In option A, \(36=6^2\) and the middle term is \(12x=2\times 6\times x\). Therefore, \(x^2+12x+36=(x+6)^2\). Option D has the constant term \(36\), but a square ending in \(6^2\) must have middle term \(12x\), not \(6x\). Exam tip: take the square root of the constant term and compare twice that value with the coefficient of the middle term.

The direct answer is option C: \(x^2+8x+15\). Multiply each term in the first binomial by each term in the second: \(x\cdot x=x^2\), \(x\cdot5=5x\), \(3\cdot x=3x\), and \(3\cdot5=15\). Combining like terms gives \(x^2+5x+3x+15=x^2+8x+15\). Option A has the correct constant but incorrectly writes \(15x\) as the middle term. Option B has the wrong sign for the middle term and would correspond to negative additions. Option C contains the correct square, middle, and constant terms. Option D uses \(2x\) instead of \(8x\), so it misses part of the cross-term addition. Memory cue: for \((x+a)(x+b)\), the result is \(x^2+(a+b)x+ab\).

Using the distributive property, \((p-2)(p+7)=p^2+7p-2p-14=p^2+5p-14\). Hence, option D is correct. In option A, the sign of the middle term is incorrect because \(7p-2p=5p\), not \(-5p\). Exam tip: multiply each term of one binomial by both terms of the other, then combine like terms.

In a perfect-square trinomial, the middle coefficient is \(2ab\) and the constant is \(b^2\). For \(x^2+10x+21\), \(2\times5=10\), but \(5^2=25\), not 21. Exam tip: check the constant term against the middle term.

\(x^2-49=x^2-7^2=(x+7)(x-7)\), so its factors are conjugates. Options C and D are perfect squares, while \(x^2+49\) is not a difference of squares. Exam tip: \(a^2-b^2\) has no middle term.

A quadratic expression in one variable contains only one variable, with highest exponent 2. In 3x² - 5x + 2, x is the only variable and its highest power is 2, so it is quadratic. x³ + x is cubic because its degree is 3, while x + y² has two variables. Exam tip: first check the number of variables, then identify the highest exponent.

A quadratic expression has degree 2, meaning that the highest power of its variable is 2 with a non-zero coefficient. In option B, \(2x^2 - 3x + 1\), the highest power of \(x\) is 2, so it is quadratic. Option A is linear, option C is cubic, and option D is a constant expression. Exam tip: determine an expression’s degree by identifying the highest exponent of the variable.

For area, multiply the dimensions: \((x+5)(x-2)=x^2-2x+5x-10=x^2+3x-10\). The student omitted the \(3x\) term. Option C also changes terms incorrectly. Exam tip: use distributive multiplication term by term.

In a quadratic expression, the coefficient of \(x^2\) must be non-zero, so \(a\ne0\) is necessary. If \(a=0\), the highest possible power becomes 1. Exam tip: first identify the highest power with a non-zero coefficient.

In \(x^2-5x+6\), the highest power of \(x\) is 2, and it is a polynomial in \(x\); therefore, it is a quadratic expression. \(4x-9\) is linear, \(2x^3+x\) is cubic, and \(\frac{3}{x}+2\) is not a polynomial because \(x\) occurs in the denominator. Exam tip: a quadratic polynomial has highest exponent 2.

In a quadratic expression, the coefficient of \(x^2\) must be non-zero; hence \(a\ne0\). If \(a=0\), the highest possible power is 1, making it linear. Exam tip: always check the coefficient of the highest power first.

Using \((a-b)^2=a^2-2ab+b^2\), take \(a=2x\) and \(b=5\). The middle term is \(-2\times2x\times5=-20x\). Writing only the squared terms is incorrect. Exam tip: always check the middle term and its sign.

(49=7^2), and a perfect square needs middle term (2\cdot7x=14x). Exam tip: take the square root of the last term.

None of the options is a difference of squares. Option B equals \((2a-3b)^2\), which is a square of a binomial, not \(A^2-B^2\). Hence the item is defective. Exam tip: a basic difference of squares has no middle term, as in \(u^2-v^2\).

\(x^2+6x+9=x^2+2\cdot x\cdot3+3^2=(x+3)^2\), so both linear factors are identical. In contrast, \(x^2+6x+8=(x+2)(x+4)\) has different factors. Exam tip: compare the middle term with \(2ab\).

\(x^2-4y^2=x^2-(2y)^2\), so it fits \(a^2-b^2=(a-b)(a+b)\) and factorises as \((x-2y)(x+2y)\). A and C are perfect-square trinomials, not differences. Exam tip: check for two squares separated by a minus sign.

Identical linear factors, such as \((x-3)(x-3)=(x-3)^2\), give a repeated root. Hence the parabola touches the x-axis at one point rather than crossing it. Exam tip: equal factors indicate a double root.

The direct answer is option B: the minimum value is \(0\). First complete the square: \(x^2+6x+9=x^2+2\cdot x\cdot3+3^2=(x+3)^2\). For every real number \(x\), a square is never negative, so \((x+3)^2\ge 0\). It becomes exactly zero when \(x+3=0\), that is, when \(x=-3\). Therefore the least possible value is 0. Option A, \(-9\), cannot be the value of a square because squares are not negative. Option B, 0, is correct because the square can become zero. Option C, 3, is the number inside the bracket and is not the minimum of the square. Option D, 9, is the constant term, but the expression can be smaller than 9 and can reach 0. The important idea is that completing the square changes the question into finding the smallest value of a square. Exam cue: whenever an expression becomes \((x-a)^2\) or \((x+a)^2\), its minimum is 0, reached when the bracket is zero.

A quadratic expression must have highest power 2, so the coefficient of \(x^2\), namely \(a\), cannot be zero. If \(a=0\), the expression becomes at most linear. Exam tip: first check the coefficient of the highest-power term.

First complete the square for \(x^2+2x\). Since \(x^2+2x+1=(x+1)^2\), we get \(x^2+2x+5=x^2+2x+1+4=(x+1)^2+4\). Therefore, option A is correct. Expanding option C gives a middle term of \(-2x\), so it cannot match the given expression. Exam tip: use the identity \(x^2+2ax+a^2=(x+a)^2\).

Dividing each term by (2) gives (x^2+4x+3). Exam tip: after taking a common factor, check all terms.