Operations on real numbers and the laws of exponents

The governing concept is extraction of perfect-square factors from a radical. Factor 128 as 64 × 2, where 64 is the greatest useful perfect-square factor. Therefore, √128 = √(64 × 2) = √64 × √2 = 8√2. Option A is correct because the remaining radicand, 2, has no square factor greater than 1. Option B is incorrect because (16√2)² = 512, not 128. Options C and D have the same numerical value as 8√2, since 4√8 = 4 × 2√2 = 8√2 and 2√32 = 2 × 4√2 = 8√2; however, they are not fully simplified because 8 and 32 still contain perfect-square factors. Thus A is the required simplest form.

The governing algebraic idea is the product of conjugate expressions: (a + b)(a − b) = a² − b². The given factors are conjugates, with a = √3 and b = 1. Hence αβ = (√3 + 1)(√3 − 1) = (√3)² − 1² = 3 − 1 = 2. If the product is expanded term by term, the middle terms +√3 and −√3 cancel, giving the same result. Therefore option A is correct. Option B can arise from an incorrect squaring or addition, option C ignores the complete multiplication, and option D does not follow from any valid simplification. The irrationality of √3 does not prevent exact calculation.

The governing concept is rationalisation of a denominator containing a surd. Multiply the numerator and denominator by the conjugate of √10−3, namely √10+3. Then 1/(√10−3)=(√10+3)/[(√10−3)(√10+3)]. Using (a−b)(a+b)=a²−b², the denominator becomes 10−9=1. Hence the expression simplifies to √10+3. Therefore option A is correct. Option B is the original denominator expression and does not result from rationalisation. Option C incorrectly leaves a factor of 19 in the denominator, while option D expands the expression incorrectly and is not algebraically equivalent. The nonzero denominator is valid because √10 is not equal to 3.

When powers with the same base are multiplied, their exponents are added: \(2^3\cdot2^{-5}\cdot2^4=2^{3+(-5)+4}=2^2\). Therefore, option A is correct. Option B results from mishandling the negative exponent, while option D incorrectly multiplies the exponents. In an exam, retain the sign of a negative exponent while adding exponents.

For powers with the same base, add exponents during multiplication and subtract the denominator's exponent during division: \(2+(-4)-(-3)=1\). Thus, \(3^1=3\), so option A is correct. Option C, \(\frac{1}{3}\), results from incorrectly treating the final exponent as \(-1\). Exam tip: apply \(a^m\div a^n=a^{m-n}\) carefully, especially when the denominator has a negative exponent.

Using the power-of-a-power rule, \((x^3)^2=x^{3\times2}=x^6\). For the remaining product and division of powers with the same base, subtract the denominator exponent: \(x^6\cdot x^{-4}\div x=x^{6-4-1}=x\). Therefore, option A is correct. Exam tip: write the denominator as \(x^1\) before combining exponents.

For any non-zero real number \(a\), \(\frac{a^m}{a^m}=a^{m-m}=a^0\). Since the value on the left is \(1\), we get \(a^0=1\), where \(a\ne0\). Option D is a close distractor, but the zero-exponent law cannot be applied when \(a=0\); in school algebra, \(0^0\) is generally treated as undefined. Exam tip: Always remember the condition \(a\ne0\) with the zero-exponent rule.

Since \(16=2^4\), we get \(16^2=(2^4)^2=2^8\). Therefore, \(\frac{16^2}{2^5}=\frac{2^8}{2^5}=2^{8-5}=2^3=8\). Exam tip: when dividing powers with the same base, subtract their exponents.

\(0.2=\frac{1}{5}\), so \((0.2)^{-2}=\left(\frac{1}{5}\right)^{-2}=25\). Converting decimals to fractions helps.

\(\left(\frac{3}{2}\right)^{-2}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\). Hence \(\frac{4}{9}\cdot\frac{9}{4}=1\).

Given \(a=2\), we have \(a^3=2^3=8\). By the negative exponent rule, \(a^{-1}=\frac{1}{a}=\frac{1}{2}\). Therefore, \(a^3+a^{-1}=8+\frac{1}{2}=\frac{17}{2}\). The value \(\frac{15}{2}\) results from evaluating one of the powers incorrectly. Exam tip: remember that \(x^{-n}=\frac{1}{x^n}\) for \(x\neq0\).

Using the exponent rule \\( (ab)^n=a^n b^n \\), we get \\( (x^2y^3)^2=(x^2)^2(y^3)^2=x^4y^6 \\). Therefore, option A is correct. In option B, the exponent of \\(y\\) has been handled incorrectly; the outside exponent must multiply 3 to give 6. Exam tip: when a product is raised to a power, apply that power to every factor.

First apply the power rule: \((2x^3)^2=2^2x^{3\times2}=4x^6\). Then \(\frac{4x^6}{4x^2}=x^{6-2}=x^4\), so option A is correct. Option D, \(x^6\), results from failing to subtract the exponent in the denominator. Exam tip: when dividing powers with the same non-zero base, subtract the exponents.

Substitute \(x=2\) into the polynomial: \(p(2)=2(2)^2-3(2)+1=2\times4-6+1=3\). Therefore, the correct value is 3. Exam tip: evaluate the power first, followed by multiplication and then addition or subtraction.

Substitute \(x=-1\) into the polynomial: \(q(-1)=(-1)^3-2(-1)^2+(-1)=-1-2-1=-4\). Thus, the correct answer is -4. Since \((-1)^2=1\), the middle term is \(-2\), not \(+2\). In exams, always use parentheses when substituting a negative value.

All the terms have the same variable and the same exponent, so they are like terms. Combine their coefficients: 3 + 5 - 2 = 6. Therefore, the simplified form is 6x^2. Option C is incorrect because it treats the negative term as positive. Exam tip: When combining like terms, operate only on the coefficients; the variable and its exponent remain unchanged.

When powers with the same non-zero base are divided, their exponents are subtracted: \(\dfrac{a^m}{a^n}=a^{m-n}\). Hence, \(\dfrac{a^7}{a^3}=a^{7-3}=a^4\). Option B incorrectly adds the exponents, a rule used for multiplication of powers with the same base. Exam tip: for division, subtract the denominator’s exponent from the numerator’s exponent.

When powers have the same base, their exponents are added, so \(a^m\times a^n=a^{m+n}\). The exponent \(m-n\) is used for division, not multiplication. Exam tip: first check whether the bases are identical.

By the distributive law, multiply 2x by each term inside the bracket: 2x·3x^2=6x^3, 2x·(-4x)=-8x^2, and 2x·1=2x. Therefore, the expanded form is 6x^3-8x^2+2x. Option B does not correctly account for multiplication by 2x, while option D has the wrong sign for the middle term. Exam tip: Multiply the term outside the bracket by every term inside it, including the constant term.

Using the distributive law, multiply each term: x·x + x·2 + 3·x + 3·2 = x² + 2x + 3x + 6. Combining the like terms 2x and 3x gives x² + 5x + 6. Option B has the coefficients of the middle and constant terms incorrect. Exam tip: for two binomials, multiply the first, outer, inner and last terms, then combine like terms.

Use the power-of-a-power law \((a^m)^n=a^{mn}\) and the quotient law for equal bases, \(a^m\div a^n=a^{m-n}\). Thus, \(\frac{(a^4)^3}{a^5}=\frac{a^{12}}{a^5}=a^{12-5}=a^7\). Therefore, option B is correct. Option C stops after applying the power-of-a-power rule, while option D incorrectly adds the exponents during division. In an exam, multiply exponents inside a power first and subtract exponents when dividing like bases.

Using the distributive property, multiply both \(2x\) and \(3\) by each term in \(x-2\): \((2x+3)(x-2)=2x^2-4x+3x-6\). Combining like terms, \(-4x+3x=-x\), gives \(2x^2-x-6\). Hence, option A is correct. Option B has an incorrect positive sign for the middle term. Exam tip: write all four products before combining like terms.

When powers with the same non-zero base are divided, their exponents are subtracted: \(a^m/a^n=a^{m-n}\). For example, \(a^7/a^3=a^4\). The \(m+n\) rule applies to multiplication. In exams, check that the base is non-zero.

Use the identity (a-b)^2=a^2-2ab+b^2. Substituting a=x and b=6 gives (x-6)^2=x^2-2(x)(6)+6^2=x^2-12x+36, so option A is correct. Option B has the wrong sign for the middle term. In exams, remember that the middle term of (a-b)^2 is -2ab.

The expression uses the difference-of-squares identity \((a+b)(a-b)=a²-b²\). Here, \(a=x
u0000\) and \(b=7
u0000\), so \((x+7)(x-7)=x²-7²=x²-49
u0000\). Option B has the wrong sign, while options C and D do not result from multiplying these conjugate binomials. In an exam, identify \((a+b)(a-b)\) directly as \(a²-b²\).