Algebraic identities

For three linear factors, the coefficient of (x^2) is (1+2+3=6). Exam tip: you can also check by expanding the first two factors.

In the product of three linear factors, the coefficient of \(x^2\) equals the sum of the constant terms. Thus, \((-2)+3+5=6\). Therefore, the correct answer is 6. Option 5 results from adding only the positive constants \(3+5\) and ignoring the negative sign of \(-2\). Exam tip: always include signs while adding constant terms.

Using \((a+b)^2=a^2+2ab+b^2\), with \(a=x\) and \(b=2y\), we get \((x+2y)^2=x^2+4xy+4y^2\). Subtracting \(x^2+4y^2\) cancels the like terms, leaving \(4xy\). \(2xy\) is a common error because the middle term is \(2\times x\times 2y=4xy\). Exam tip: calculate the coefficient of the middle term carefully while expanding a square.

Using \((a+b)^2=a^2+2ab+b^2\), take \(a=2x\) and \(b=3y\). The middle term is \(2(2x)(3y)=12xy\), so A is correct. B has 6xy instead. Exam tip: verify a perfect square by checking twice the product of the square roots of the end terms.

\(x^4-81y^4=(x^2)^2-(9y^2)^2\). Applying \(a^2-b^2=(a-b)(a+b)\) gives option A. Option C has a sum, not a difference. Exam tip: first rewrite both terms as perfect squares.

The first term is \(64x^2=(8x)^2\) and the last term is \(25=5^2\). Their middle term is \(2\times 8x\times 5=80x\), so \(64x^2+80x+25=(8x+5)^2\). Option B would give a middle term of \(-80x\). Exam tip: always verify the middle term using \(a^2+2ab+b^2=(a+b)^2\).

The first term is \(25m^2=(5m)^2\), and the last term is \(49n^2=(7n)^2\). Using \((a-b)^2=a^2-2ab+b^2\), with \(a=5m\) and \(b=7n\), the middle term becomes \(-2(5m)(7n)=-70mn\). Hence the expression is \((5m-7n)^2\). In contrast, \((5m+7n)^2\) would have the positive middle term \(+70mn\). Exam tip: while identifying a perfect square, always check the sign of the middle term and verify \(2ab\).

First use the identity \((x+a)(x-a)=x^2-a^2\). The expression becomes \((x^2-a^2)(x^2+a^2)\). Applying \((p-q)(p+q)=p^2-q^2\), with \(p=x^2\) and \(q=a^2\), gives \(x^4-a^4\). Option D is the expansion of \((x^2-a^2)^2\), so it is not applicable here. Exam tip: identify conjugate factors step by step and use the difference-of-squares identity.

Use the identity \((a+b)^2+(a-b)^2=2a^2+2b^2\). Here, \(a=2x\) and \(b=3y\), so \((2x+3y)^2+(2x-3y)^2=2(2x)^2+2(3y)^2=8x^2+18y^2\). Subtracting \(8x^2\) gives \(18y^2\). The \(12xy\) and \(24xy\) options are incorrect because the mixed \(xy\) terms cancel when the two squares are added. Exam tip: When \((a+b)^2\) and \((a-b)^2\) occur together, look for cancellation of the mixed terms.

\((x+5)(x-3)=x^2+2x-15\) and \((x-5)(x+3)=x^2-2x-15\). On adding, \(+2x\) and \(-2x\) cancel, giving \(2x^2-30\). \(2x^2+30\) results from incorrectly adding the constant terms. Exam tip: expand each product separately before combining like terms.

Expanding the first product gives \((x+5)(x-3)=x^2+2x-15\). Similarly, \((x-5)(x+3)=x^2-2x-15\). Subtracting the second expression gives \(x^2+2x-15-(x^2-2x-15)=4x\). Hence, the correct answer is \(4x\). The option \(8x\) may result from handling the negative signs while subtracting incorrectly. Exam tip: when a minus sign precedes brackets, change the sign of every term inside the brackets.

The direct answer is B: the missing terms are 2ab+2bc+2ca. Begin with the square of a sum:
\((a+b+c)^2=(a+b+c)(a+b+c)\). Multiplying each term by every term gives a², b², c², ab twice, bc twice, and ca twice. Therefore,
\((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\). Option A has the pair products but misses their required factor 2, so it is incomplete. Option B is correct because every pair appears twice in the multiplication. Option C, 3abc, is a three-variable product and belongs to neither the ordinary square expansion nor the required degree-two result. Option D is only the original sum, not the extra cross terms produced by squaring. A useful exam cue is: for three terms, write the three individual squares, then twice each of the three pair products. Check that every term has total degree two.

\(x^2+12x+36=(x+6)^2\) because the middle term is \(2\times x\times6=12x\). In option B, the constant term should be \(36\), not \(18\). Exam tip: check whether \(b^2=4ac\).

Here, \(51=50+1\) and \(49=50-1\). Using the identity \((a+b)^2+(a-b)^2=2a^2+2b^2\), with \(a=50\) and \(b=1\), we get \(51^2+49^2=2(50)^2+2(1)^2=5000+2=5002\). The value 5000 includes only \(2\times 50^2\); \(2\times 1^2\) must also be added. Exam tip: apply this identity directly when two numbers are equally spaced from a middle number.

Use the identity \(a^2-b^2=(a+b)(a-b)\). Here, \(51^2-49^2=(51+49)(51-49)=100\times2=200\). Option 100 is only the sum of the two numbers, while 2 is their difference; their product gives the required value. Exam tip: for the difference of squares of nearby numbers, apply this identity directly.

Using \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\), we get \(a^2+b^2+c^2=12^2-2\times35=144-70=74\). Thus, the claim is correct. \(109\) results from forgetting the factor \(2\). Exam tip: always check the coefficient of mixed terms.

First, use the identity \((x-1)(x+1)=x^2-1\). The expression then becomes \((x^2-1)(x^2+1)\), which again has the form \((a-b)(a+b)=a^2-b^2\). Therefore, it simplifies to \(x^4-1\). \(x^4+1\) is not the result of this difference-of-squares identity. Exam tip: identify conjugate binomial pairs and apply identities step by step.

This is a difference of squares: \(a^2-b^2=(a+b)(a-b)\). Here, \(a=2x+3\) and \(b=2x+1\). Thus, \(a+b=4x+4\) and \(a-b=2\), so the expression becomes \((4x+4)\times2=8x+8\). The expression \(4x+4\) is only \(a+b\), not the final simplified form. Exam tip: first identify the two bases of the squares, then apply the difference-of-squares identity.

Use the identity \(a^2-b^2=(a+b)(a-b)\). Here, \(a=3x+5\) and \(b=3x-1\). Thus, \(a+b=6x+4\) and \(a-b=6\). Therefore, \((6x+4)\times6=36x+24\). Option B results from an incorrect multiplication of the \(6x\) term by 6. Exam tip: identify the two squared expressions first and apply the identity directly instead of expanding both squares.

Here \(9p^2=(3p)^2\) and \(16q^2=(4q)^2\). The middle term is \(-24pq=-2(3p)(4q)\), so the expression is \((3p-4q)^2\). Exam tip: check the sign of the middle term carefully.

Given \(a+b+c=10\), we get \((a+b+c)^2=10^2=100\). Option 10 is only the value of the sum, not its square. Exam tip: when a power is outside brackets, substitute the bracket's value first and then apply the power.

Here \(4a^2=(2a)^2\) and \(9b^2=(3b)^2\). The required middle term is \(2\times2a\times3b=12ab\), so the expression is \((2a+3b)^2\). Exam tip: always check both the sign and coefficient of the middle term.

Squaring a sum gives the square of the first term, twice their product, and the square of the second term. Hence \((a+b)^2=a^2+2ab+b^2\). Exam tip: the middle term is \(+2ab\) for a sum, not \(-2ab\).

Group the expression as \(x^2+6x+9-y^2=(x^2+6x+9)-y^2=(x+3)^2-y^2\). This is a difference of squares: \((x+3)^2-y^2=[(x+3)-y][(x+3)+y]=(x+3-y)(x+3+y)\).

This uses ( (a+b)^2-(a-b)^2=4ab ) where (a=4x) and (b=3y). Exam tip: take full terms as (a) and (b).