Quadratic expressions

A quadratic expression has highest power 2 and is generally of the form \(ax^2+bx+c\), where \(a\ne0\). In \(3x^2+2x+1\), the highest power of \(x\) is 2, so it is quadratic. \(5x+7\) is linear because its highest power is 1, while \(4x^3+x\) is cubic. Exam tip: identify the degree by checking the highest power of the variable.

The degree of a polynomial is the highest power of its variable. In \(5x^2-3x+7\), the powers of \(x\) are 2, 1, and 0, so the highest is 2. The term \(-3x\) has power 1, but it does not determine the degree. Exam tip: check the greatest exponent.

A quadratic expression must have 2 as the highest power of x, so the coefficient a of \(x^2\) must be non-zero. If \(a=0\), the expression becomes linear or constant. Exam tip: check the \(x^2\) coefficient first.

In a quadratic expression, the coefficient of \(x^2\) must be non-zero, so \(a\ne0\). If \(a=0\), the highest power is at most 1, making it linear or constant. Exam tip: check the highest power with a non-zero coefficient.

The identity \((a-b)^2=a^2-2ab+b^2\) shows that option A is correct. Option B represents \((a+b)^2\). In exams, identify the sign of the middle term first.

\(x^2-25=x^2-5^2\), so using \(a^2-b^2=(a-b)(a+b)\), it factorises as \((x-5)(x+5)\). \(x^2+25\) is a sum of squares. Exam tip: first check whether both terms are perfect squares.

Area = length × breadth = \((x+5)(x+3)\) = \(x^2+3x+5x+15=x^2+8x+15\). The option \(x^2+15\) misses the middle terms. Exam tip: add both cross terms carefully.

For a perfect-square trinomial, \(x^2+px+q\) must be expressible as \((x+r)^2\). On expanding, \((x+r)^2=x^2+2rx+r^2\), so \(p=2r\) and \(q=r^2\). Therefore, \(p^2=(2r)^2=4r^2=4q\). The condition \(p^2=q\) misses the required factor of 4. Exam tip: for \(x^2+px+q\), compare the square of the middle coefficient with \(4q\).

Using the distributive property, \((x+3)(x+2)=x\cdot x+x\cdot2+3\cdot x+3\cdot2=x^2+2x+3x+6=x^2+5x+6\). Therefore, option C is correct. In option D, the like terms \(2x\) and \(3x\) have not been combined to get \(5x\). Exam tip: while multiplying two binomials, write all four products first and then combine like terms.

\(4a^2-25b^2=(2a)^2-(5b)^2\), so it is a difference of squares and factors as \((2a-5b)(2a+5b)\). Options A and C are perfect squares, not differences. Exam tip: check whether both terms are squares with a minus sign.

A quadratic expression must have highest power 2, so the coefficient \(a\) of \(x^2\) must be non-zero. If \(a=0\), it becomes \(bx+c\), which is linear or constant. Exam tip: check the coefficient of the highest power first.

A quadratic expression must have 2 as the highest power of \(x\), so the coefficient \(a\) of \(x^2\) must be non-zero. If \(a=0\), it may become linear or constant. Exam tip: check the highest non-zero power first.

In a quadratic expression, the coefficient of \(x^2\) must be non-zero; hence \(a\ne0\). If \(a=0\), the \(x^2\) term disappears and the expression is at most linear. Exam tip: check the coefficient of the highest power first.

Using distribution, \((x+5)(x-2)=x^2-2x+5x-10=x^2+3x-10\). Thus the student is correct. Exam tip: add the two cross-products carefully and check their signs before combining like terms.

\(x^2-49=x^2-7^2=(x-7)(x+7)\), so it is a difference of two squares. \(x^2+49\) has addition, not subtraction. Exam tip: look for \(a^2-b^2\).

\(x^2-49=x^2-7^2\) is a difference of squares. Using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-7)(x+7)\). \(x^2+49\) is a sum of squares. Exam tip: conjugate factors have the same terms but opposite signs.

By the factor theorem, if \(x-5\) is a factor of \(p(x)\), then \(p(5)=0\). Hence, 5 is a zero of the polynomial. The degree is the highest power of \(x\), not a value of \(x\). Exam tip: from \(x-a\), identify the zero as \(a\).

\(x^2-49=x^2-7^2\), so it is a difference of two squares and factorises as \((x-7)(x+7)\). \(x^2+49\) is a sum, not a difference, of squares. Exam tip: first rewrite constants such as \(49\) as \(7^2\).

For a quadratic expression ax² + bx + c, if c = 0, then ax² + bx = x(ax + b); hence x is a factor. The leading coefficient a cannot be zero for it to remain quadratic. Exam tip: inspect the constant term first to spot a common factor.

A quadratic expression must have 2 as the highest power of \(x\), so the coefficient \(a\) of \(x^2\) must be non-zero. If \(a=0\), the expression becomes at most linear. Exam tip: check the coefficient of the highest power first.