Operations on real numbers and the laws of exponents

Here, \(81x^2=(9x)^2\) and \(100y^2=(10y)^2\). Thus, the expression becomes \((9x)^2-(10y)^2\). Using the difference of squares identity \(a^2-b^2=(a-b)(a+b)\), we get \((9x-10y)(9x+10y)\). Options C and D are squares whose expansions introduce an additional \(90xy\) term, so they are incorrect. Exam tip: in the difference of two squares, the two factors differ only in the sign between the terms.

Using the power-of-a-power rule, \((3^2)^4=3^{2\times4}=3^8\). For multiplication of powers with the same base, add the exponents: \(3^5\cdot3^{-2}=3^{5+(-2)}=3^3\). Therefore, \(\frac{3^8}{3^3}=3^{8-3}=3^5\), so option A is correct. Option B gives only the denominator’s power, not the simplified value of the whole fraction. Exam tip: add exponents when multiplying like bases and subtract them when dividing.

Rewrite every quantity using base 2. We have \(4^3=(2^2)^3=2^6\u0005, \(2^{-1}\u0005 remains as it is, and \(8=2^3\u0005. Therefore the expression becomes \(\frac{2^6\cdot2^{-1}}{2^3}\u0005. Using exponent laws, multiplication adds exponents and division subtracts them, so the total exponent is \(6+(-1)-3=2\u0005.

Thus the value is \(2^2=4\u0005, making option B correct. The negative exponent means reciprocal, since \(2^{-1}=\frac12\u0005; it does not mean that the final answer is negative. Direct calculation also gives \(64\cdot\frac12\div8=4\u0005.

For powers with the same nonzero base, exponents are added during multiplication and subtracted during division. Thus, \(\frac{x^7\cdot x^{-3}}{x^2}=x^{7+(-3)-2}=x^2\). Option C results from simplifying only \(x^7\cdot x^{-3}\) and ignoring the division by \(x^2\). Exam tip: subtract the denominator’s exponent when dividing like bases.

When a quotient is raised to a power, the exponent applies to both numerator and denominator: \(\left(\frac{x}{y}\right)^n=\frac{x^n}{y^n}\). In B, the denominator lacks the exponent. Exam tip: apply the power to every factor.

To simplify a quotient of monomials, divide the numerical coefficients and subtract the exponent of each variable in the denominator from its exponent in the numerator. The coefficient becomes \(18\div6=3\u0005. For \(x\u0005, the exponent is \(5-2=3\u0005, and for \(y\u0005, it is \(3-1=2\u0005. The conditions \(x\ne0\u0005 and \(y\ne0\u0005 allow this cancellation.

Therefore the simplified expression is \(3x^3y^2\u0005, so option A is correct. Exponents are subtracted because \(a^m/a^n=a^{m-n}\u0005 for nonzero \(a\u0005. Adding exponents would apply to multiplication, not division, which is why the other forms are incorrect.

\(\left(\frac{1}{5}\right)^{-2}=25\) and \(\left(\frac{1}{3}\right)^{-2}=9\). Therefore the sum is (34).

For every non-zero real number, the zero-exponent law gives \(a^0=1\). In contrast, \(a^{-1}=1/a\), which is not generally 1. Exam tip: always check that the base is non-zero before applying this law.

By the law of exponents, \(a^{-1}=\frac{1}{a}\). Therefore, \(5^2-5^{-1}=25-\frac{1}{5}=\frac{125-1}{5}=\frac{124}{5}\). Option B results from incorrectly adding 1 in the numerator instead of subtracting it. In an exam, remember that a negative exponent represents the reciprocal of the base.

Substitute \(x=3\) into the polynomial: \(p(3)=4(3)^2-5(3)+2=4\times9-15+2=36-15+2=23\). Therefore, the correct value is 23. Exam tip: evaluate the exponent before multiplication, and do not interpret \(4(3)^2\) as \((4\times3)^2\).

Substituting \(x=-2\), we get \(q(-2)=(-2)^3+2(-2)^2-4(-2)=-8+8+8=8\). Note that \((-2)^2=4\) and \(-4(-2)=+8\); always use parentheses when substituting a negative value.

When powers with the same base are multiplied, their exponents are added: \(x^4\times x^3=x^{4+3}=x^7\). In option D, exponents are multiplied, giving \(x^{12}\). Exam tip: first check whether the bases are identical.

When powers with the same non-zero base are multiplied, their exponents are added, so \(a^m \times a^n=a^{m+n}\). In division, exponents are subtracted, not multiplied. Exam tip: check whether the bases are the same first.

Because a minus sign precedes the second bracket, the signs of both terms inside it change: (9x+2)-(4x-6)=9x+2-4x+6. Combining like terms gives (9x-4x)+(2+6)=5x+8. Exam tip: When removing a bracket preceded by a minus sign, change the sign of every term inside the bracket.

Using the distributive law, multiply \(4x\) by every term in the bracket: \(4x\cdot3x^2=12x^3\), \(4x\cdot(-2x)=-8x^2\), and \(4x\cdot5=20x\). Therefore, the expansion is \(12x^3-8x^2+20x\). Option B fails to multiply each term by the outside factor \(4x\). Exam tip: when multiplying a monomial by a polynomial, distribute it to every term and combine the powers of the same variable.

Using the distributive property,
(x+6)(x+4)=x^2+4x+6x+24
. Combining like terms gives
4x+6x=10x
, so the expansion is
x^2+10x+24
. Option B incorrectly interchanges the middle-term coefficient and the constant term. Exam tip: Check with
(x+a)(x+b)=x^2+(a+b)x+ab
; here the middle coefficient is
6+4=10
and the constant term is
6\times4=24
.

Using the distributive property, (x − 7)(x + 3) = x² + 3x − 7x − 21 = x² − 4x − 21, so option A is correct. Option B results from adding the middle terms with the wrong sign. In an exam, multiply each term of one binomial by both terms of the other and then combine like terms.

Apply the distributive property: \((4x+5)(x-3)=4x\cdot x+4x\cdot(-3)+5\cdot x+5\cdot(-3)\). This gives \(4x^2-12x+5x-15=4x^2-7x-15\), so option A is correct. In option B, the middle terms \(-12x+5x\) have been combined with the wrong sign. In an exam, check the signs of the two middle terms carefully.

Apply the identity 8a+b9^2=a^2+2ab+b^2. Here, a=x and b=10, so 8x+109^2=x^2+2cdot xcdot10+10^2=x^2+20x+100. Therefore, option A is correct. Option B is incorrect because it omits the middle term 20x. In an exam, always check the middle term 2ab when squaring a binomial.

Use the identity \((a-b)^2=a^2-2ab+b^2\). Here, \(a=x\) and \(b=11\), so \((x-11)^2=x^2-2(x)(11)+11^2=x^2-22x+121\). Exam tip: the middle term is negative when the binomial is a difference.

This expression uses the identity \\(a+b)(a-b)=a^2-b^2\\). Here, \\(a=x\\) and \\(b=12\\), so the result is \\(x^2-12^2=x^2-144\\). Option B incorrectly gives the sum of the squares; opposite signs in the two binomials produce a difference of squares. Exam tip: recognise \\(a+b)(a-b)\\) directly as \\(a^2-b^2\\).

Here, \(121x^2=(11x)^2\) and \(144=12^2\). Therefore, \(121x^2-144=(11x)^2-12^2\). Using the difference-of-squares identity \(a^2-b^2=(a-b)(a+b)\), the factorized form is \((11x-12)(11x+12)\). Options C and D are squares of binomials, so they do not represent this difference. Exam tip: whenever an expression is the difference of two perfect squares, apply \(a^2-b^2=(a-b)(a+b)\).

\(x^2+22x+121=x^2+2\cdot x\cdot11+11^2\). Using the perfect-square identity \(a^2+2ab+b^2=(a+b)^2\), it equals \((x+11)^2\). Option B would produce the middle term \(-22x\), so it is incorrect. Exam tip: take the square roots of the first and last terms and verify the middle term.

x^2-24x+144 can be written as x^2-2\cdot x\cdot12+12^2. Using the identity a^2-2ab+b^2=(a-b)^2, it becomes (x-12)^2. Option B would produce a positive middle term, so it is incorrect. Exam tip: take the square roots of the first and last terms and then check the sign of the middle term.

Use the identity \\(m^2+n^2=(m+n)^2-2mn\\). Thus, \\(m^2+n^2=13^2-2(36)=169-72=97\\), so the correct answer is 97. The value 169 results from using only \\((m+n)^2\\) and omitting the \\-2mn\\) term. Exam tip: When the sum and product of two numbers are given, apply this identity directly.