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same expression. Therefore, only factors multiplying the whole numerator and denominator can be cancelled."},{"text":"Common terms (expressions that are added or subtracted in the numerator/denominator).","correct":false,"explanation":"Terms within a sum or difference cannot be cancelled individually. The numerator and denominator must first be factorised, because only common factors multiplying the whole expressions can cancel."},{"text":"Common variables (letters that appear anywhere in both numerator/denominator).","correct":false,"explanation":"A shared variable cannot be cancelled merely because it appears in both expressions. It must be a factor multiplying the whole numerator and denominator."},{"text":"Common coefficients (numbers attached to terms in both numerator/denominator).","correct":false,"explanation":"Coefficients attached to separate terms cannot automatically be cancelled. Only a common factor of the entire numerator and entire denominator may be divided out."}]},{"id":"7aecf411-4407-4d2a-94ed-74a4467c9915","cardType":"cloze","points":null,"front":"In algebra, a [ ] is something being multiplied, whereas a term is part of a sum or difference.","back":"In algebra, a factor is something being multiplied, whereas a term is part of a sum or difference.","mcqOptions":[{"text":"factor","correct":true,"explanation":"A factor is an expression or quantity involved in multiplication. For example, $x$ and $x+2$ are the two factors in $x(x+2)$."},{"text":"term","correct":false,"explanation":"A term is a component separated by addition or subtraction, rather than something necessarily being multiplied. The required word is factor."},{"text":"coefficient","correct":false,"explanation":"A coefficient is the numerical multiplier of a variable, such as $3$ in $3x$. The general name for something being multiplied is a factor."},{"text":"variable","correct":false,"explanation":"A variable is a letter representing a value, such as $x$. Something involved in multiplication is called a factor, whether or not it contains a variable."}]},{"id":"6b5b0862-1aaf-4317-8725-9e8b04c58de6","cardType":"cloze","points":null,"front":"In algebra, a factor is something being multiplied, whereas a [ ] is part of a sum or difference.","back":"In algebra, a factor is something being multiplied, whereas a term is part of a sum or difference.","mcqOptions":[{"text":"term","correct":true,"explanation":"A term is one part of an expression separated from other parts by addition or subtraction, such as $3x$ and $4$ in $3x+4$."},{"text":"factor","correct":false,"explanation":"A factor is something being multiplied, not a part separated by addition or subtraction. A component of a sum or difference is called a term."},{"text":"coefficient","correct":false,"explanation":"A coefficient is the number multiplying a variable within a term, such as $3$ in $3x$. The whole component of a sum is a term."},{"text":"expression","correct":false,"explanation":"An expression can contain several terms and operations, so it is not specifically one part of a sum or difference. That individual part is called a term."}]},{"id":"eeb5c1f1-8295-4288-a489-92f80cfc6164","cardType":"basic","points":2,"front":"Why is it incorrect to cancel the $x$ in the expression $\\frac{x+4}{x^2-16}$?","back":"The $x$ is a term in a sum, not a factor multiplying the whole numerator.","mcqOptions":[{"text":"The $x$ is a term in a sum, not a factor multiplying the whole numerator.","correct":true,"explanation":"In $x+4$, the $x$ is joined to $4$ by addition and does not multiply the whole numerator. Only common factors of complete expressions may cancel."},{"text":"The $x$ is a factor in the numerator, but does not appear anywhere in the denominator.","correct":false,"explanation":"The denominator does contain $x$, but mere appearance in both parts is insufficient. In the numerator, $x$ is a term in the sum $x+4$, not a factor of the whole numerator."},{"text":"The $x$ can only be cancelled when it has a coefficient of $1$ in both parts.","correct":false,"explanation":"Cancellation does not depend on coefficients being $1$. It fails here because $x$ is a term in $x+4$, rather than a factor multiplying the entire numerator."},{"text":"The $x$ is part of a quadratic, and quadratic expressions can never have factors cancelled.","correct":false,"explanation":"Quadratic expressions can have factors cancelled after factorisation. The problem here is that $x$ is a term in the numerator’s sum, not a factor of the complete numerator."}]},{"id":"3dff86be-c5c4-41cf-9ccf-c7d565eb6c27","cardType":"cloze","points":null,"front":"To simplify an algebraic fraction, you must [ ] the numerator and denominator before cancelling common factors.","back":"To simplify an algebraic fraction, you must factorise the numerator and denominator before cancelling common factors.","mcqOptions":[{"text":"factorise","correct":true,"explanation":"Factorising rewrites sums or differences as products, revealing factors shared by the numerator and denominator. Those common factors can then be cancelled."},{"text":"expand","correct":false,"explanation":"Expanding turns products into sums and can hide common factors rather than reveal them. The numerator and denominator should be factorised before cancellation."},{"text":"calculate","correct":false,"explanation":"An algebraic fraction containing variables usually cannot simply be calculated to one number. It must be factorised to identify common factors that can cancel."},{"text":"rearrange","correct":false,"explanation":"Rearranging terms does not convert a sum into a product or expose cancellable factors. The required process is to factorise the numerator and denominator."}]},{"id":"180af324-a6a0-4d84-982e-b4eeb58b8126","cardType":"cloze","points":null,"front":"To simplify an algebraic fraction, you must factorise the numerator and denominator before [ ] common factors.","back":"To simplify an algebraic fraction, you must factorise the numerator and denominator before cancelling common factors.","mcqOptions":[{"text":"cancelling","correct":true,"explanation":"Once both expressions are factorised, any identical factor multiplying the whole numerator and denominator can be cancelled by dividing both by that factor."},{"text":"expanding","correct":false,"explanation":"Expanding common factors would turn products back into sums and prevent direct cancellation. After factorising, the next step is cancelling identical factors."},{"text":"adding","correct":false,"explanation":"Common factors are not combined by addition. They are cancelled because dividing both the numerator and denominator by the same non-zero factor preserves the fraction’s value."},{"text":"subtracting","correct":false,"explanation":"Subtracting common factors would change the value of the fraction. After factorisation, shared factors are cancelled by division from the numerator and denominator."}]},{"id":"54ece75e-7cb6-4156-b5d8-a6291af8f86c","cardType":"basic","points":1,"front":"What is the general factorisation for a difference of two squares $a^2 - b^2$?","back":"$$(a - b)(a + b)$","mcqOptions":[{"text":"$$(a - b)(a + b)$","correct":true,"explanation":"Multiplying the conjugate factors gives $a^2+ab-ab-b^2=a^2-b^2$, because the two middle terms cancel."},{"text":"$$(a - b)^2$","correct":false,"explanation":"Squaring a difference produces $a^2-2ab+b^2$, which includes a middle term and has a positive $b^2$. The correct factorisation is $(a-b)(a+b)$."},{"text":"$$(a + b)^2$","correct":false,"explanation":"Squaring a sum produces $a^2+2ab+b^2$, not a difference of two squares. The required conjugate factors are $(a-b)(a+b)$."},{"text":"$$(a + b)(b - a)$","correct":false,"explanation":"Reversing the subtraction in the second factor changes the sign: this product equals $b^2-a^2$. The correct order is $(a-b)(a+b)$."}]},{"id":"8826b692-695a-4720-849a-ba4bd9ee23a0","cardType":"basic","points":1,"front":"Factorise the denominator of $\\frac{4x^2+16x}{x^2-16}$ using the difference of two squares.","back":"$$(x - 4)(x + 4)$","mcqOptions":[{"text":"$$(x - 4)(x + 4)$","correct":true,"explanation":"The denominator is $x^2-4^2$, a difference of two squares. Applying $a^2-b^2=(a-b)(a+b)$ gives $(x-4)(x+4)$."},{"text":"$$(x - 4)^2$","correct":false,"explanation":"This treats the expression as a squared difference and expands to $x^2-8x+16$. The difference of squares $x^2-16$ factorises as $(x-4)(x+4)$."},{"text":"$$(x + 4)^2$","correct":false,"explanation":"This treats the expression as a squared sum and expands to $x^2+8x+16$. Using conjugate factors instead gives $(x-4)(x+4)$."},{"text":"$$(x - 8)(x + 2)$","correct":false,"explanation":"Although the constants multiply to $-16$, these factors produce an unwanted $-6x$ term. Opposite constants cancel the middle terms, giving $(x-4)(x+4)$."}]},{"id":"70cdc45e-bc82-4ca2-be4c-a327f9bc4df0","cardType":"cloze","points":null,"front":"The [ ] of a fraction is the fraction turned upside down, such as $\\frac{x-3}{x+1}$ for $\\frac{x+1}{x-3}$.","back":"The reciprocal of a fraction is the fraction turned upside down, such as $\\frac{x-3}{x+1}$ for $\\frac{x+1}{x-3}$.","mcqOptions":[{"text":"reciprocal","correct":true,"explanation":"The reciprocal is formed by interchanging a fraction’s numerator and denominator, so the reciprocal of $\\frac{x+1}{x-3}$ is $\\frac{x-3}{x+1}$."},{"text":"denominator","correct":false,"explanation":"The denominator is only the expression below the fraction bar, not the result of turning a fraction upside down. The required term is reciprocal."},{"text":"numerator","correct":false,"explanation":"The numerator is the expression above the fraction bar and does not describe swapping the two parts. A fraction turned upside down is its reciprocal."},{"text":"quotient","correct":false,"explanation":"A quotient is the result of division, rather than a fraction with its numerator and denominator interchanged. The correct mathematical term is reciprocal."}]},{"id":"b528e1f1-79f6-4878-b1c7-47ce0a4c440f","cardType":"basic","points":3,"front":"Outline the 3-step process for dividing algebraic fractions.","back":"

Keep the first fraction
Change division to multiplication
Flip the second fraction (reciprocal)

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

Keep the first fraction
Change division to multiplication
Flip the second fraction (reciprocal)

","correct":true,"explanation":"Dividing by a fraction is equivalent to multiplying by its reciprocal. Therefore, keep the first fraction, replace division with multiplication, and flip only the second fraction."},{"text":"

Flip the first fraction
Change division to multiplication
Keep the second fraction unchanged

","correct":false,"explanation":"Flipping the first fraction changes the dividend rather than converting the divisor into its reciprocal. Keep the first fraction unchanged and flip only the second fraction."},{"text":"

Keep the first fraction
Change multiplication to division
Flip the second fraction (reciprocal)

","correct":false,"explanation":"Changing multiplication to division reverses the required procedure. For fraction division, keep the first fraction, change division to multiplication, and flip the second fraction."},{"text":"

Keep the first fraction
Change division to multiplication
Keep the second fraction unchanged

","correct":false,"explanation":"Replacing division with multiplication without taking the reciprocal gives a different value. The second fraction must be flipped after the operation is changed to multiplication."}]},{"id":"1586e028-a4e6-43f4-9d00-d2c4d029d028","cardType":"basic","points":1,"front":"When simplifying a division problem like $\\frac{A}{B} \\div \\frac{C}{D}$, which fraction is flipped?","back":"The second fraction ($\\frac{C}{D}$) only.","mcqOptions":[{"text":"The second fraction ($\\frac{C}{D}$) only.","correct":true,"explanation":"To divide by a fraction, multiply by its reciprocal. Thus $\\frac{A}{B}$ remains unchanged, while the divisor $\\frac{C}{D}$ becomes $\\frac{D}{C}$."},{"text":"The first fraction ($\\frac{A}{B}$) only.","correct":false,"explanation":"The first fraction is the dividend and must remain unchanged. It is the second fraction, $\\frac{C}{D}$, that is flipped to form its reciprocal."},{"text":"The first and second fractions ($\\frac{A}{B}$ and $\\frac{C}{D}$).","correct":false,"explanation":"Flipping both fractions alters the dividend as well as the divisor, producing the wrong expression. Only the second fraction, $\\frac{C}{D}$, is flipped."},{"text":"The neither fraction; both remain unchanged.","correct":false,"explanation":"Neither fraction remains unflipped: division is converted into multiplication by taking the reciprocal of the divisor. Therefore, only the second fraction is flipped."}]},{"id":"18f915f0-4336-48b5-9270-f68aa62af0c5","cardType":"cloze","points":null,"front":"When solving an equation with algebraic fractions, the primary goal is to [ ] by multiplying every term by the common denominator.","back":"When solving an equation with algebraic fractions, the primary goal is to remove the denominators by multiplying every term by the common denominator.","mcqOptions":[{"text":"remove the denominators","correct":true,"explanation":"Multiplying every term by a common denominator cancels each fractional denominator, converting the equation into one without algebraic fractions and making it easier to solve."},{"text":"expand the numerators","correct":false,"explanation":"Multiplication by a common denominator cancels denominators rather than expanding numerators. The primary goal is to remove the denominators and obtain an equation without fractions."},{"text":"factor the numerators","correct":false,"explanation":"Factoring numerators may help with later simplification, but it is not achieved simply by multiplying every term by the common denominator. The goal is to remove the denominators."},{"text":"cross-multiply the fractions","correct":false,"explanation":"Cross-multiplication applies directly to an equation containing one fraction on each side, not generally to every term. Multiplying by the common denominator removes the denominators."}]},{"id":"f08b774a-6674-4416-ace3-e20daaf0c8c8","cardType":"cloze","points":null,"front":"When solving an equation with algebraic fractions, the primary goal is to remove the denominators by multiplying every term by the [ ].","back":"When solving an equation with algebraic fractions, the primary goal is to remove the denominators by multiplying every term by the common denominator.","mcqOptions":[{"text":"common denominator","correct":true,"explanation":"A common denominator contains all denominator factors required to cancel the fractions. Multiplying every term by it produces an equivalent equation with no fractional denominators."},{"text":"greatest common factor","correct":false,"explanation":"A greatest common factor identifies shared factors but may not contain everything needed to cancel each denominator. Every term must instead be multiplied by the common denominator."},{"text":"common numerator","correct":false,"explanation":"Numerators do not determine which multiplier clears the fractions. The required multiplier is the common denominator, which contains the necessary factors from all original denominators."},{"text":"largest denominator","correct":false,"explanation":"Choosing whichever denominator appears largest does not guarantee that all denominator factors will cancel. The multiplier must be a common denominator of every fraction."}]},{"id":"a931f9a1-51c7-4f1f-9e31-345931cd03a5","cardType":"basic","points":1,"front":"If solving $\\frac{3}{x+2} + \\frac{4}{x+7} = 1$, what is the common denominator used to clear the fractions?","back":"$$(x+2)(x+7)$","mcqOptions":[{"text":"$$(x+2)(x+7)$","correct":true,"explanation":"The denominators $x+2$ and $x+7$ have no common algebraic factor, so their product $(x+2)(x+7)$ is the common denominator used to clear both fractions."},{"text":"$$(x+2)+(x+7)$","correct":false,"explanation":"Adding the denominators is a common slip, but $2x+9$ is not divisible by either original denominator. The required common denominator is $(x+2)(x+7)$."},{"text":"$$(x+2)(x-7)$","correct":false,"explanation":"Changing $x+7$ to $x-7$ introduces a sign error and would not cancel the original denominator. The common denominator is $(x+2)(x+7)$."},{"text":"$$x+9$","correct":false,"explanation":"Adding only the constants ignores the variable terms and does not produce a multiple of either denominator. The common denominator needed is $(x+2)(x+7)$."}]},{"id":"9624b99c-ffad-4080-89f4-5415bac59cc4","cardType":"basic","points":2,"front":"Why must you identify excluded values (e.g., $x \\neq -2$) before solving an algebraic fraction equation?","back":"To ensure the final solution does not make any original denominator equal to zero.","mcqOptions":[{"text":"To ensure the final solution does not make any original denominator equal to zero.","correct":true,"explanation":"Division by zero is undefined, so any value that makes an original denominator zero cannot be a solution, even if it appears after the equation is rearranged."},{"text":"To ensure the final solution does not make any original numerator equal to zero.","correct":false,"explanation":"A numerator may equal zero without making a fraction undefined; it usually makes the fraction equal zero. Exclusions prevent an original denominator from equalling zero."},{"text":"To ensure the final solution does not make any original denominator equal to one.","correct":false,"explanation":"Denominators are allowed to equal one, since division by one is defined. Excluded values specifically prevent any original denominator from becoming zero."},{"text":"To ensure the final solution does not make every original denominator have the same value.","correct":false,"explanation":"Different denominators may take the same value without making the equation undefined. Excluded values are identified because no original denominator is permitted to equal zero."}]},{"id":"af2ef845-6b2f-43c9-a356-81deabb5c4c3","cardType":"cloze","points":null,"front":"In an equation with three terms like $\\frac{5}{x-1} + \\frac{2}{x+3} = 1$, you must multiply [ ] (including the 1) by the common denominator.","back":"In an equation with three terms like $\\frac{5}{x-1} + \\frac{2}{x+3} = 1$, you must multiply every term (including the 1) by the common denominator.","mcqOptions":[{"text":"every term","correct":true,"explanation":"Multiplying every term preserves the equation’s balance and allows all fractional denominators to cancel. This includes the constant $1$ on the right-hand side."},{"text":"only the fractional terms","correct":false,"explanation":"Multiplying only the fractions changes one side without applying the same operation to the constant $1$. Every term in the equation must be multiplied by the common denominator."},{"text":"only the left-hand side","correct":false,"explanation":"Applying the multiplier only to the left-hand side breaks the equality because the right-hand side is unchanged. Every term, including $1$, must receive the multiplier."},{"text":"only the denominators","correct":false,"explanation":"Multiplying denominators in isolation changes the fractions rather than performing an equivalent operation on the equation. The common denominator must multiply every complete term."}]},{"id":"c01bcfee-444c-4403-adb8-4dab9fdce5b0","cardType":"basic","points":1,"front":"How do you convert the ratio statement $A:B = C:D$ into an equation?","back":"$$\\frac{A}{B} = \\frac{C}{D}$","mcqOptions":[{"text":"$$\\frac{A}{B} = \\frac{C}{D}$","correct":true,"explanation":"A ratio $A:B$ is written as the fraction $\\frac{A}{B}$, while $C:D$ becomes $\\frac{C}{D}$. Since the ratios are equal, the fractions are equal."},{"text":"$$\\frac{A}{B} = \\frac{D}{C}$","correct":false,"explanation":"This reverses the second ratio, changing $C:D$ into $D:C$. The order of each ratio must be preserved, giving $\\frac{A}{B} = \\frac{C}{D}$."},{"text":"$$\\frac{B}{A} = \\frac{C}{D}$","correct":false,"explanation":"This reverses only the first ratio, treating $A:B$ as $B:A$. Keeping both quantities in their original order gives $\\frac{A}{B} = \\frac{C}{D}$."},{"text":"$$\\frac{A}{D} = \\frac{C}{B}$","correct":false,"explanation":"This incorrectly pairs terms from different ratios as numerators and denominators. Each ratio is converted separately, so the equation is $\\frac{A}{B} = \\frac{C}{D}$."}]},{"id":"5e07a512-abcb-43e2-ae5a-10e8ca106681","cardType":"basic","points":2,"front":"What is the technique of \"cross-multiplication\" when solving $\\frac{A}{B} = \\frac{C}{D}$?","back":"Multiplying each numerator by the opposite denominator ($AD = BC$).","mcqOptions":[{"text":"Multiplying each numerator by the opposite denominator ($AD = BC$).","correct":true,"explanation":"Cross-multiplication forms diagonal products across the equals sign: $A$ multiplies $D$, and $C$ multiplies $B$. This produces the equivalent equation $AD = BC$."},{"text":"Multiplying each numerator by its own denominator ($AB = CD$).","correct":false,"explanation":"This multiplies vertically within each fraction rather than diagonally across the equation. Cross-multiplication uses the opposite denominators, so the resulting equation is $AD = BC$."},{"text":"Multiplying both numerators and both denominators together ($AC = BD$).","correct":false,"explanation":"This groups like positions instead of forming diagonal products. Cross-multiplication pairs each numerator with the denominator on the other side, producing $AD = BC$."},{"text":"Dividing each numerator by the opposite denominator ($A \\div D = C \\div B$).","correct":false,"explanation":"The word “cross” describes the diagonal pairing, but the operation is multiplication rather than division. The two diagonal products are equated to give $AD = BC$."}]},{"id":"5f1ce4e3-2feb-4fad-8730-16970713f0f8","cardType":"cloze","points":null,"front":"If an answer must be in the form $\\frac{ax+b}{x+c}$, the letters $a$, $b$, and $c$ represent [ ] (whole numbers).","back":"If an answer must be in the form $\\frac{ax+b}{x+c}$, the letters $a$, $b$, and $c$ represent integers (whole numbers).","mcqOptions":[{"text":"integers","correct":true,"explanation":"Integers are numbers with no fractional or decimal part, including positive numbers, negative numbers and zero. They are the required type of coefficients and constants here."},{"text":"fractions","correct":false,"explanation":"Fractions can represent non-whole values, whereas the stated restriction requires whole-number values. The letters $a$, $b$ and $c$ therefore represent integers."},{"text":"irrationals","correct":false,"explanation":"Irrational numbers cannot be written as exact fractions and include values such as $\\sqrt{2}$. The form instead specifies that $a$, $b$ and $c$ are integers."},{"text":"variables","correct":false,"explanation":"Although letters often denote variables, here $a$, $b$ and $c$ are fixed numerical coefficients or constants. The stated whole-number condition means they represent integers."}]},{"id":"7fe02c19-bdcf-4158-8d03-3819560c51f9","cardType":"basic","points":1,"front":"How can you verify your factorisation is correct before cancelling factors in a fraction?","back":"Quickly expand the brackets to see if they return to the original expression.","mcqOptions":[{"text":"Quickly expand the brackets to see if they return to the original expression.","correct":true,"explanation":"Expanding reverses factorisation. If multiplying out the brackets reproduces every term of the original expression with the correct coefficients and signs, the factorisation is correct."},{"text":"Quickly add the brackets to see if they return to the original expression.","correct":false,"explanation":"Bracketed factors are multiplied, not added, so adding them cannot reverse factorisation. Expanding the brackets correctly checks whether their product returns to the original expression."},{"text":"Quickly cancel the brackets to see if they return to the original expression.","correct":false,"explanation":"Cancelling before checking risks removing factors based on an incorrect factorisation. The brackets should first be expanded to confirm that their product matches the original expression."},{"text":"Quickly differentiate the brackets to see if they return to the original expression.","correct":false,"explanation":"Differentiation changes an expression and is not the reverse of factorisation. Expanding multiplies the factors back together, allowing direct comparison with the original expression."}]},{"id":"bb25ba5a-f790-4f0a-80c0-26cd98a94e89","cardType":"basic","points":2,"front":"Factorise $2x^2 + 11x + 5$.","back":"$$(2x + 1)(x + 5)$","mcqOptions":[{"text":"$$(2x + 1)(x + 5)$","correct":true,"explanation":"Expanding gives $2x^2 + 10x + x + 5 = 2x^2 + 11x + 5$. Therefore these brackets factorise the quadratic exactly."},{"text":"$$(2x + 5)(x + 1)$","correct":false,"explanation":"These brackets expand to $2x^2 + 7x + 5$, so the middle coefficient is too small. The correct factorisation is $(2x + 1)(x + 5)$."},{"text":"$$(2x - 1)(x - 5)$","correct":false,"explanation":"Using two negative signs keeps the constant positive but produces a negative middle term: $2x^2 - 11x + 5$. The required factors are $(2x + 1)(x + 5)$."},{"text":"$$(2x + 1)(x + 11)$","correct":false,"explanation":"This treats the middle coefficient as a bracket constant and expands to $2x^2 + 23x + 11$. The correct factor pair gives $(2x + 1)(x + 5)$."}]},{"id":"48e0cc96-7ca9-4a02-9355-6c0c224746ca","cardType":"cloze","points":null,"front":"Factorising $3x^2 + 9x$ by taking out a common factor results in $\\text{[...]}$.","back":"Factorising $3x^2 + 9x$ by taking out a common factor results in $\\boxed{3x(x+3)}$.","mcqOptions":[{"text":"$$3x(x+3)$","correct":true,"explanation":"Both terms share the greatest common factor $3x$. Dividing $3x^2$ and $9x$ by $3x$ leaves $x$ and $3$, giving $3x(x+3)$."},{"text":"$$3x(x+6)$","correct":false,"explanation":"Dividing $9x$ by the common factor $3x$ gives $3$, not $6$. Taking out $3x$ correctly produces $3x(x+3)$."},{"text":"$$3(x+3)$","correct":false,"explanation":"This loses a factor of $x$ and expands to $3x+9$, not the original quadratic expression. The greatest common factor is $3x$, giving $3x(x+3)$."},{"text":"$$x(3x+3)$","correct":false,"explanation":"The second term inside the brackets should be $9$ if only $x$ is extracted, so this expands incorrectly. Fully extracting $3x$ gives $3x(x+3)$."}]},{"id":"d404994f-d67c-48f7-8cc0-44a6e535cafb","cardType":"basic","points":2,"front":"What quadratic factors are found when simplifying $\\frac{x^2+6x}{x^2+8x+12}$?","back":"

Numerator: $x(x+6)$
Denominator: $(x+2)(x+6)$

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

Numerator: $x(x+6)$
Denominator: $(x+2)(x+6)$

","correct":true,"explanation":"The numerator has common factor $x$, while two numbers adding to $8$ and multiplying to $12$ are $2$ and $6$. This reveals the common factor $(x+6)$."},{"text":"

Numerator: $x(x+6)$
Denominator: $(x+3)(x+4)$

","correct":false,"explanation":"Although $3$ and $4$ multiply to $12$, they add to $7$, not $8$. The denominator factors as $(x+2)(x+6)$, while the numerator is $x(x+6)$."},{"text":"

Numerator: $x(x+3)$
Denominator: $(x+2)(x+6)$

","correct":false,"explanation":"Taking $x$ out of $x^2+6x$ leaves $x+6$, not $x+3$. The correct numerator is $x(x+6)$ and the denominator is $(x+2)(x+6)$."},{"text":"

Numerator: $(x+2)(x+3)$
Denominator: $(x+2)(x+6)$

","correct":false,"explanation":"The numerator has no constant term, so it must contain a factor of $x$; these brackets produce a constant of $6$. Its correct factorisation is $x(x+6)$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Algebraic Fractions","title":"Algebraic Fractions Flashcards","topbar":{"backHref":"/gcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"526fa632-4547-4730-9ed5-00a435ba4f38","walled":"$L39"}],"$L3a","$L3b"]