Quadratic expressions

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.

In option A, the highest power of \(y\) is 2, so it is quadratic. B is cubic and C is linear, while D has the variable in the denominator. Exam tip: identify an expression’s degree by checking the highest power of the variable.

A quadratic expression has highest power 2, with a non-zero coefficient of \(x^2\). In \(x^3+x^2+1\), the highest power of \(x\) is 3, so it is a cubic expression, not a quadratic one. Each of the other expressions has highest power 2. Exam tip: to find the degree of a polynomial, check the highest power of the variable after simplification.

Write it as ((x+4)^2-y^2) and apply the difference of squares identity. Exam tip: form the perfect square first.

Use the distributive property: \((2x+1)(x-4)=2x\cdot x+2x\cdot(-4)+1\cdot x+1\cdot(-4)\). Thus, \(2x^2-8x+x-4=2x^2-7x-4\), so option A is correct. In option B, \(-8x+x\) has incorrectly been treated as \(+7x\). Exam tip: write all four products first, then combine only like terms.

Using the distributive property, \((3x-2)(x+5)=3x\cdot x+3x\cdot5-2\cdot x-2\cdot5=3x^2+15x-2x-10=3x^2+13x-10\). Therefore, option A is correct. Option C does not combine the like terms \(15x\) and \(-2x\). Exam tip: after expanding, combine only terms with the same variable and exponent.

This is the form \((a-b)(a+b)=a^2-b^2\). Exam tip: quickly identify conjugate binomials.

Such a trinomial has the form \(a^2-2ab+b^2\), which expands from \((a-b)^2\). In \((a+b)^2\), the middle term is \(+2ab\). Exam tip: use the sign of the middle term to identify the square.

Using the identity
\((a-b)^2=a^2-2ab+b^2\), we get
\((4x-3)^2=16x^2-24x+9\). The term without x is the constant term, so it is 9. Here, -24 is the coefficient of x, while 16 is the coefficient of \(x^2\). Exam tip: in the square of a binomial, the square of the numerical term gives the constant term.

\((x+5)^2=x^2+10x+25\). The given expression has constant term 16, so subtract 9 from 25: \(x^2+10x+16=(x+5)^2-9\). Exam tip: halve the coefficient of \(x\), then add and subtract its square.

A quadratic expression has degree 2. Here, the term \(5x^2\) gives the highest power of \(x\) as 2, so it is quadratic. The constant term and number of terms do not decide the degree. Exam tip: check the highest exponent.

Expanding \((x+a)^2\) gives \(x^2+2ax+a^2\), so it is the required perfect-square form. In \((x-a)^2\), the middle term is \(-2ax\). Exam tip: use the sign of the middle term to identify the square.

Use \((a+b)^2=a^2+2ab+b^2\). Here, \(2\times x\times4=8x\), so the expansion is \(x^2+8x+16\). Choosing \(4x\) misses the factor 2. Exam tip: always check the middle term while expanding a square.

(-6x-2x=-8x), and ((-6)(-2)=12), which equals (3\cdot4). Exam tip: the product of the two numbers must be (ac).

In \(7p^2-4p+1\), the highest power of \(p\) is 2, so it is quadratic. Option B is cubic and option C is linear. Exam tip: classify a polynomial by its highest exponent.

The identity \((m-n)(m+n)=m^2-n^2\) represents a difference of squares, so A is correct. Options B and C are \((m+n)^2\) and \((m-n)^2\). Exam tip: look for a minus sign between two squares.

Area = \((x+5)(x-2)\). Expanding gives \(x^2-2x+5x-10=x^2+3x-10\). Option D has the wrong sign for the constant term. Exam tip: track signs carefully while multiplying binomials.

(x^2-16) has two terms and highest power (2). Exam tip: check both binomial and quadratic conditions.