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Since $7x$ means $7 \\times x$, its coefficient is 7."},{"text":"-7","correct":false,"explanation":"A negative coefficient would make the term $-7x$, not $7x$. The numerical factor multiplying $x$ here is positive, so the coefficient is 7."},{"text":"1","correct":false,"explanation":"This confuses the implicit coefficient of $x$ alone with the coefficient in $7x$. The written numerical factor multiplying $x$ is 7."},{"text":"49","correct":false,"explanation":"This incorrectly squares the numerical factor rather than identifying it. The coefficient is simply the number multiplying $x$, which is 7."}]},{"id":"0c0c7f82-4ea4-4155-b5b2-c2cbc8305f93","cardType":"basic","points":2,"front":"Why are $x$ and $x^2$ not like terms?","back":"They do not have the exact same variable part ($x^2$ is $x$ multiplied by itself).","mcqOptions":[{"text":"They do not have the exact same variable part ($x^2$ is $x$ multiplied by itself).","correct":true,"explanation":"Like terms must contain identical variables raised to identical powers. Here, $x$ has exponent 1 while $x^2$ has exponent 2, so they cannot be collected."},{"text":"They do not have the exact same coefficient ($x^2$ has a coefficient of $2$).","correct":false,"explanation":"Both terms have an implicit coefficient of 1, so their coefficients are not different. They are unlike because their variable parts, $x$ and $x^2$, have different powers."},{"text":"They do not contain the exact same variable ($x^2$ uses a different variable).","correct":false,"explanation":"Both terms use the same variable, $x$. They are not like terms because $x$ and $x^2$ contain different powers of that variable."},{"text":"They do not have a numerical coefficient ($x$ and $x^2$ contain no numbers).","correct":false,"explanation":"Both terms do have numerical coefficients: each coefficient is implicitly 1. The reason they are unlike is that their variable parts have different exponents."}]},{"id":"bbd91fc1-d3b4-4d96-9724-030995d59871","cardType":"basic","points":2,"front":"Simplify $7x + 5 + 4x - 2 - 3x$ by collecting like terms.","back":"$$8x + 3$","mcqOptions":[{"text":"$$8x + 3$","correct":true,"explanation":"The variable terms combine to $(7+4-3)x=8x$, while the constants combine to $5-2=3$. Therefore the simplified expression is $8x+3$."},{"text":"$$11x + 3$","correct":false,"explanation":"This omits the subtraction of $3x$ when collecting variable terms. Including every like term gives $(7+4-3)x=8x$, and the constants give 3."},{"text":"$$8x + 7$","correct":false,"explanation":"This adds 5 and 2 instead of respecting the negative sign before 2. The constants total $5-2=3$, while the variable terms total $8x$."},{"text":"$$8x - 3$","correct":false,"explanation":"This reverses the sign of the constant total. Since $5-2=3$, not $-3$, and the variable terms total $8x$, the result is $8x+3$."}]},{"id":"e48977fe-02db-4b20-a426-95c2d49c916c","cardType":"cloze","points":null,"front":"The distributive law for expanding brackets is $k(a + b) = $ $\\text{[...]}$.","back":"The distributive law for expanding brackets is $k(a + b) = $ $\\boxed{ka + kb}$.","mcqOptions":[{"text":"$$ka + kb$","correct":true,"explanation":"The factor $k$ multiplies each term inside the bracket. Multiplying it by both $a$ and $b$ gives $ka+kb$."},{"text":"$$kab$","correct":false,"explanation":"Multiplying $a$ and $b$ together changes addition into multiplication. Distribution requires multiplying $k$ by each separate term, giving $ka+kb$."},{"text":"$$ka + b$","correct":false,"explanation":"This distributes $k$ to $a$ but leaves $b$ unchanged. The outside factor must multiply both terms, so the expansion is $ka+kb$."},{"text":"$$k + a + b$","correct":false,"explanation":"This treats multiplication by $k$ as adding $k$ to the bracket’s terms. Distribution instead multiplies both terms by $k$, producing $ka+kb$."}]},{"id":"c06bbbd6-5871-4308-8227-fb35eacd3329","cardType":"basic","points":2,"front":"What is the result of expanding $8(4s - 3)$?","back":"$$32s - 24$","mcqOptions":[{"text":"$$32s - 24$","correct":true,"explanation":"Distribute 8 across both terms: $8\\times4s=32s$ and $8\\times(-3)=-24$. Combining these products gives $32s-24$."},{"text":"$$32s - 3$","correct":false,"explanation":"This multiplies only the first term by 8 and leaves $-3$ unchanged. The factor 8 must multiply both terms, making the constant $-24$."},{"text":"$$12s - 24$","correct":false,"explanation":"This adds 8 and 4 instead of multiplying them. The variable product is $8\\times4s=32s$, and distributing to $-3$ gives $-24$."},{"text":"$$32s + 24$","correct":false,"explanation":"This loses the negative sign when multiplying the second term. A positive 8 multiplied by negative 3 gives $-24$, so the expansion is $32s-24$."}]},{"id":"44dfb85a-dffa-4c7b-983d-3622db0210a4","cardType":"basic","points":2,"front":"Expand $5x(3x - 2)$.","back":"$$15x^2 - 10x$","mcqOptions":[{"text":"$$15x^2 - 10x$","correct":true,"explanation":"Multiply $5x$ by each bracketed term: $5x\\times3x=15x^2$ and $5x\\times(-2)=-10x$. This gives $15x^2-10x$."},{"text":"$$15x^2 - 10$","correct":false,"explanation":"This drops the factor $x$ when multiplying $5x$ by $-2$. That product is $-10x$, so the complete expansion is $15x^2-10x$."},{"text":"$$15x^2 - 2x$","correct":false,"explanation":"This fails to multiply the second term by the numerical factor 5. Since $5x\\times(-2)=-10x$, the correct expansion is $15x^2-10x$."},{"text":"$$15x^2 + 10x$","correct":false,"explanation":"This loses the negative sign from $-2$. Multiplying positive $5x$ by negative 2 gives $-10x$, so the second term must remain negative."}]},{"id":"e16ac0bc-6834-455b-bcc9-3256eedbae1e","cardType":"basic","points":1,"front":"What is a common mistake when expanding a bracket like $8(4s - 3)$?","back":"Only multiplying the first term (e.g. writing $32s - 3$).","mcqOptions":[{"text":"Only multiplying the first term (e.g. writing $32s - 3$).","correct":true,"explanation":"The factor outside a bracket must multiply every term inside it. Multiplying only $4s$ leaves the constant undistributed; 8 must also multiply $-3$ to give $-24$."},{"text":"Multiplying both the first and second terms (e.g. writing $32s - 24$).","correct":false,"explanation":"Multiplying both bracketed terms is the required use of the distributive law, not a mistake. It correctly produces $8\\times4s=32s$ and $8\\times(-3)=-24$."},{"text":"Preserving the negative sign (e.g. writing $32s - 24$).","correct":false,"explanation":"Preserving the negative sign is correct because a positive factor times a negative term remains negative. The actual common mistake is failing to multiply the second term by 8."},{"text":"Calculating both products correctly (e.g. writing $32s - 24$).","correct":false,"explanation":"Calculating both products accurately is the correct expansion process. The common error is calculating only $8\\times4s$ and leaving the bracket’s second term unchanged."}]},{"id":"80fd29fd-6ef5-4876-b9e7-adc0d914b52d","cardType":"basic","points":3,"front":"Expand and simplify $6(g + 4) + 2(g - 7)$.","back":"$$8g + 10$","mcqOptions":[{"text":"$$8g + 10$","correct":true,"explanation":"Expanding gives $6g+24+2g-14$. Collecting like terms then produces $8g$, while the constants give $24-14=10$, so the result is $8g+10$."},{"text":"$$8g + 38$","correct":false,"explanation":"This treats the negative term in $2(g-7)$ as positive, producing $+14$ instead of $-14$. The constants should total $24-14=10$."},{"text":"$$8g + 17$","correct":false,"explanation":"This fails to multiply $-7$ by the outside factor 2, using $-7$ instead of $-14$. Correct distribution makes the constant total $24-14=10$."},{"text":"$$8g - 10$","correct":false,"explanation":"This reverses the subtraction when combining the constants. They total $24-14=10$, not $-10$, while the variable terms correctly combine to $8g$."}]},{"id":"60c6d171-bb29-4871-af60-f51b4db46a80","cardType":"basic","points":2,"front":"How does a minus sign before a bracket (e.g. $-2(5y + 1)$) affect the expansion?","back":"It changes the signs of every term inside the bracket when multiplied.","mcqOptions":[{"text":"It changes the signs of every term inside the bracket when multiplied.","correct":true,"explanation":"The negative multiplier applies to every term in the bracket. For example, $-2(5y + 1) = -10y - 2$, so both terms change sign."},{"text":"It changes the sign of only the first term inside the bracket when multiplied.","correct":false,"explanation":"Applying the minus sign only to the first term ignores the distributive law. The multiplier acts on every term, so $-2(5y + 1) = -10y - 2$."},{"text":"It keeps the signs of every term inside the bracket when multiplied.","correct":false,"explanation":"Ignoring the negative sign treats the multiplier as positive. Every term must be multiplied by $-2$, giving $-10y - 2$ rather than positive terms."},{"text":"It changes the operation between the terms inside the bracket when multiplied.","correct":false,"explanation":"The operation inside the bracket is not changed before multiplication. Instead, the negative multiplier is distributed to both terms, changing each resulting term’s sign."}]},{"id":"c97c317a-8b32-40b4-bec2-cd16990c58cf","cardType":"basic","points":3,"front":"Expand and simplify $4(3y - 5) - 2(5y + 1)$.","back":"$$2y - 22$","mcqOptions":[{"text":"$$2y - 22$","correct":true,"explanation":"Expanding gives $12y - 20 - 10y - 2$. Collecting like terms then gives $(12y - 10y) + (-20 - 2) = 2y - 22$."},{"text":"$$2y - 18$","correct":false,"explanation":"This comes from treating $-2 \\times 1$ as $+2$. The negative multiplier gives $-2$, so the constants total $-20 - 2 = -22$."},{"text":"$$22y - 22$","correct":false,"explanation":"This adds $12y$ and $10y$, overlooking that the second bracket is multiplied by $-2$. The variable terms are $12y - 10y = 2y$."},{"text":"$$2y - 12$","correct":false,"explanation":"The variable terms have been combined correctly, but the constants have not. Expanding produces $-20$ and $-2$, which combine to $-22$, giving $2y - 22$."}]},{"id":"7f7da1dc-3672-4cac-8134-006a5ec1fcad","cardType":"cloze","points":null,"front":"Factorising is the [ ] of expanding.","back":"Factorising is the opposite of expanding.","mcqOptions":[{"text":"opposite","correct":true,"explanation":"Factorising rewrites an expression as a product of factors, while expanding multiplies out those factors. The two processes therefore undo one another."},{"text":"result","correct":false,"explanation":"An expanded expression can be the result of expanding, but factorising itself is not that result. Factorising is the opposite process because it reconstructs a product of factors."},{"text":"purpose","correct":false,"explanation":"Factorising is not the purpose of expanding. It is the opposite operation: expanding removes brackets, whereas factorising rewrites an expression using brackets and common factors."},{"text":"continuation","correct":false,"explanation":"Factorising does not continue the process of expanding; it reverses it. Expanding multiplies out brackets, while factorising places an expression back into a product form."}]},{"id":"5e1bb099-fa9d-4443-93d7-02f7d125dd2e","cardType":"basic","points":2,"front":"What are the two requirements to factorise fully?","back":"

Take out the highest common factor (HCF) of the numbers
Take out the HCF of the variables

","mcqOptions":[{"text":"

Take out the highest common factor (HCF) of the numbers
Take out the HCF of the variables

","correct":true,"explanation":"A fully factorised expression has the greatest possible numerical factor and every common variable factor outside the bracket. No further common factor should remain inside."},{"text":"

Take out the lowest common multiple (LCM) of the numbers
Take out the HCF of the variables

","correct":false,"explanation":"The LCM is used to find common multiples, not common factors. Full factorisation requires the HCF of the numerical coefficients and the HCF of the variables."},{"text":"

Take out the highest common factor (HCF) of the numbers
Take out the highest power of each variable

","correct":false,"explanation":"Taking the highest variable power can introduce factors that are not present in every term. Use only the greatest power common to all terms, alongside the numerical HCF."},{"text":"

Take out the smallest coefficient from the numbers
Take out every variable in the expression

","correct":false,"explanation":"The smallest coefficient is not necessarily a factor of every coefficient, and not every variable is necessarily common. Full factorisation uses only the numerical and variable HCFs."}]},{"id":"af31cf12-3e75-48f6-bbbb-3dde89986019","cardType":"basic","points":2,"front":"Factorise $20y + 35$.","back":"$$5(4y + 7)$","mcqOptions":[{"text":"$$5(4y + 7)$","correct":true,"explanation":"The HCF of $20$ and $35$ is $5$. Dividing both terms by $5$ gives $4y$ and $7$, so $20y + 35 = 5(4y + 7)$."},{"text":"$$5(4y + 35)$","correct":false,"explanation":"Only the first term has been divided by the outside factor. Both terms must be divided by $5$, so $35 \\div 5 = 7$ and the answer is $5(4y + 7)$."},{"text":"$$10(2y + 7)$","correct":false,"explanation":"Although $20y \\div 10 = 2y$, $35$ is not divisible by $10$. Expanding this gives $20y + 70$, whereas the correct factorisation is $5(4y + 7)$."},{"text":"$$5(4y + 6)$","correct":false,"explanation":"This uses an incorrect division for the constant term: $35 \\div 5 = 7$, not $6$. Expanding the correct answer $5(4y + 7)$ returns $20y + 35$."}]},{"id":"6ea25999-6924-42a0-ae62-bbf28a63059c","cardType":"basic","points":2,"front":"What is the highest common factor of $12m^2n$ and $18mn^2$?","back":"$$6mn$","mcqOptions":[{"text":"$$6mn$","correct":true,"explanation":"The numerical HCF of $12$ and $18$ is $6$. The lowest powers common to both terms are $m^1$ and $n^1$, giving the HCF $6mn$."},{"text":"$$6m^2n$","correct":false,"explanation":"This incorrectly takes the higher power of $m$. Since $18mn^2$ contains only one factor of $m$, the common variable factor is $m$, making the HCF $6mn$."},{"text":"$$6mn^2$","correct":false,"explanation":"This incorrectly takes the higher power of $n$. Since $12m^2n$ contains only one factor of $n$, the greatest variable factor common to both terms is $mn$."},{"text":"$$3mn$","correct":false,"explanation":"This is a common factor, but it is not the highest one because both coefficients are divisible by $6$. Including the common variables gives the HCF $6mn$."}]},{"id":"25175dce-05e7-4085-9d14-a17e8aaf9ab9","cardType":"basic","points":2,"front":"Factorise fully $12m^2n + 18mn^2$.","back":"$$6mn(2m + 3n)$","mcqOptions":[{"text":"$$6mn(2m + 3n)$","correct":true,"explanation":"The HCF is $6mn$. Dividing $12m^2n$ by it gives $2m$, and dividing $18mn^2$ by it gives $3n$, producing $6mn(2m + 3n)$."},{"text":"$$3mn(4m + 6n)$","correct":false,"explanation":"This expands to the original expression, but it is not fully factorised because $2$ remains common inside the bracket. Taking it out gives $6mn(2m + 3n)$."},{"text":"$$6m(2mn + 3n^2)$","correct":false,"explanation":"This is equivalent to the original expression, but it leaves a common factor of $n$ inside the bracket. Taking out the full HCF gives $6mn(2m + 3n)$."},{"text":"$$6mn(2m + 2n)$","correct":false,"explanation":"The second term has been divided incorrectly: $18mn^2 \\div 6mn = 3n$, not $2n$. The fully factorised expression is therefore $6mn(2m + 3n)$."}]},{"id":"667424e0-e5e3-4bba-926f-2991827623eb","cardType":"basic","points":1,"front":"How can you check if your factorisation is correct?","back":"Expand the brackets; it should return to the original expression.","mcqOptions":[{"text":"Expand the brackets; it should return to the original expression.","correct":true,"explanation":"Factorising and expanding are opposite processes. Multiplying the factors back out should reproduce every term and sign in the original expression exactly."},{"text":"Substitute one value; matching results prove the expressions are identical.","correct":false,"explanation":"Two different expressions can produce the same result for one chosen value, so this does not prove equivalence. Expanding the brackets must reproduce the original expression term by term."},{"text":"Factorise the brackets again; they should contain no numerical terms.","correct":false,"explanation":"Factorising again does not directly verify the multiplication, and correctly factorised brackets may contain numbers. The reliable check is to expand and compare with the original expression."},{"text":"Count the terms; there should be the same number in both expressions.","correct":false,"explanation":"Factorisation usually changes the visible number of terms, so term counting cannot establish equivalence. Expanding the factors should instead return exactly to the original expression."}]},{"id":"b190456e-d164-40cd-9f81-9255145ee98d","cardType":"basic","points":1,"front":"What is the most efficient first step to solve $5(b + 4) = 45$?","back":"Divide both sides by 5.","mcqOptions":[{"text":"Divide both sides by 5.","correct":true,"explanation":"Dividing by 5 immediately removes the outer multiplication, giving $b + 4 = 9$. Subtracting 4 then gives $b = 5$."},{"text":"Subtract 4 from both sides.","correct":false,"explanation":"Subtracting 4 does not undo the addition inside the brackets because the entire bracket is multiplied by 5. Divide both sides by 5 first."},{"text":"Divide both sides by 45.","correct":false,"explanation":"Dividing by 45 does not cancel the coefficient outside the brackets. The efficient first step is to divide both sides by 5, producing $b + 4 = 9$."},{"text":"Multiply both sides by 5.","correct":false,"explanation":"Multiplying by 5 increases the coefficient rather than cancelling it. Division by 5 is the inverse operation needed to isolate the bracket efficiently."}]},{"id":"97f834ee-37e9-4f0b-a9ca-e1552643afad","cardType":"basic","points":2,"front":"Solve $5(b + 4) = 45$.","back":"$$b = 5$","mcqOptions":[{"text":"$$b = 5$","correct":true,"explanation":"Dividing both sides by 5 gives $b + 4 = 9$. Subtracting 4 from both sides then gives the solution $b = 5$."},{"text":"$$b = 13$","correct":false,"explanation":"This results from adding 4 after obtaining $b + 4 = 9$. The correct inverse operation is subtraction, so $b = 9 - 4 = 5$."},{"text":"$$b = 8.2$","correct":false,"explanation":"This treats $5(b + 4)$ as $5b + 4$, failing to multiply 4 by 5. Dividing first gives $b + 4 = 9$, so $b = 5$."},{"text":"$$b = 41$","correct":false,"explanation":"This subtracts 4 from 45 but ignores that the whole bracket is multiplied by 5. First divide by 5, then subtract 4 to obtain $b = 5$."}]},{"id":"10b7e02c-70db-4bfc-ad1c-b84ea8551e8a","cardType":"basic","points":2,"front":"Solve $4(c - 3) = 18$.","back":"$$c = 7.5$","mcqOptions":[{"text":"$$c = 7.5$","correct":true,"explanation":"Dividing both sides by $4$ gives $c - 3 = 4.5$. Adding $3$ to both sides then gives $c = 7.5$."},{"text":"$$c = 1.5$","correct":false,"explanation":"Subtracting $3$ after dividing reverses the required operation. From $c - 3 = 4.5$, add $3$, rather than subtracting it, to obtain $c = 7.5$."},{"text":"$$c = 5.25$","correct":false,"explanation":"This results from treating $4(c - 3)$ as $4c - 3$. Correct expansion gives $4c - 12 = 18$, so $4c = 30$ and $c = 7.5$."},{"text":"$$c = 18$","correct":false,"explanation":"This incorrectly treats the bracket as though $4(c - 3)$ meant $(4 - 3)c$. Dividing by $4$ first and then adding $3$ gives $c = 7.5$."}]},{"id":"a8b229e3-5680-4882-970f-7cad726e1c67","cardType":"cloze","points":null,"front":"In algebra, a [ ] is a letter that represents a number, while a term is a part of an expression separated by plus or minus signs.","back":"In algebra, a variable is a letter that represents a number, while a term is a part of an expression separated by plus or minus signs.","mcqOptions":[{"text":"variable","correct":true,"explanation":"A variable is a letter or symbol used to represent a number whose value may be unknown or may change within an algebraic expression."},{"text":"constant","correct":false,"explanation":"A constant is a fixed numerical value, not a letter representing a number that may vary. The defined algebraic object is a variable."},{"text":"coefficient","correct":false,"explanation":"A coefficient is the numerical factor multiplying a variable, such as 5 in $5x$. A letter representing a number is called a variable."},{"text":"expression","correct":false,"explanation":"An expression is a mathematical combination of terms and operations, rather than a single letter representing a number. Such a letter is called a variable."}]},{"id":"0b94316d-a9dd-4c04-bfd4-deb86d33baa9","cardType":"cloze","points":null,"front":"In algebra, a variable is a letter that represents a number, while a [ ] is a part of an expression separated by plus or minus signs.","back":"In algebra, a variable is a letter that represents a number, while a term is a part of an expression separated by plus or minus signs.","mcqOptions":[{"text":"term","correct":true,"explanation":"A term is one part of an algebraic expression. Terms are separated from one another by addition or subtraction signs, including their associated signs."},{"text":"expression","correct":false,"explanation":"An expression consists of one or more terms linked by operations, so it is not an individual part separated by plus or minus signs. That part is a term."},{"text":"equation","correct":false,"explanation":"An equation states that two expressions are equal and therefore contains an equals sign. A part of an expression separated by plus or minus signs is a term."},{"text":"factor","correct":false,"explanation":"A factor is a quantity multiplied by another quantity, not a part identified by addition or subtraction boundaries. Such a part of an expression is a term."}]}]}],"examCountdown":null,"heading":"Expanding and Factorising","title":"Expanding and Factorising Flashcards","topbar":{"backHref":"/igcse/maths/oxford-aqa/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"b2d1991d-eb04-4af5-8bef-94a692a17001","walled":"$L39"}],"$L3a","$L3b"]