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sign","correct":false,"explanation":"An equals sign would turn the expression into an equation, such as $4x+7=15$. The given expression has no equals sign."},{"text":"an inequality sign","correct":false,"explanation":"An inequality sign would create an inequality, such as $4x+7>15$. The given algebraic expression contains no equals sign."},{"text":"a fixed solution","correct":false,"explanation":"A fixed solution belongs to an equation or inequality after a variable is solved. The expression $4x+7$ has no equals sign."}]},{"id":"57a25cd2-0b58-d313-f140-52704c21c004","cardType":"cloze","points":null,"front":"Like terms have the same [ ].","back":"Like terms have the same variable part.","mcqOptions":[{"text":"variable part","correct":true,"explanation":"Terms are like terms when their variables and corresponding powers match exactly, so they have the same variable part."},{"text":"coefficient","correct":false,"explanation":"Like terms may have different coefficients: for example, $3x$ and $8x$ are like terms. They must instead have the same variable part."},{"text":"constant term","correct":false,"explanation":"A constant term has no variable, whereas many like terms do. Like terms are identified by having the same variable part."},{"text":"numerical value","correct":false,"explanation":"The numerical values of like terms can differ because their coefficients can differ. Their variable part, including each exponent, must be the same."}]},{"id":"12193e4b-3054-1e33-5235-3d6a08b963a8","cardType":"basic","points":1,"front":"How do you collect like terms?","back":"Add or subtract their coefficients.","mcqOptions":[{"text":"Add or subtract their coefficients.","correct":true,"explanation":"Like terms share an identical variable part, so their coefficients can be added or subtracted while that variable part remains unchanged."},{"text":"Add or subtract their variables.","correct":false,"explanation":"The variables are not added or subtracted separately; they remain unchanged. Collect like terms by adding or subtracting their coefficients."},{"text":"Multiply or divide their coefficients.","correct":false,"explanation":"Multiplying coefficients finds a product rather than collecting a sum or difference. Like terms are collected by adding or subtracting their coefficients."},{"text":"Add or subtract their exponents.","correct":false,"explanation":"Exponents remain unchanged when like terms are collected. The correct method is to add or subtract the coefficients of terms with matching variable parts."}]},{"id":"deec6cb4-d35b-f037-b435-fae0da116462","cardType":"basic","points":1,"front":"Simplify $8x+5x-4$.","back":"$13x-4$","mcqOptions":[{"text":"$13x-4$","correct":true,"explanation":"The like terms $8x$ and $5x$ have coefficients that add to $13$, while the constant $-4$ remains unchanged."},{"text":"$3x-4$","correct":false,"explanation":"Subtracting the coefficients instead of adding them gives $3x$. Since both variable terms are positive, $8x+5x=13x$, giving $13x-4$."},{"text":"$9x$","correct":false,"explanation":"The constant $-4$ is not a like term with the $x$-terms, so it cannot be added to their coefficient. The result is $13x-4$."},{"text":"$13x+4$","correct":false,"explanation":"Treating $-4$ as positive changes its sign. Collecting only $8x$ and $5x$ gives $13x$, while the constant remains $-4$."}]},{"id":"9a138a31-9416-c202-6d59-7b4c07964ebc","cardType":"cloze","points":null,"front":"To expand one bracket, multiply the outside factor by [ ].","back":"To expand one bracket, multiply the outside factor by every term inside.","mcqOptions":[{"text":"every term inside","correct":true,"explanation":"The distributive law requires the factor outside a bracket to multiply every term inside, including terms separated by addition or subtraction."},{"text":"only the first term","correct":false,"explanation":"Multiplying only the first term leaves the bracket only partly expanded. The outside factor must multiply every term inside the bracket."},{"text":"only the final term","correct":false,"explanation":"Multiplying only the final term omits the other products required by the distributive law. The outside factor must multiply every term inside."},{"text":"every term outside","correct":false,"explanation":"Terms outside the bracket are not the contents being distributed over. To expand one bracket, the outside factor multiplies every term inside."}]},{"id":"496c5aa3-9f24-9b6d-1156-9f0b6aae2d34","cardType":"basic","points":1,"front":"Expand $6(3x+4)$.","back":"$18x+24$","mcqOptions":[{"text":"$18x+24$","correct":true,"explanation":"Distributing $6$ across both terms gives $6\\times3x=18x$ and $6\\times4=24$, so the expanded expression is $18x+24$."},{"text":"$18x+4$","correct":false,"explanation":"The constant $4$ has been left unchanged instead of being multiplied by $6$. Distributing to both terms gives $18x+24$."},{"text":"$9x+10$","correct":false,"explanation":"Adding $6$ to each term is not how brackets are expanded. Multiplying both terms by $6$ gives $18x+24$."},{"text":"$18x+10$","correct":false,"explanation":"Multiplying $3x$ by $6$ correctly gives $18x$, but $6\\times4$ is $24$, not $10$. The correct expansion is $18x+24$."}]},{"id":"74e8fd92-56be-763b-3b21-6f3f8627a23e","cardType":"basic","points":1,"front":"What multiplication is missed in the incorrect expansion $6(3x+4)=18x+4$?","back":"Multiplying $6$ by $4$.","mcqOptions":[{"text":"Multiplying $6$ by $4$.","correct":true,"explanation":"The $6$ was distributed to $3x$ but not to the constant $4$. The missing product is $6\\times4=24$."},{"text":"Multiplying $6$ by $3x$.","correct":false,"explanation":"The product $6\\times3x$ was included correctly as $18x$. The omitted multiplication is $6$ by $4$, which should produce $24$."},{"text":"Multiplying $3x$ by $4$.","correct":false,"explanation":"The two terms inside the bracket are added, not multiplied together. Expansion requires multiplying the outside $6$ by $4$ as well as by $3x$."},{"text":"Multiplying $18x$ by $4$.","correct":false,"explanation":"The term $18x$ is already the result of multiplying $6$ by $3x$ and should not then multiply $4$. The missing product is $6\\times4$."}]},{"id":"91e53ec6-5531-67a4-80ea-6efe391f1e6c","cardType":"basic","points":2,"front":"Expand $3p(p-5)$.","back":"$3p^2-15p$","mcqOptions":[{"text":"$3p^2-15p$","correct":true,"explanation":"Distributing $3p$ gives $3p\\times p=3p^2$ and $3p\\times(-5)=-15p$, producing $3p^2-15p$."},{"text":"$3p^2-15$","correct":false,"explanation":"The outside factor is $3p$, so multiplying it by $-5$ gives $-15p$, not $-15$. The correct expansion is $3p^2-15p$."},{"text":"$3p-15p$","correct":false,"explanation":"Multiplying $3p$ by $p$ gives $3p^2$, not $3p$. The second product is $-15p$, so the correct result is $3p^2-15p$."},{"text":"$3p^2+15p$","correct":false,"explanation":"The negative sign in $p-5$ must be retained when multiplying by $3p$. The second product is $-15p$, giving $3p^2-15p$."}]},{"id":"acdc3ce3-a385-5bed-2247-54f8782487a4","cardType":"cloze","points":null,"front":"Factorising is the reverse of [ ].","back":"Factorising is the reverse of expanding.","mcqOptions":[{"text":"expanding","correct":true,"explanation":"Expanding removes brackets by multiplying terms, while factorising reverses this process by rewriting an expression as a product containing brackets."},{"text":"simplifying","correct":false,"explanation":"Simplifying combines or reduces parts of an expression but is not the reverse process. Factorising is the reverse of expanding because it introduces factors and brackets."},{"text":"solving","correct":false,"explanation":"Solving finds values that satisfy an equation, rather than rewriting an expression as a product. Factorising is specifically the reverse of expanding."},{"text":"substituting","correct":false,"explanation":"Substituting replaces variables with known values and does not reverse bracket multiplication. Factorising reverses expanding by placing an expression into a product form."}]},{"id":"adaf43a1-3a5c-8eb7-015a-00fb1c661012","cardType":"basic","points":1,"front":"What HCF should be taken out first when factorising $8r+20$?","back":"$4$","mcqOptions":[{"text":"$4$","correct":true,"explanation":"The highest number dividing both $8$ and $20$ is $4$. There are no variables common to both terms, so the HCF is $4$."},{"text":"$2$","correct":false,"explanation":"Although $2$ divides both coefficients, it is not their highest common factor. The greatest number dividing both $8$ and $20$ is $4$."},{"text":"$8$","correct":false,"explanation":"Using the first coefficient alone overlooks whether it divides the other coefficient. Since $8$ does not divide $20$, the HCF is $4$."},{"text":"$20$","correct":false,"explanation":"Using the constant term alone overlooks whether it divides the other coefficient. Since $20$ does not divide $8$, the HCF is $4$."}]},{"id":"4322b5da-fd68-908e-3358-d6e2d28067ef","cardType":"basic","points":1,"front":"Factorise $8r+20$.","back":"$4(2r+5)$","mcqOptions":[{"text":"$4(2r+5)$","correct":true,"explanation":"Dividing both terms by their HCF, $4$, gives $8r ÷ 4=2r$ and $20 ÷ 4=5$, producing the stated factorisation."},{"text":"$4(2r+4)$","correct":false,"explanation":"The constant was incorrectly divided, since expanding this expression gives $8r+16$. Dividing $20$ by $4$ gives $5$, so the correct result is $4(2r+5)$."},{"text":"$4(4r+5)$","correct":false,"explanation":"The coefficient of $r$ was not divided by the outside factor, so expansion gives $16r+20$. The correct factorisation is $4(2r+5)$."},{"text":"$8(r+5)$","correct":false,"explanation":"The constant inside the bracket was not divided by $8$, so expansion gives $8r+40$. Taking out the HCF $4$ gives $4(2r+5)$."}]},{"id":"4cfd637e-cf4a-436c-8297-2da09f1f0dc9","cardType":"cloze","points":null,"front":"To check a factorised answer, [ ] it.","back":"To check a factorised answer, expand it.","mcqOptions":[{"text":"expand","correct":true,"explanation":"Multiplying the outside factor by every term inside the brackets reproduces the original expression, so expanding is the appropriate way to check a factorisation."},{"text":"factorise","correct":false,"explanation":"Factorising again does not directly verify that the brackets reproduce the original expression. The answer should be expanded and compared with the expression given."},{"text":"simplify","correct":false,"explanation":"Simplifying may tidy an expression, but it does not specifically check the multiplication represented by a factorised form. Expand the brackets to recover the original expression."},{"text":"solve","correct":false,"explanation":"Solving applies to an equation with an unknown value, not to checking an equivalent expression. Expand the factorised form and compare it with the original."}]},{"id":"c260a2cd-2b40-eae5-7358-4609e8cfba5f","cardType":"basic","points":1,"front":"What full common factor should be taken out of $14a^2b+21ab^2$?","back":"$7ab$","mcqOptions":[{"text":"$7ab$","correct":true,"explanation":"The coefficient HCF is $7$, while both terms contain at least one $a$ and one $b$. Therefore, the full common factor is $7ab$."},{"text":"$7a$","correct":false,"explanation":"This includes the coefficient HCF and the common $a$, but misses the common factor $b$. The full common factor is therefore $7ab$."},{"text":"$7b$","correct":false,"explanation":"This includes the coefficient HCF and the common $b$, but misses the common factor $a$. Both variables are shared, so the full common factor is $7ab$."},{"text":"$14ab$","correct":false,"explanation":"The variable part is common, but $14$ is not a factor of $21$. The coefficient HCF is $7$, making the full common factor $7ab$."}]},{"id":"18f1ec92-8495-d4b7-9956-fcb9f305b966","cardType":"basic","points":2,"front":"Fully factorise $14a^2b+21ab^2$.","back":"$7ab(2a+3b)$","mcqOptions":[{"text":"$7ab(2a+3b)$","correct":true,"explanation":"Dividing each term by $7ab$ gives $2a$ and $3b$. Expanding the result reproduces $14a^2b+21ab^2$, confirming the full factorisation."},{"text":"$7ab(2a+3ab)$","correct":false,"explanation":"An extra $a$ has been left in the second bracket term, so expansion gives $21a^2b^2$ instead of $21ab^2$. The correct result is $7ab(2a+3b)$."},{"text":"$7ab(2a^2+3b)$","correct":false,"explanation":"The first term retains one too many factors of $a$, producing $14a^3b$ when expanded. Dividing $14a^2b$ by $7ab$ gives $2a$."},{"text":"$7ab(2a+3b^2)$","correct":false,"explanation":"The second term retains one too many factors of $b$, producing $21ab^3$ when expanded. Dividing $21ab^2$ by $7ab$ gives $3b$."}]},{"id":"6e57d4a5-f7c4-8dbb-db43-52516f9c06b1","cardType":"basic","points":2,"front":"Why is $7(2a^2b+3ab^2)$ not fully factorised?","back":"Both $a$ and $b$ can still be taken out.","mcqOptions":[{"text":"Both $a$ and $b$ can still be taken out.","correct":true,"explanation":"Each term inside the brackets contains both $a$ and $b$, so these variables remain common factors. Taking out $7ab$ gives the fully factorised form."},{"text":"Both $2$ and $3$ can still be taken out.","correct":false,"explanation":"The coefficients $2$ and $3$ have an HCF of $1$, so neither can be taken from both terms. The remaining common factors are $a$ and $b$."},{"text":"Both $a^2$ and $b^2$ can still be taken out.","correct":false,"explanation":"Each term contains only one guaranteed common factor of each variable, not their squares. Both $a$ and $b$ can still be taken out once."},{"text":"Both $14$ and $21$ can still be taken out.","correct":false,"explanation":"Neither original coefficient divides both terms: the numerical HCF is only $7$. The incomplete factorisation instead leaves both $a$ and $b$ common inside the brackets."}]},{"id":"f4610c6d-7603-4fed-ad68-4bceae1709c7","cardType":"cloze","points":null,"front":"After expanding brackets, simplify by collecting [ ].","back":"After expanding brackets, simplify by collecting like terms.","mcqOptions":[{"text":"like terms","correct":true,"explanation":"Terms with identical variable parts and powers can be combined by adding or subtracting their coefficients, so like terms should be collected after expansion."},{"text":"unlike terms","correct":false,"explanation":"Terms with different variable parts or powers cannot be combined by addition or subtraction. After expanding, only like terms should be collected."},{"text":"common factors","correct":false,"explanation":"Common factors are taken out during factorising rather than collected after expansion. The simplification step combines like terms that have matching variables and powers."},{"text":"numerical terms","correct":false,"explanation":"Limiting collection to constants would leave matching algebraic terms unsimplified. All like terms, including terms with identical variable parts, should be collected."}]},{"id":"f73834fe-e9ae-b78e-8af3-9ae1d6bb91a7","cardType":"basic","points":2,"front":"Expand and simplify $5(s+3)+2(s-4)$.","back":"$7s+7$","mcqOptions":[{"text":"$7s+7$","correct":true,"explanation":"Expanding gives $5s+15+2s-8$. Collecting like terms produces $7s$, while the constants combine to give $15-8=7$."},{"text":"$7s+23$","correct":false,"explanation":"Adding $8$ instead of subtracting it ignores the negative term from $2(s-4)$. The constants are $15-8=7$, so the correct result is $7s+7$."},{"text":"$7s-7$","correct":false,"explanation":"The constant terms have been combined in the wrong order. Expanding gives constants of $15$ and $-8$, which total $7$, so the correct result is $7s+7$."},{"text":"$10s+7$","correct":false,"explanation":"The coefficients should be added after expansion, not multiplied. Since $5s+2s=7s$ and $15-8=7$, the correct simplified expression is $7s+7$."}]},{"id":"f5149a2a-dc79-6a84-ea3a-a1aea9cda9ce","cardType":"basic","points":1,"front":"Expand $-3(y+5)$.","back":"$-3y-15$","mcqOptions":[{"text":"$-3y-15$","correct":true,"explanation":"The multiplier $-3$ applies to both terms inside the bracket. Therefore $-3\\times y=-3y$ and $-3\\times5=-15$, giving $-3y-15$."},{"text":"$-3y+15$","correct":false,"explanation":"A negative multiplied by a positive remains negative, so the constant term cannot be $+15$. Distributing $-3$ correctly gives $-3y-15$."},{"text":"$3y-15$","correct":false,"explanation":"The negative multiplier must also make the $y$ term negative. Multiplying both terms in the bracket by $-3$ gives the correct expansion $-3y-15$."},{"text":"$-3y-5$","correct":false,"explanation":"The $5$ has not been multiplied by $-3$. Every term inside the bracket must be multiplied, so the constant becomes $-15$ and the answer is $-3y-15$."}]},{"id":"c1817931-8e1b-a75c-e2f0-b508a4bb5a32","cardType":"basic","points":2,"front":"Expand and simplify $4(2y-3)-3(y+5)$.","back":"$5y-27$","mcqOptions":[{"text":"$5y-27$","correct":true,"explanation":"Expanding gives $8y-12-3y-15$. Collecting the variable terms and constants separately gives $8y-3y=5y$ and $-12-15=-27$."},{"text":"$5y+3$","correct":false,"explanation":"Treating $-3\\times5$ as positive loses the negative multiplier outside the second bracket. It produces $-15$, so the correct simplified result is $5y-27$."},{"text":"$11y-27$","correct":false,"explanation":"The term $-3y$ must be subtracted from $8y$, not added to it. Thus the coefficient is $8-3=5$, giving $5y-27$."},{"text":"$5y-17$","correct":false,"explanation":"The constant $5$ has effectively been subtracted without being multiplied by $3$. The second bracket contributes $-15$, so the constants total $-27$ and the answer is $5y-27$."}]},{"id":"e5489a9d-0708-b8ca-21d8-215d08d1908a","cardType":"basic","points":1,"front":"If a whole bracket is multiplied by a number in an equation, what is usually quicker to do first?","back":"Divide both sides by the outside number.","mcqOptions":[{"text":"Divide both sides by the outside number.","correct":true,"explanation":"Dividing both sides by the outside multiplier preserves the equation and removes that multiplier immediately, leaving the bracketed expression ready to solve."},{"text":"Subtract the outside number from both sides.","correct":false,"explanation":"Subtracting the multiplier does not undo multiplication. The inverse operation is division, so both sides should be divided by the outside number first."},{"text":"Divide one side by the outside number.","correct":false,"explanation":"Changing only one side breaks the equality of the equation. To cancel the multiplier correctly, divide both sides by the outside number."},{"text":"Multiply both sides by the outside number.","correct":false,"explanation":"Multiplying again increases the coefficient instead of cancelling it. Division is the inverse of multiplication, so divide both sides by the outside number."}]},{"id":"d1cd3d1f-7f20-60c4-a8a3-a5840c65374f","cardType":"basic","points":2,"front":"Solve $6(d+2)=48$.","back":"$d=6$","mcqOptions":[{"text":"$d=6$","correct":true,"explanation":"Dividing both sides by $6$ gives $d+2=8$. Subtracting $2$ from both sides then gives the solution $d=6$."},{"text":"$d=8$","correct":false,"explanation":"This stops after dividing $48$ by $6$, leaving $d+2=8$ rather than isolating $d$. Subtracting $2$ gives the correct solution $d=6$."},{"text":"$d=10$","correct":false,"explanation":"Adding $2$ reverses the required final operation. After division gives $d+2=8$, subtracting $2$, rather than adding it, produces $d=6$."},{"text":"$d=46$","correct":false,"explanation":"Subtracting $2$ directly from $48$ ignores that the whole bracket is multiplied by $6$. Divide by $6$ first, then subtract $2$ to obtain $d=6$."}]},{"id":"d35e1722-9f12-1ebc-7e1f-f3f20aeddeae","cardType":"basic","points":2,"front":"Solve $2(v-6)=15$.","back":"$v=13.5$","mcqOptions":[{"text":"$v=13.5$","correct":true,"explanation":"Dividing both sides by $2$ gives $v-6=7.5$. Adding $6$ to both sides then produces the solution $v=13.5$."},{"text":"$v=1.5$","correct":false,"explanation":"Subtracting $6$ after dividing uses the wrong inverse operation. Since the equation becomes $v-6=7.5$, adding $6$ gives the correct answer $v=13.5$."},{"text":"$v=10.5$","correct":false,"explanation":"This comes from expanding as $2v-6$ instead of $2v-12$. The multiplier applies to both bracket terms, and solving correctly gives $v=13.5$."},{"text":"$v=21$","correct":false,"explanation":"Adding $6$ to $15$ but failing to divide by the outside multiplier does not isolate the bracket. Divide by $2$ first, then add $6$ to get $v=13.5$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Expanding and Factorising","title":"Expanding and Factorising Flashcards","topbar":{"backHref":"/gcse/maths/aqa/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"afce9c8b-9520-447a-8de8-45cce756bb7d","walled":"$L39"}],"$L3a","$L3b"]