Quadratic expressions

\(9x^2+12x+4=(3x)^2+2(3x)(2)+2^2=(3x+2)^2\), so A is a perfect-square trinomial. D is also a square form, but it has a negative middle term: \((3x-2)^2\). Exam tip: compare the middle term with \(2ab\).

First recognise the first three terms as a perfect square: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=(x+2)^2\). The full expression is therefore \((x+2)^2-y^2\). This has the form \(A^2-B^2\), whose factorisation is \((A-B)(A+B)\). Taking \(A=x+2\) and \(B=y\) gives the complete factorisation.

Thus \(x^2+4x+4-y^2=(x+2-y)(x+2+y)\), so option A is correct. Option C is only an intermediate form and is not completely factorised. Option B changes the sign of the 4x term, so it represents a different expression. Option D incorrectly treats the whole expression as one square. Applying the two identities in sequence gives the answer safely.

Direct answer: option A, \((x-3-z)(x-3+z)\). First group the first three terms: \(x^2-6x+9=(x-3)^2\). The expression becomes \((x-3)^2-z^2\). Use the difference-of-squares identity \(a^2-b^2=(a-b)(a+b)\), with \(a=x-3\) and \(b=z\). This gives \((x-3-z)(x-3+z)\). Option A is exactly this factorisation. Option B uses \(x+3\), whose square would give \(+6x\), not \(-6x\). Option C is only an intermediate form, not fully factorised. Option D squares the whole difference incorrectly and does not represent \(a^2-b^2\). Memory cue: complete the square, then use difference of squares.

\((y-p)(y-q)=y^2-(p+q)y+pq\)

. Comparing it with \(y^2-7y+12\)

gives p+q=7 and pq=12. A sum of -7 would produce +7y. Exam tip: expand the signs once to verify the middle term.

Direct answer: Option C is correct. Expand the given factors: \((x-3)(x-6)=x^2-6x-3x+18=x^2-9x+18\). Compare this with \(x^2-bx+18\). The coefficient next to \(x\) is \(-9\) in the expansion and \(-b\) in the question, so \(b=9\). Option A, 3, is only the first constant used in one factor. Option B, 6, is only the second constant used in the other factor. Option C is correct because the two constants add to 9 while their product is 18: \(3+6=9\) and \(3\times6=18\). Option D, 18, is the product of the constants and the constant term, not the coefficient \(b\). Be careful with signs: two factors containing negative constants produce a negative middle coefficient, written as \(-bx\), while \(b\) itself is positive. Memory cue: for \((x-p)(x-q)\), the \(x\) coefficient is \(-(p+q)\).

Expanding \((2x+5)(x+2)\) gives \(2x^2+4x+5x+10\). Therefore, the middle term is \(4x+5x=9x\). Comparing it with \(2x^2+mx+10\), we get \(m=9\). The values 4 and 5 are coefficients of the separate cross terms, whereas \(m\) is their sum. Exam tip: add both cross terms when finding the middle term in a product of two binomials.

(81=9^2), so the negative middle term is (-2\cdot x\cdot9=-18x). In exams do not forget the given sign.

The zeroes are \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). Two distinct real zeroes require a positive quantity inside the square root, so \(b^2-4ac>0\). If it equals zero, the two zeroes coincide. Exam tip: check the discriminant first to identify the nature of zeroes.

Completed-square form rewrites a quadratic expression as a square of a binomial plus or minus a remaining constant. For an expression of the form \(x^2+bx+c\), take half of the coefficient of x, square it, and use that value to create the perfect square. The added amount must then be balanced so that the original expression remains unchanged.

Here, half of 6 is 3, and \(x^2+6x+9=(x+3)^2\). Since the original constant is 11, split it as \(11=9+2\). Therefore, \(x^2+6x+11=(x+3)^2+2\), which is option A. The plus sign inside the bracket is needed because the middle term is positive. Option D has the wrong remaining constant, while options B and C do not correctly reproduce the original expression.

Since \(25=5^2\), we have \(x^2-25=x^2-5^2\). Using \(a^2-b^2=(a-b)(a+b)\), it factorises as \((x-5)(x+5)\). Exam tip: check whether both terms are squares.

Every variable term in \(3x^2-2xy+y^2+5\) has total degree 2; for example, the degree of \(xy\) is \(1+1=2\). Hence A is quadratic. In B, each term has degree 3. Exam tip: add exponents in each term to identify its degree.

The square of a difference identity is (a-b)² = a² - 2ab + b². Rearranging it gives a²+b² = (a-b)² + 2ab. The given difference is 5, so its square is 25. The product ab is 6, so twice the product is 12. Adding these values gives 25 + 12 = 37. Therefore choice C is correct.

The plus sign before 2ab is essential because the negative middle term has already been moved to the other side. A common mistake is to report 25, which is only the square of the difference and omits the contribution of the product. Choice 37 includes both pieces required by the identity, so it is the complete value of a²+b².

In \((x+p)(x-p)=x^2-p^2\), the middle terms cancel. Hence \(x^2-81=x^2-9^2=(x+9)(x-9)\). \(x^2+18x+81\) is a perfect square, not a product of conjugate binomials. Exam tip: opposite constants give a zero middle term.

On simplifying option B, \((x+3)(x-3)+9=x^2-9+9=x^2\). Its highest power is 2, so it is a quadratic expression. Options A and D simplify to 4 and 1 respectively, while option C simplifies to \(5x\), which is linear. Exam tip: simplify an expression completely before identifying its degree.

A quadratic expression must have degree 2, so the coefficient of \(x^2\), namely \(a\), must be non-zero. The coefficients \(b\) and \(c\) may be zero. Exam tip: check that the highest-power coefficient is non-zero first.

The cross terms (10x) and (3x) combine to (13x). In exams check options using both leading coefficient and constant term.

\(4x^2-25y^2=(2x)^2-(5y)^2\). Using \(a^2-b^2=(a+b)(a-b)\), it becomes \((2x+5y)(2x-5y)\). Options C and D are perfect squares; first check for a difference of squares.

Let \(x^2+bx+c=(x+a)^2\). Since \((x+a)^2=x^2+2ax+a^2\), we get \(b=2a\) and \(c=a^2\). Hence \(b^2=4a^2=4c\). In exams, square the middle coefficient and compare it with \(4c\).

In standard form, the (x^2) term comes first, then the (x) term, then the constant term. In exams arrange terms in descending powers.

In option A, \(9x^2=(3x)^2\) and \(16y^2=(4y)^2\). The required middle term for these squares is \(-2(3x)(4y)=-24xy\). Hence, \(9x^2-24xy+16y^2=(3x-4y)^2\), so it is the perfect square of a binomial. In option B, the last term is \(12y^2\), not \(16y^2\), so it cannot form this perfect square. Exam tip: take the square roots of the first and last terms, then check whether the middle term equals \(\pm2ab\).

The direct answer is option C: \(s=12\). Multiply the two binomials term by term: \((x+5)(x+7)=x^2+7x+5x+35=x^2+12x+35\). Comparing this result with \(x^2+sx+35\), the coefficient of \(x\) is \(s=12\). Option A, 5, is only the first constant part and does not include the contribution from 7. Option B, 7, is only the second constant part and is also incomplete. Option C, 12, is correct because the middle coefficient is the sum \(5+7\). Option D, 35, is the product \(5\cdot7\), so it belongs to the constant term rather than the middle coefficient. This follows from the general identity \((x+a)(x+b)=x^2+(a+b)x+ab\). The signs matter: for two plus signs, both the middle coefficient and constant are positive. Exam cue: in a monic quadratic factored as \((x+a)(x+b)\), the middle coefficient is the sum and the last term is the product.

(64=8^2) and (-16x=-2\cdot x\cdot8). In exams check both the last term and the middle term in a perfect square.

The number attached to (x^2) is (3). In exams, identify a coefficient from the numerical part attached to the term.

In standard form, write the (x^2) term, then the (x) term, then the constant. In exams, follow descending powers.

When \(a\ne0\), the highest power of \(x\) is 2, so the expression is quadratic. \(b\) and \(c\) may be zero; if \(a=0\), it becomes linear or constant. Exam tip: check the non-zero coefficient of the highest power first.