Factorisation

Answer: A. A trinomial that is the square of a binomial follows \((u+v)^2=u^2+2uv+v^2\) or \((u-v)^2=u^2-2uv+v^2\). In A, \(9x^2=(3x)^2\), \(16=4^2\), and the middle term is \(2(3x)(4)=24x\). Hence \(9x^2+24x+16=(3x+4)^2=(3x+4)(3x+4)\). Option B has 15 instead of the required 16. Option C has an incorrect middle coefficient; it should be 24x if the outer terms are 9x² and 16. Option D has the wrong sign pattern: \((3x-4)^2\) would have a positive 16, not −16. The discriminant check also confirms A: \(24^2=4\cdot9\cdot16\).

Answer: A. Use \((u+v)^2=u^2+2uv+v^2\). In A, \(9p^2=(3p)^2\), \(4q^2=(2q)^2\), and the middle term is \(2(3p)(2q)=12pq\). Hence \(9p^2+12pq+4q^2=(3p+2q)^2\). Option B has 5q², but the square of 2q must give 4q². Option C has a negative final term, so it does not match a binomial square with these outer terms; the sign pattern is also inconsistent. Option D has equal-looking coefficients but is not the square of a binomial: the first and last terms would suggest \(\sqrt6p\) and \(\sqrt6q\), whose cross-term would be 12pq, but the resulting form is not the stated standard integer binomial square. Check the outer squares and exactly twice their product.

The correct answer is A: \(7a-6b\). For a square, area equals side multiplied by itself, so we must express the given trinomial as a perfect square. Use the identity \((u-v)^2=u^2-2uv+v^2\). Taking \(u=7a\) and \(v=6b\), we get \((7a-6b)^2=49a^2-84ab+36b^2\). Therefore the side has magnitude \(|7a-6b|\). The condition \(7a>6b\) makes this quantity positive, so the side is \(7a-6b\). Option B would produce a positive middle term, \(+84ab\), not the given negative term. Option C is not the square root of the whole trinomial, and option D has squared variables where a length expression is required. Memory cue: in a perfect square, the middle term is twice the product of the two square-root terms, with its sign determined by the sign inside the bracket.

The correct answer is A. A trinomial is a perfect square when it has the form \(u^2\pm2uv+v^2=(u\pm v)^2\). In option A, the first term is \((3p)^2\) and the last term is \((4q)^2\). Twice their product is \(2(3p)(4q)=24pq\), and the middle term is \(-24pq\). Hence \(9p^2-24pq+16q^2=(3p-4q)^2\). Option B has the same first and last terms, but its middle term is \(-20pq\), not \(-24pq\). Option C has an incorrect last term and also a positive middle term. Option D has a negative final square term, so it cannot match the standard square identity over the usual school algebra setting. The safest method is to take square roots of the first and last terms and compare twice their product with the middle term. A minus middle term means the binomial uses subtraction.

The correct answer is D. The difference-of-squares identity is \(A^2-B^2=(A-B)(A+B)\), so the expression must contain two square terms separated by a minus sign. Option A is \(x^2-4^2=(x-4)(x+4)\). Option B is \((3a)^2-b^2=(3a-b)(3a+b)\). Option C is \((p^2)^2-(q^2)^2=(p^2-q^2)(p^2+q^2)\), and the first factor can be factored further if complete factorisation is wanted. Option D is \((2m)^2+(5n)^2\), a sum of squares, not a difference of squares, so this particular identity does not apply. Option D might tempt a learner because both terms are squares, but the sign between them is essential. The memory cue is: ‘same square forms, minus sign, then split into difference and sum.’ Do not replace a plus sign by a minus sign while applying the identity.

The correct answer is B: \(x(x+1)(x+2)\). First identify the common factor in all three terms: \(x^3+3x^2+2x=x(x^2+3x+2)\). Now factor the quadratic trinomial. We need two numbers whose product is \(2\) and whose sum is \(3\); they are \(1\) and \(2\). Therefore \(x^2+3x+2=(x+1)(x+2)\), giving \(x(x+1)(x+2)\). Option A would produce \(x(x^2+5x+6)=x^3+5x^2+6x\). Option C would produce \(x^3+4x^2+3x\). Option D takes out only part of the common factor and expands to \(x^3+3x^2\), missing \(2x\). ‘Complete’ means continue factoring until no factor can be split further over the intended integers. Memory cue: take out the common factor first, then factor the remaining quadratic by product and sum.

The correct answer is A. Use the identity \((a-b)^2=a^2-2ab+b^2\). Here \(a=3p\) and \(b=4q\). Squaring the first term gives \(9p^2\), the cross term is \(-2(3p)(4q)=-24pq\), and the square of the second term is \(16q^2\). Thus \((3p-4q)^2=9p^2-24pq+16q^2\). Option B has the wrong final coefficient: the square of \(4q\) is \(16q^2\), not \(12q^2\). Option C has an incorrect middle coefficient; it should be \(-24pq\), not \(-18pq\). Option D has both the wrong sign and wrong middle coefficient. The minus sign inside the binomial creates a negative cross term, but the last term remains positive because a square is nonnegative in the algebraic identity. Memory cue: square first, double product with the same sign as the bracket’s cross term, then square the second term.

The correct answer is C: x(x − 3)(x + 3). Begin by taking the common factor x from both terms: x³ − 9x = x(x² − 9). The remaining expression is a difference of squares because 9 = 3². Therefore x² − 9 = x² − 3² = (x − 3)(x + 3). Combining the common factor with these two factors gives x(x − 3)(x + 3). Option A is incomplete and incorrect: x(x − 9) expands to x² − 9x, not x³ − 9x. Option B is incorrect because (x − 3)² = x² − 6x + 9, which introduces an unwanted middle term when multiplied by x. Option C is correct; multiplying (x − 3)(x + 3) gives x² − 9, and then multiplying by x gives x³ − 9x. Option D omits the common factor x, so it expands only to x² − 9 and cannot equal the cubic expression. A reliable routine is to remove the common factor first, then look for a familiar identity, and finally expand the answer to verify it. Memory cue: factor out x before recognising x² − 9 as a difference of squares.

The correct answer is C. The first term \(16x^2\) is \((4x)^2\), and the last term \(9y^2\) is \((3y)^2\). For the identity \((a-b)^2=a^2-2ab+b^2\), the middle term must be \(-2(4x)(3y)=-24xy\). Option C has exactly this value, so \(16x^2-24xy+9y^2=(4x-3y)^2\). Option A has the wrong final coefficient: \(8y^2\) is not \((3y)^2\). Option B has a positive middle term, which would correspond to addition, but then the middle coefficient would be positive in a different identity. Option D has \(-12xy\), not the required \(-24xy\). The key check is not only the first and last terms; the cross term must be twice the product of their square roots. Memory cue: square roots 4x and 3y give 24xy when doubled, so the correct negative middle term is -24xy.

The correct answer is A: (a + d)(b − c). Begin by grouping terms that have a similar part: ab − ac + db − dc = (ab − ac) + (db − dc). Take the common factor from each group: a(b − c) + d(b − c). Now (b − c) is common to both terms, so take it outside: (a + d)(b − c). Verification is useful: multiplying gives ab − ac + db − dc, exactly the original expression. Option B gives ab − ac − db + dc, so the signs of the last two terms are wrong. Option C gives ab + ac + db + dc, so the signs involving c are wrong. Option D gives ab + ac − db − dc, which also has incorrect signs. Memory cue: group in pairs and preserve every sign; a common bracket must be identical in both groups.

The correct answer is A: \((3x+2)\) and \((2x+5)\). For a rectangle, area equals length multiplied by breadth, so we need two factors whose product is \(6x^2+19x+10\). Multiply option A: \((3x+2)(2x+5)=3x(2x)+3x(5)+2(2x)+2(5)=6x^2+15x+4x+10=6x^2+19x+10\). Therefore, these two expressions can be the length and breadth. Option B gives \((3x+5)(2x+2)=6x^2+16x+10\), so its middle term is wrong. Option C gives \((6x+5)(x+2)=6x^2+17x+10\), also not the required polynomial. Option D gives \((3x+10)(2x+1)=6x^2+23x+10\). A reliable method is to multiply every proposed pair and compare all three parts: the coefficient of \(x^2\), the coefficient of \(x\), and the constant term. Matching only the first and last terms is not enough.

The correct answer is A. Conjugate binomials have the form (U − V) and (U + V); their product is U² − V². In A, 49x² = (7x)², so the expression is (7x)² − (3y + 2)². Applying the identity gives [(7x) − (3y + 2)][(7x) + (3y + 2)] = (7x − 3y − 2)(7x + 3y + 2). Option B is (7x + 3y + 2)², a square of a binomial. Option C is a sum of squares, not a difference. Option D is (7x − 3y − 2)². Neither square trinomial is a product of conjugates in the required direct form. Memory cue: conjugates differ only in the sign between the same two parts.

The correct answer is A. Use the identity (u − v)² = u² − 2uv + v². Let u = 7a and v = 5b. Then u² = 49a², −2uv = −2(7a)(5b) = −70ab, and v² = 25b². Therefore (7a − 5b)² expands exactly to 49a² − 70ab + 25b², so the student’s claim is correct. Option B forgets the factor 2 in the middle term. Option C is wrong because squaring a negative term gives a positive final square. Option D confuses 7²a² with 2 × 7a². A reliable check is to identify the square roots of the first and last terms, then calculate twice their product. Memory cue: in a squared binomial, the middle term is always ±2 times the two parts.

The direct answer is A: \(9a^2-24ab+16b^2=(3a-4b)^2\). A perfect-square trinomial follows \(X^2-2XY+Y^2=(X-Y)^2\) or \(X^2+2XY+Y^2=(X+Y)^2\). Here the first term is \((3a)^2\), the last term is \((4b)^2\), and twice their product is \(2(3a)(4b)=24ab\). The middle sign is negative, so we use \((3a-4b)^2\). Option A therefore matches all three parts. B has \(15b^2\), not the required \(16b^2\), so it cannot be the square of \(3a-4b\). C has a negative final square term, which does not fit this real perfect-square pattern, and its signs do not match the required identity. D has only two terms, so it is a difference of squares, not a trinomial. A useful check is: take square roots of the first and last terms, then verify that the middle term is plus or minus twice their product.

The direct answer is A. Expand the proposed factors carefully: \((x+p)(x+q)=x^2+qx+px+pq=x^2+(p+q)x+pq\). Now compare this with \(x^2+mx+n\). The coefficient of \(x\) must be the same, so \(p+q=m\). The constant term must also be the same, so \(pq=n\). Option A states exactly these two relations. Option B interchanges the roles of the middle coefficient and the constant term; addition of the numbers belongs to the coefficient of \(x\), while multiplication belongs to the constant term. Option C uses a difference, but expansion produces a sum \(p+q\), not generally \(p-q\). Option D incorrectly associates a difference with the constant term. The safest method is always to multiply the two binomials and compare like terms. Memory cue: “sum makes the middle term; product makes the last term.”

The direct answer is A: \(x^4+2x^2+1=(x^2+1)^2\). Treat \(x^2\) as one algebraic quantity, say \(X=x^2\). The expression becomes \(X^2+2X+1\), which has the identity \(X^2+2X+1=(X+1)^2\). Replacing \(X\) by \(x^2\) gives \((x^2+1)^2\). Option A is therefore correct. B would expand to \(x^4+4x^3+6x^2+4x+1\), containing terms that are not present. C expands to \(x^4-2x^2+1\), so its middle sign is wrong. D is equal to \((x^2-1)^2\), and therefore also has the negative middle term \(-2x^2\). The key idea is to notice the first and last terms as squares: \(x^4=(x^2)^2\) and \(1=1^2\), then check that the middle term is twice their product, namely \(2x^2\).

The direct answer is A: \(9x^2+12xy+4y^2=(3x+2y)^2\). For two identical binomial factors, the expression must have the form \((U+V)^2=U^2+2UV+V^2\), or \((U-V)^2=U^2-2UV+V^2\). The first term gives \(U=3x\), and the last term gives \(V=2y\). Twice their product is \(2(3x)(2y)=12xy\), exactly the middle term in A. B has a negative last term, so it cannot be the square of these real binomial terms. C has a middle coefficient 6 instead of the required 12. D has final term \(2y^2\), not \((2y)^2=4y^2\). Thus only A passes both checks. Memory cue: square the first and last parts mentally, then calculate “twice the product” to test the middle term.

The direct answer is A: \(3x^2y-12xy+12y=3y(x-2)^2\). First identify the common factor in all three terms. Each term contains \(3y\), so taking it outside gives \(3y(x^2-4x+4)\). The remaining trinomial is a perfect square because \(x^2-4x+4=x^2-2(x)(2)+2^2=(x-2)^2\). Hence the complete factorisation is \(3y(x-2)^2\). Option B would expand to \(3y(x^2+4x+4)\), giving a positive middle term, so its sign is wrong. Option C does not preserve the terms correctly: expanding it gives \(3x^2y-12xy\) and loses the constant term \(12y\). Option D stops before completing the factorisation; its inner trinomial is still factorisable. The important habit is to continue after taking out a common factor and check whether what remains matches an identity.

The direct answer is A: \(x^2-11x+30=(x-5)(x-6)\). For a monic quadratic \(x^2+bx+c\), find two numbers whose product is \(c\) and whose sum is \(b\). Here the product must be 30 and the sum must be −11. Since the product is positive, the numbers have the same sign; since the sum is negative, both are negative. The pair −5 and −6 works: \((-5)(-6)=30\) and \(-5+(-6)=-11\). Expanding A gives \(x^2-6x-5x+30=x^2-11x+30\). B gives a positive middle term, \(+11x\), so its sign is wrong. C has product 30 but sum −13, not −11. D has product 30 but sum +13. The useful warning is to check both conditions, not only the product. The signs of the constant and middle terms quickly tell whether the pair should be positive or negative.

The direct answer is A: \(x^2+14x+49=(x+7)^2\). A repeated-binomial factorisation has the form \((x+r)^2=x^2+2rx+r^2\) or \((x-r)^2=x^2-2rx+r^2\). In A, the first term is \(x^2\), the constant term is \(7^2=49\), and twice the corresponding number is \(2(7)=14\), exactly the middle coefficient. Option B has the correct first two terms but 48 is not \(7^2\), so it is not a perfect square. C has constant 49 but the middle coefficient should be 14, not 7. D has a negative middle term but a negative constant term; a square of a real binomial has a non-negative squared constant part in this pattern, and \((x-7)^2=x^2-14x+49\), not D. Another quick test for \(ax^2+bx+c\) is \(b^2=4ac\) when it is a perfect-square trinomial; for A, \(14^2=4(1)(49)=196\).

The direct answer is A: \(x^2-10x+25=(x-5)^2=(x-5)(x-5)\). A trinomial is a perfect-square trinomial when its first and last terms are squares and its middle term is twice the product of their square roots. Here, \(x^2=(x)^2\), \(25=(-5)^2\), and \(2(x)(-5)=-10x\). Therefore all three terms match the identity \(a^2+2ab+b^2=(a+b)^2\). Option B has constant term 20, which is not the required square 25. Option C has the wrong sign and also has 20 instead of 25. Option D has constant term \(-25\), so it cannot be the square of a real linear binomial with these terms. Memory cue: for \(x^2+bx+c\), take the square root of \(c\) and check whether twice that number gives the middle coefficient.

The direct answer is D: \(x^2+5x+7\) cannot be factorised into two linear factors with integer coefficients. For a monic trinomial \(x^2+bx+c\), we need two integers whose product is \(c\) and whose sum is \(b\). For D, the factor pairs of 7 are 1 and 7, or their negative versions. Their positive sum is 8, not 5, so no suitable integer pair exists. The other options work: \(x^2+5x+6=(x+2)(x+3)\), \(x^2-7x+12=(x-3)(x-4)\), and \(x^2+4x+3=(x+1)(x+3)\). Thus D is the only choice that fails both required conditions. A useful warning is that a quadratic may still factorise over non-integer or irrational numbers; the question specifically restricts the factors to integer coefficients.

The direct answer is A: \(x^2+x-12=(x+4)(x-3)\). The constant terms of the two binomials are 4 and -3, so their signs are opposite. Their product is \(4\times(-3)=-12\), which agrees with the constant term of the trinomial; their sum, 4+(-3)=1, agrees with the coefficient of \(x\). This gives a complete check. In B, the constant product is positive, so the constants have the same sign. The same is true for C and D, whose constant terms are positive. The key rule is that a negative constant term in a monic quadratic requires one positive and one negative constant in its integer factors. Do not decide from the middle-term sign alone; use the constant-term product first, then verify the sum.

The direct answer is A: \((2x+3)(3x+4)\). Expand it carefully: the first terms give \(2x\cdot3x=6x^2\), the cross-products give \(2x\cdot4=8x\) and \(3\cdot3x=9x\), and the last terms give \(3\cdot4=12\). Adding the cross-products gives \(8x+9x=17x\), so the expansion is exactly \(6x^2+17x+12\). Option B gives \(6x^2+27x+12\), because its cross-products are 24x and 3x. Option C gives \(6x^2+14x+12\). Option D gives \(6x^2+22x+12\). Thus only A matches the leading term, middle term, and constant term simultaneously. For \((px+q)(rx+s)\), remember that the middle coefficient comes from \(ps+qr\), not from just one cross-product.

The direct answer is A: \(49a^2-b^2=(7a)^2-b^2\). This has the exact form \(u^2-v^2\), so the difference-of-squares identity gives \((7a-b)(7a+b)\). Expanding confirms it: \((7a-b)(7a+b)=49a^2+7ab-7ab-b^2=49a^2-b^2\). Option B is a sum of squares, not a difference, so this identity does not apply. Option C is \((7a-b)^2\), a perfect-square trinomial, while D is \((7a+b)^2\); neither is a difference of two squares in the required direct form. The important distinction is between a minus sign separating two square terms and a squared binomial containing a middle term. Memory cue: difference of squares produces two factors with the same first term and opposite signs in the second term.