Quadratic expressions

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.

The first three terms form a perfect square: \(x^2+8x+16=(x+4)^2\). The original expression can therefore be rewritten as \( (x+4)^2-y^2\). This is a difference of two squares, using \(A^2-B^2=(A-B)(A+B)\), with A equal to \(x+4\) and B equal to y.

Applying the identity gives \( (x+4-y)(x+4+y)\), which is option D. Multiplying the factors produces \( (x+4)^2-y^2=x^2+8x+16-y^2\), so the result is verified. Option A contains only one of the two required factors and is incomplete. Option C changes the structure into a square, while option B does not reproduce the original expression. First forming the perfect square makes the factorisation clear.

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.

The product has two binomials that are identical except for the signs between x and 7. Such a pair is called a difference-and-sum pair, or conjugate binomials. When these are multiplied, the two middle terms cancel each other. This gives the identity \((a-b)(a+b)=a^2-b^2\), which is also called the difference of two squares.

Here, take \(a=x\) and \(b=7\). Applying the identity gives \((x-7)(x+7)=x^2-7^2=x^2-49\). Direct multiplication gives the same result: \(x^2+7x-7x-49\), and the middle terms disappear. Therefore option C is correct. Option A would result from incorrectly treating the product as a sum of squares.

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.

A quadratic expression must retain an \(x^2\) term, so its coefficient must satisfy \(a\ne0\). The coefficients \(b\) and \(c\) may be zero. Exam tip: check the coefficient of the highest power first.

In \(4x^2-7x\), the highest power of x is 2, so it is quadratic, and its constant term is 0. \(4x^2-7\) has constant term −7. Exam tip: the constant term contains no x.

For an expression of the form \(x^2+bx+c\), look for two numbers whose sum is \(b\) and whose product is \(c\). Here the middle coefficient is \(-5\), while the constant term is \(6\). Thus the required numbers must add to \(-5\) and multiply to \(6\). Both numbers must be negative because their product is positive and their sum is negative.

The numbers are \(-2\) and \(-3\). Therefore the factors are \( (x-2)(x-3) \). Expanding gives \(x^2-3x-2x+6=x^2-5x+6\), confirming choice B. Choices with plus signs give a positive middle coefficient, so they cannot match the expression.

A quadratic expression has the form \(ax^2+bx+c\), where \(a\ne0\). Its graph is a parabola, opening upward or downward according to the sign of \(a\). A straight line represents a linear expression. Exam tip: an \(x^2\) term indicates a parabola.

(x^2+4x+4=(x+2)^2) and (x^2-4=(x-2)(x+2)). Exam tip: distinguish similar-looking quadratics by identity.

Here (-2-3=-5) and ((-2)(-3)=6). Exam tip: for a positive constant and negative middle term, take both numbers negative.

The direct answer is option A:
\((2x-1)(x-3)\). To factorise, we look for two binomials whose first terms multiply to \(2x^2\), whose constant terms multiply to \(3\), and whose middle terms add to \(-7x\). Expand A step by step: \((2x-1)(x-3)=2x^2-6x-x+3=2x^2-7x+3\), so it matches exactly. Option A is correct because all three terms agree. Option B gives \((2x+1)(x-3)=2x^2-5x-3\), so both the middle and constant terms differ. Option C gives \((2x-3)(x-1)=2x^2-5x+3\), so the middle coefficient is wrong. Option D gives \((x-1)(x-3)=x^2-4x+3\), so even the coefficient of \(x^2\) is wrong. A useful exam cue is to expand every proposed factorisation and compare all three terms, not only the first and last terms.

Since (36=6^2), the middle term will be (2\cdot6\cdot x=12x). Exam tip: double the square root of the constant term.