Operations on real numbers and the laws of exponents

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.

Use the identity \(a^2+b^2=(a-b)^2+2ab\). Thus, \(a^2+b^2=8^2+2(12)=64+24=88\). The value 76 results from adding only \(ab\) instead of \(2ab\), so it is incorrect. In exams, carefully identify the sign and the coefficient of \(2ab\) in the relevant identity.

Use the identity \\(a-b)^2=a^2-2ab+b^2. Since \\(97=100-3, \\(97)^2=(100-3)^2=10000-600+9=9409, so option A is correct. The nearby numerical distractors result from small place-value or middle-term errors; in an exam, check the term \\(-2ab carefully.

\(104^2=(100+4)^2=100^2+2(100)(4)+4^2=10000+800+16=10816\). Therefore, the correct answer is 10816. The nearby distractor 10840 is incorrect because the sum of the middle term 800 and the last term 16 is 816, not 840. Exam tip: Always include the middle term \(2ab\) when using \((a+b)^2=a^2+2ab+b^2\).

Here, \(72=70+2\) and \(68=70-2\). Therefore, \((70+2)(70-2)=70^2-2^2=4900-4=4896\). The identity used is \((a+b)(a-b)=a^2-b^2\). Exam tip: When two numbers are equally spaced around a central number, use the difference-of-squares identity.

Use the difference-of-squares identity \(a^2-b^2=(a-b)(a+b)\), which is valid for real numbers, including decimals. Here \(a=3.5\) and \(b=2.5\). Thus \(3.5^2-2.5^2=(3.5-2.5)(3.5+2.5)=1\times6=6\). A direct check gives \(3.5^2=12.25\) and \(2.5^2=6.25\), and \(12.25-6.25=6\), confirming the result. Therefore option B is correct. Option A confuses the difference with one of the factors, while options C and D arise from incorrect addition or squaring. Factoring is especially efficient here because the two numbers differ by exactly 1.