Operations on real numbers and the laws of exponents

Apply the identity \\(a+b\\)^2=a^2+2ab+b^2\\). With \\(a=x\\) and \\(b=8\\), we get \\(x^2+2(x)(8)+8^2=x^2+16x+64\\). Option B omits the middle term, while option C uses an incorrect coefficient for it. In an exam, always check that the middle term is \\(2ab\\).

Use the identity \\((a-b)^2=a^2-2ab+b^2\\). Here, \\(a=x\\) and \\(b=9\\), so \\(x^2-2(x)(9)+9^2=x^2-18x+81\\). Option B has the wrong sign in the middle term because this is the square of a difference, not a sum. In an exam, remember that the middle term in \\((a-b)^2\\) is always \\(-2ab\\).

When powers with the same base are divided, their exponents are subtracted: \(a^m\div a^n=a^{m-n}\), where \(a\ne0\). Option B is the multiplication rule, not the quotient rule. In exams, first check that the bases match and the base is non-zero.

\(64x^2-81=(8x)^2-9^2\). Using the difference of squares identity \(a^2-b^2=(a-b)(a+b)\), we get \((8x-9)(8x+9)\). Options C and D are squares of binomials, which would contain a middle term, so they are not correct. Exam tip: Whenever two perfect squares are being subtracted, check for the difference of squares identity first.

Use the identity \((a+b)^2=a^2+2ab+b^2\). Taking \(a=x\) and \(b=9\), we get \((x+9)^2=x^2+2\cdot x\cdot9+9^2=x^2+18x+81\). Therefore, option A is correct. Option B would give the middle term \(-18x\), while option D omits the \(18x\) term. Exam tip: Take the square root of the constant term and use the sign of the middle term to choose between \((x+a)^2\) and \((x-a)^2\).

Using the perfect-square identity (a-b)^2=a^2-2ab+b^2, we get x^2-20x+100=x^2-2\cdot x\cdot10+10^2=(x-10)^2. Option B would produce a middle term of +20x, so it is incorrect. Exam tip: take the square roots of the first and last terms and check whether twice their product matches the middle term.

Use the identity \(m^2+n^2=(m+n)^2-2mn\). Thus, \(m^2+n^2=11^2-2(24)=121-48=73\). Option B results from subtracting \(mn\) only once, whereas the identity requires subtracting \(2mn\). In an exam, writing the relevant algebraic identity first is the quickest method.

Use the identity \(a^2+b^2=(a-b)^2+2ab\). Thus, \(a^2+b^2=7^2+2(10)=49+20=69\). Therefore, the correct answer is 69. Remember not to omit the term \(2ab\); using only \((a-b)^2+ab\) would incorrectly give 59.

Since \(98=100-2\), use the identity \((a-b)^2=a^2-2ab+b^2\): \(98^2=(100-2)^2=100^2-2(100)(2)+2^2=10000-400+4=9604\). Therefore, option A is correct. Exam tip: For squaring a number close to a convenient base such as 100, use the identity \((a\pm b)^2\).

(103)^2=(100+3)^2=100^2+2(100)(3)+3^2=10000+600+9=10609
, so option A is correct. In exams, remember the middle term
(2ab)
in
(a+b)^2=a^2+2ab+b^2
; omitting it can lead to the incorrect answer 10009.

64 × 56 = (60 + 4)(60 − 4). Using the identity (a + b)(a − b) = a² − b², we get 60² − 4² = 3600 − 16 = 3584. Therefore, 3584 is correct. Exam tip: Use the difference-of-squares identity when two factors are equally spaced around the same middle number.

The governing algebraic concept is the difference-of-squares identity, \(a^2-b^2=(a-b)(a+b)\). Taking \(a=2.5\) and \(b=1.5\), the expression becomes \((2.5-1.5)(2.5+1.5)\). The first factor is \(1\), and the second factor is \(4\), so their product is \(1\times4=4\). Direct checking gives \(2.5^2=6.25\) and \(1.5^2=2.25\), whose difference is also \(6.25-2.25=4\). Thus option C is correct. Options A, B and D can result from an arithmetic error or from applying the identity incorrectly; the decimal values do not change the validity of the identity.

\(\left(\frac{1}{4}\right)^{-2}=16\) and \(\left(\frac{1}{5}\right)^{-2}=25\). The sum is (41).

Substituting the given values, \(a^2b+ab^2=4^2\times1+4 imes1^2=16+4=20\). Alternatively, factorising gives \(ab(a+b)=4 imes1 imes(4+1)=20\). In an exam, evaluate the powers before carrying out the multiplication.

The greatest common factor of \(3x^4\) and \(9x^3\) is \(3x^3\). Taking it outside gives \(3x^4+9x^3=3x^3(x+3)\), since \(3x^3\times x=3x^4\) and \(3x^3 imes3=9x^3\). Option C is not a valid factorisation because factoring out 9 would require a fractional coefficient in the bracket, not \(x+1\). Exam tip: multiply the factor outside the bracket by every term inside to verify the result.

Dividing the numerical coefficients gives \(24\div6=4\). For powers with the same base, subtract the exponents: \(a^{5-2}=a^3\) and \(b^{3-2}=b\). Hence, \(\frac{24a^5b^3}{6a^2b^2}=4a^3b\), so option A is correct. In option B, the exponent of \(b\) has not been reduced correctly. Exam tip: for division of like bases, use \(x^m\div x^n=x^{m-n}\).

Multiply the numerical coefficients to get \(5\times2=10\). For like bases, add the exponents: \(x^{3+1}=x^4\) and \(y^{1+2}=y^3\). Therefore, the product is \(10x^4y^3\). Option C omits one power of \(x\), while option D combines the exponents incorrectly. Exam tip: when multiplying monomials, multiply the coefficients and add the exponents of like variables.

Substituting \(x=3\) gives \(x^3=27\) and \(x^2=9\). Therefore, \(2x^3-5x^2+4=2(27)-5(9)+4=54-45+4=13\), so option C is correct. As an exam tip, evaluate powers first and then perform multiplication, addition, and subtraction in the proper order.

\\((-3)^4=81\\) because an even power gives a positive result, while \\((-3)^3=-27\\) because an odd power retains the negative sign. Therefore, \\(x^4+x^3=81+(-27)=54\\), so option B is correct. Exam tip: Always check the sign when raising a negative number to an even or odd power.

For division with like bases, subtract the exponent in the denominator from the exponent in the numerator: \(x^{6-3}y^{-3-(-5)}=x^3y^2\). Therefore, \(x^3y^2\) is correct. Option B mishandles the exponents of \(y\), while C and D incorrectly add the exponents of \(x\). Exam tip: when dividing powers with the same base, subtract the denominator exponent from the numerator exponent.