Algebraic identities

Use the identity \((x+y)(x-y)=x^2-y^2\), where \(x=5a\) and \(y=2b\). Therefore, \((5a+2b)(5a-2b)=(5a)^2-(2b)^2=25a^2-4b^2\). Option A incorrectly uses the sum of squares; the product of two binomials with opposite signs gives a difference of squares. Exam tip: whenever you see \((p+q)(p-q)\), apply \(p^2-q^2\) directly.

Since \(98=100-2\), writing \(98^2=(100-2)^2\) is the most convenient form. Using \((a-b)^2=a^2-2ab+b^2\) with \(a=100\) and \(b=2\) makes the calculation quick. Although \((90+8)^2\) and \((50+48)^2\) also equal \(98^2\), they do not use the advantage of the nearby round number 100. Exam tip: express a number in terms of a nearby multiple of 10 or 100 when finding its square.

Use the identity \((a+b)^2=a^2+2ab+b^2\). Here, \(x^2+16x+64=x^2+2\times x\times 8+8^2\), so it equals \((x+8)^2\). If it were \((x-8)^2\), the middle term would be \(-16x\), not \(+16x\). Exam tip: match the middle term with \(2ab\) and the last term with \(b^2\).

Using the identity \((a-b)^2=a^2-2ab+b^2\), put \(a=p\) and \(b=9\). Then \((p-9)^2=p^2-18p+81\). Hence, the correct factor form is \((p-9)^2\). The expression \((p+9)^2\) has middle term \(+18p\), so it is not correct. Exam tip: for a perfect-square trinomial, halve the coefficient of the middle term and check whether its square equals the constant term.

Here, \(103=100+3\) and \(97=100-3\). Thus the product is of the form \((100+3)(100-3)\), so we use \((a+b)(a-b)=a^2-b^2\). Hence, \(103\times97=100^2-3^2=10000-9=9991\). The identity \((a-b)^2\) is for the square of one binomial, so it does not fit here. Exam tip: when two numbers are equally spaced from the same number, look for the difference-of-squares identity.

The square of three terms includes all squares and twice each pair product. Exam tip: do not forget (2ab+2bc+2ca).

In \((2m+n-1)^2\), the \(mn\)-term is formed only by multiplying \(2m\) and \(n\). In the expansion of a square, this cross-term is \(2\times(2m)\times n=4mn\). Therefore, the correct answer is \(4mn\). It cannot be \(-4mn\) because both \(2m\) and \(n\) are positive terms. Exam tip: for a particular mixed term in the square of three terms, use \(2ab\) only for the relevant pair.

Expand: \((x-3)(x+8)=x^2+8x-3x-24=x^2+5x-24\). The constant term is \(-24\).

\((x+4)(x+9)=x^2+9x+4x+36=x^2+13x+36\). Therefore, the coefficient of \(x\) is \(13\). The number \(36\) is the constant term, not the coefficient of \(x\). Exam tip: In \((x+a)(x+b)\), the coefficient of \(x\) is always \(a+b\).

The coefficient of (x) is (-6-5=-11) and the constant term is (30). Exam tip: product of two negative numbers is positive.

Use \((a-b)^2=a^2-2ab+b^2\). Here \(a=2p\) and \(b=5q\). So \((2p-5q)^2=(2p)^2-2(2p)(5q)+(5q)^2=4p^2-20pq+25q^2\).

The direct answer is option B, 20000. Use the difference-of-squares identity \(a^2-b^2=(a+b)(a-b)\), with \(a=1005\) and \(b=995\). First find the sum: \(1005+995=2000\). Then find the difference: \(1005-995=10\). Multiply: \(2000\times10=20000\). Option A, 10000, is not the product of the required sum and difference. Option B, 20000, follows exactly from the identity and is correct. Option C, 5000, does not use the correct sum and difference calculation. Option D, 2000, is only the sum of the two numbers, not the final value. This method avoids calculating two large squares separately. Memory cue: square difference means add the numbers, subtract the numbers, then multiply.

The difference-of-squares identity is \(p^2-q^2=(p+q)(p-q)\), so option C is correct. Options A and B represent squares of binomials. Exam tip: on seeing a difference of squares, look for conjugate factors \((p+q)(p-q)\).

Use the identity \((a+b)^2+(a-b)^2=2a^2+2b^2\). Here, \(a=x\) and \(b=5\), so \((x+5)^2+(x-5)^2=2x^2+2(5)^2=2x^2+50\). Option D represents only one square-type result, whereas the sum of both squares is required. Exam tip: in such expressions, the middle terms with opposite signs cancel out.

Writing \(49\) as \(50-1\) uses the nearby round number \(50\). Using \((a-b)^2=a^2-2ab+b^2\), \(49^2=(50-1)^2=2500-100+1=2401\). Although \(48+1\) is also a valid split, calculating \(48^2\) is less convenient than calculating \(50^2\). Exam tip: for quick squaring, express a number in terms of a nearby multiple of 10 or 100.

Use the identity \((a+b)^2-(a-b)^2=4ab\). Here, \(a=2x\) and \(b=3y\), so the value is \(4\times 2x\times 3y=24xy\). The expression \(4x^2-9y^2\) is the product \((2x+3y)(2x-3y)\), not the difference of the two squares given here. Exam tip: first identify the pattern \((a+b)^2-(a-b)^2\) before expanding.

This expression is a difference of two squares. The identity is \\(a^2-b^2=(a+b)(a-b)\\). It is useful because it changes the subtraction of two large squares into two simpler factors. Here, take \(a=73\) and \(b=27\). Then \(a+b=100\) and \(a-b=46\). Their product is \(100\times46=4600\), so the value is 4600 and option A is correct.

The same result can be checked directly: \(73^2=5329\) and \(27^2=729\), and \(5329-729=4600\). However, the identity is faster and reduces the chance of making a large multiplication error. The other numerical choices do not equal the product of the sum and difference. Therefore option A follows from the difference-of-squares identity.

Using the distributive property, \((3x+2)(3x+5)=3x\cdot3x+3x\cdot5+2\cdot3x+2\cdot5\). Thus, \(9x^2+15x+6x+10=9x^2+21x+10\), so option B is correct. Option D misses the cross term \(2\cdot3x=6x\). Exam tip: write all four products first, then combine like terms.

Take 2 as a common factor: \(2x^2+12x+18=2(x^2+6x+9)\). Here, \(x^2+6x+9=x^2+2\cdot x\cdot3+3^2=(x+3)^2\). Therefore, the correct factor form is \(2(x+3)^2\). In contrast, \(2(x+6)^2\) would produce a middle term of \(24x\), so it is incorrect. Exam tip: for a perfect-square trinomial, check whether the middle term equals \(2ab\).

The expression is a difference of two perfect squares: \(9a^2=(3a)^2\) and \(25b^2=(5b)^2\). Using \(x^2-y^2=(x+y)(x-y)\), we get \(9a^2-25b^2=(3a+5b)(3a-5b)\). Option B is incorrect because squaring it also produces a middle term, \(-30ab\). Exam tip: first rewrite each term as a perfect square, then apply the difference-of-squares identity.

When a sum of several terms is squared, every term is multiplied by every term. For three terms, the identity is \((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\). The product of \(a\) and \(b\) appears twice: once as \(a\times b\) and once as \(b\times a\). Since both are like terms, they combine to give \(2ab\). Thus option B identifies the source of the ab-term.

The terms \(a^2+b^2\) are square terms, not the cross-product term, while \(abc\) does not occur in a square of a sum. Likewise, \(b^2+c^2\) contains only square terms for those two variables. Expanding the brackets or recalling the identity leads directly to the same result. Therefore the ab-term is formed from twice the product of a and b, namely \(2ab\), so option B is correct.

The governing concept is identifying terms without variables. When (2x + y + 1)² is expanded, the only term formed without x or y is the square of the constant part: 1² = 1. Terms such as 4x², y² and the cross-products contain variables, so they cannot be constant terms. Therefore option C is correct; 2 and 4 are not produced as the required constant term, and 0 is absent.

Expanding \((a-b)^2=(a-b)(a-b)\) gives \(a^2-ab-ab+b^2=a^2-2ab+b^2\). Hence, option A is true for every real value of \(a\) and \(b\). Option D is the expansion of \((a+b)^2\), while B has the wrong sign for \(b^2\) and C has an incorrect coefficient in the middle term. Exam tip: the middle term in \((x-y)^2\) is always \(-2xy\).

Using the identity

(a-b)^2=a^2-2ab+b^2,

with a=3x and b=1, we get (3x-1)^2=9x^2-6x+1. Therefore,

9x^2-6x+1-(9x^2+1)=-6x.

Hence, -6x is the correct answer. The option 6x results from missing the negative sign of the middle term. Exam tip: in (a-b)^2, the middle term is always -2ab.

The standard identity gives the middle term as \(2ab\). The given expression has middle term \(10ab\). So the difference (error) is \(10ab-2ab=8ab\).