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5x + 7$, which term is the constant term?","back":"7","mcqOptions":[{"text":"7","correct":true,"explanation":"The constant term is the term containing no variable. In $3x^2 - 5x + 7$, only 7 has no factor of $x$."},{"text":"3","correct":false,"explanation":"The number 3 is the coefficient of $x^2$, not the constant term. The constant term is 7 because it does not contain $x$."},{"text":"-5","correct":false,"explanation":"The number -5 is the coefficient of $x$, not the constant term. The correct answer is 7 because that term contains no variable."},{"text":"0","correct":false,"explanation":"A constant term is present, so it is not implicitly zero. The correct answer is 7 because 7 is the term without a variable."}]},{"id":"8f862e7d-1898-6a04-c298-6b92c5a5b8d5","cardType":"cloze","points":null,"front":"A [ ] is the number multiplying a variable, such as 3 in $3x$.","back":"A coefficient is the number multiplying a variable, such as 3 in $3x$.","mcqOptions":[{"text":"coefficient","correct":true,"explanation":"A coefficient is the numerical factor multiplying a variable. For example, the coefficient of $x$ in $3x$ is 3."},{"text":"constant","correct":false,"explanation":"A constant is a term without a variable, rather than a number multiplying one. The correct term is coefficient, as 3 is the coefficient in $3x$."},{"text":"exponent","correct":false,"explanation":"An exponent shows the power to which a variable is raised, such as 2 in $x^2$. The number multiplying a variable is its coefficient."},{"text":"variable","correct":false,"explanation":"A variable is the letter representing an unknown or changing value. The number multiplying that variable is called its coefficient, such as 3 in $3x$."}]},{"id":"65e50139-5f31-a475-24f9-25f63c1cb129","cardType":"cloze","points":null,"front":"To [ ] means to put an expression back into brackets, while to expand means to remove them.","back":"To factorise means to put an expression back into brackets, while to expand means to remove them.","mcqOptions":[{"text":"factorise","correct":true,"explanation":"Factorising rewrites an expression as a product of factors, usually shown using brackets. It is the reverse process of expanding brackets."},{"text":"simplify","correct":false,"explanation":"Simplifying makes an expression more concise but does not necessarily put it into brackets. Factorise is the correct term for reversing expansion and forming a product."},{"text":"expand","correct":false,"explanation":"Expanding removes brackets by multiplication, so it is the opposite process. Factorise is the correct term for putting an expression back into brackets."},{"text":"rearrange","correct":false,"explanation":"Rearranging changes the order or position of terms, often within an equation. Factorise specifically means rewriting an expression as factors inside brackets."}]},{"id":"a288abbe-25cc-860b-3d35-496ce13bb2f9","cardType":"cloze","points":null,"front":"To factorise means to put an expression back into brackets, while to [ ] means to remove them.","back":"To factorise means to put an expression back into brackets, while to expand means to remove them.","mcqOptions":[{"text":"expand","correct":true,"explanation":"Expanding removes brackets by multiplying every relevant term and then collecting like terms. It is the reverse process of factorising an expression."},{"text":"simplify","correct":false,"explanation":"Simplifying may involve collecting like terms or reducing an expression, but it does not specifically mean removing brackets. The required term is expand."},{"text":"factorise","correct":false,"explanation":"Factorising puts an expression into brackets rather than removing them. The opposite process, which multiplies out the brackets, is called expanding."},{"text":"solve","correct":false,"explanation":"Solving finds the value of an unknown in an equation; it does not mean removing brackets. The correct term is expand, which means multiplying out brackets."}]},{"id":"9666af6d-6c17-82d0-3aff-c5192f116852","cardType":"basic","points":2,"front":"Simplify the expression $x^2 + 7x - 3x - 21$ by collecting like terms.","back":"$$x^2 + 4x - 21$","mcqOptions":[{"text":"$$x^2 + 4x - 21$","correct":true,"explanation":"The like terms are $7x$ and $-3x$, which combine to give $4x$. The $x^2$ and constant terms remain unchanged."},{"text":"$$x^2 + 10x - 21$","correct":false,"explanation":"This treats $-3x$ as positive and adds 7 and 3. Keeping the negative sign gives $7x-3x=4x$, so the result is $x^2+4x-21$."},{"text":"$$x^2 + 4x + 21$","correct":false,"explanation":"The variable terms have been collected correctly, but the constant’s sign has changed. The original constant is $-21$, so the answer is $x^2+4x-21$."},{"text":"$$5x^2 - 21$","correct":false,"explanation":"This incorrectly combines $x^2$ and $x$ terms, which are not like terms. Only $7x$ and $-3x$ combine, giving $x^2+4x-21$."}]},{"id":"31116f6a-e00c-2a66-f579-5bba9a69f344","cardType":"basic","points":1,"front":"When expanding two brackets, how many multiplications must you perform to avoid missing terms?","back":"4","mcqOptions":[{"text":"4","correct":true,"explanation":"Each of the two terms in the first bracket must multiply each of the two terms in the second bracket, giving $2\\times2=4$ multiplications."},{"text":"2","correct":false,"explanation":"This counts only one multiplication for each term in the first bracket. Each must multiply both terms in the second bracket, so 4 multiplications are required."},{"text":"3","correct":false,"explanation":"Three multiplications usually means one cross-term has been missed. Two terms multiplied by two terms produce four separate multiplications before like terms are collected."},{"text":"6","correct":false,"explanation":"This overcounts the required products. With two terms in each bracket, every first-bracket term pairs with two second-bracket terms, giving exactly 4 multiplications."}]},{"id":"68bd9584-c992-1ea8-af93-3601b8e01506","cardType":"cloze","points":null,"front":"A [ ] expression has a squared variable as its highest power, such as $x^2 + 4x - 12$.","back":"A quadratic expression has a squared variable as its highest power, such as $x^2 + 4x - 12$.","mcqOptions":[{"text":"quadratic","correct":true,"explanation":"A quadratic expression has highest variable power 2. Since the leading power in $x^2+4x-12$ is $x^2$, the expression is quadratic."},{"text":"linear","correct":false,"explanation":"A linear expression has highest variable power 1, such as $3x+2$. An expression whose highest power is 2 is called quadratic."},{"text":"cubic","correct":false,"explanation":"A cubic expression has highest variable power 3, such as $x^3+2x$. A highest power of 2 makes the expression quadratic."},{"text":"exponential","correct":false,"explanation":"In an exponential expression, the variable appears in the exponent, such as $2^x$. A polynomial with highest power $x^2$ is quadratic."}]},{"id":"ed5681d5-32a1-af98-698a-343002ac7bed","cardType":"basic","points":2,"front":"Expand and simplify $(x + 5)(x - 4)$.","back":"$$x^2 + x - 20$","mcqOptions":[{"text":"$$x^2 + x - 20$","correct":true,"explanation":"The four products are $x^2$, $-4x$, $5x$, and $-20$. Combining the middle terms gives $-4x+5x=x$, producing $x^2+x-20$."},{"text":"$$x^2 + 9x - 20$","correct":false,"explanation":"This treats the cross-term $-4x$ as positive, giving $5x+4x$. Keeping its negative sign gives $5x-4x=x$, so the result is $x^2+x-20$."},{"text":"$$x^2 - x - 20$","correct":false,"explanation":"This reverses the subtraction when combining the cross-terms. Since $5x-4x=x$, not $-x$, the correct expansion is $x^2+x-20$."},{"text":"$$x^2 + x + 20$","correct":false,"explanation":"The constant product has the wrong sign. Multiplying $5$ by $-4$ gives $-20$, so the correct expansion is $x^2+x-20$."}]},{"id":"a816dea4-e560-7316-3775-3a1471dfe523","cardType":"basic","points":1,"front":"What is the result of multiplying $3x$ by $2x$?","back":"$$6x^2$","mcqOptions":[{"text":"$$6x^2$","correct":true,"explanation":"Multiply the coefficients to get $3 \\times 2 = 6$, then multiply $x \\times x = x^2$, giving $6x^2$."},{"text":"$$6x$","correct":false,"explanation":"This omits the index produced when the variables are multiplied. Since $x \\times x = x^2$, the correct product is $6x^2$."},{"text":"$$5x^2$","correct":false,"explanation":"This adds the coefficients instead of multiplying them. The coefficients give $3 \\times 2 = 6$, while $x \\times x = x^2$, so the answer is $6x^2$."},{"text":"$$6x^3$","correct":false,"explanation":"This incorrectly adds an extra power of $x$. Multiplying $x^1$ by $x^1$ gives $x^{1+1}=x^2$, so the correct result is $6x^2$."}]},{"id":"566222c1-bc85-4fa8-146e-b2c0dd21c097","cardType":"basic","points":2,"front":"Expand and simplify $(4x - 1)(2x + 3)$.","back":"$$8x^2 + 10x - 3$","mcqOptions":[{"text":"$$8x^2 + 10x - 3$","correct":true,"explanation":"The four products are $8x^2$, $12x$, $-2x$, and $-3$. Combining the middle terms gives $12x-2x=10x$, producing $8x^2+10x-3$."},{"text":"$$8x^2 + 14x - 3$","correct":false,"explanation":"This treats the product $(-1)(2x)$ as $+2x$. It is $-2x$, so the middle terms combine to $12x-2x=10x$ and the answer is $8x^2+10x-3$."},{"text":"$$8x^2 + 10x + 3$","correct":false,"explanation":"The constant sign is wrong: multiplying $-1$ by $+3$ gives $-3$, not $+3$. The correctly expanded expression is $8x^2+10x-3$."},{"text":"$$8x^2 + 12x - 3$","correct":false,"explanation":"This misses the inner product $(-1)(2x)=-2x$. All four products must be included, so the middle coefficient is $12-2=10$ and the result is $8x^2+10x-3$."}]},{"id":"1c984e17-3cb4-1e3d-eade-56433d9bea92","cardType":"cloze","points":null,"front":"When expanding brackets, a negative multiplied by a negative results in a [ ] value.","back":"When expanding brackets, a negative multiplied by a negative results in a positive value.","mcqOptions":[{"text":"positive","correct":true,"explanation":"Multiplying two numbers with the same sign gives a positive result. Therefore, a negative multiplied by another negative produces a positive value."},{"text":"negative","correct":false,"explanation":"A negative result comes from multiplying numbers with different signs. Two negative factors have the same sign, so their product is positive."},{"text":"zero","correct":false,"explanation":"A product is zero only when at least one factor is zero. Two negative factors instead multiply to produce a positive value."},{"text":"unchanged","correct":false,"explanation":"Multiplication does not leave the sign unchanged automatically. The sign rule states that two negative factors produce a positive value."}]},{"id":"9198dda5-0b72-1c0a-be3e-34631b03c75c","cardType":"basic","points":1,"front":"What is the first step to expanding a squared bracket like $(b - 6)^2$?","back":"Rewrite it as two brackets: $(b - 6)(b - 6)$.","mcqOptions":[{"text":"Rewrite it as two brackets: $(b - 6)(b - 6)$.","correct":true,"explanation":"A squared bracket means the entire bracket is multiplied by itself. Therefore, $(b-6)^2$ must first be written as $(b-6)(b-6)$."},{"text":"Rewrite it as two brackets: $(b - 6)(b + 6)$.","correct":false,"explanation":"Changing the second sign creates a difference of two squares, not a squared bracket. Both factors must be identical, so the correct rewrite is $(b-6)(b-6)$."},{"text":"Rewrite it as two terms: $b^2 - 6^2$.","correct":false,"explanation":"Squaring a bracket does not mean squaring each term separately; this misses the middle term. The correct first step is to write $(b-6)(b-6)$."},{"text":"Rewrite it as two brackets: $(b - 6)(b - 12)$.","correct":false,"explanation":"Doubling the constant inside one bracket misinterprets the square. A squared bracket is multiplied by an identical copy, so it becomes $(b-6)(b-6)$."}]},{"id":"627f0864-0710-8e48-0a68-5b18c3155dfc","cardType":"basic","points":2,"front":"Expand and simplify $(b - 6)^2$.","back":"$$b^2 - 12b + 36$","mcqOptions":[{"text":"$$b^2 - 12b + 36$","correct":true,"explanation":"Writing $(b-6)^2$ as $(b-6)(b-6)$ gives $b^2-6b-6b+36$. Combining the middle terms produces $b^2-12b+36$."},{"text":"$$b^2 - 36$","correct":false,"explanation":"This incorrectly treats the expression as a difference of two squares and omits the middle term. Expanding two identical brackets gives $b^2-12b+36$."},{"text":"$$b^2 - 12b - 36$","correct":false,"explanation":"The constant sign is wrong because $(-6)(-6)=+36$. Including the two middle products gives the correct expansion $b^2-12b+36$."},{"text":"$$b^2 - 6b + 36$","correct":false,"explanation":"This includes only one of the two middle products. Both are $-6b$, so they combine to $-12b$, giving $b^2-12b+36$."}]},{"id":"d6e789c6-67ff-4cb6-c6ae-324bae253be0","cardType":"basic","points":2,"front":"When factorising $x^2 + bx + c$, what two properties must the chosen number pair satisfy?","back":"

Multiply to give $c$ (the product)
Add to give $b$ (the sum)

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

Multiply to give $c$ (the product)
Add to give $b$ (the sum)

","correct":true,"explanation":"In $(x+m)(x+n)$, the constant is $mn$ and the middle coefficient is $m+n$. Therefore, the numbers must multiply to $c$ and add to $b$."},{"text":"

Add to give $c$ (the sum)
Multiply to give $b$ (the product)

","correct":false,"explanation":"This reverses the product and sum conditions. Expanding $(x+m)(x+n)$ shows that $mn=c$ and $m+n=b$, so the pair multiplies to $c$ and adds to $b$."},{"text":"

Multiply to give $b$ (the product)
Add to give $c$ (the sum)

","correct":false,"explanation":"This assigns the coefficients to the wrong operations. The bracket constants create $c$ when multiplied and create $b$ when added, so the pair multiplies to $c$ and adds to $b$."},{"text":"

Multiply to give $c$ (the product)
Subtract to give $b$ (the difference)

","correct":false,"explanation":"The middle coefficient comes from adding the two cross terms, not subtracting the bracket constants. The chosen numbers must multiply to $c$ and add to $b$."}]},{"id":"dd7a5862-82a8-693a-938f-37a74c89624b","cardType":"basic","points":2,"front":"Factorise $x^2 + 9x + 20$.","back":"$$(x + 4)(x + 5)$","mcqOptions":[{"text":"$$(x + 4)(x + 5)$","correct":true,"explanation":"The numbers $4$ and $5$ multiply to $20$ and add to $9$. Therefore, $x^2+9x+20$ factorises as $(x+4)(x+5)$."},{"text":"$$(x + 2)(x + 10)$","correct":false,"explanation":"Although $2$ and $10$ multiply to $20$, they add to $12$, not $9$. The required pair is $4$ and $5$, giving $(x+4)(x+5)$."},{"text":"$$(x - 4)(x - 5)$","correct":false,"explanation":"The constants multiply to $20$, but their sum is $-9$, producing a negative middle term. A positive $9x$ requires $(x+4)(x+5)$."},{"text":"$$(x + 1)(x + 20)$","correct":false,"explanation":"Although $1$ and $20$ multiply to $20$, they add to $21$, not $9$. The numbers $4$ and $5$ satisfy both conditions, giving $(x+4)(x+5)$."}]},{"id":"d2dac6d7-a65a-2481-0c81-0b6c66536a47","cardType":"basic","points":1,"front":"If a quadratic has a positive constant and a negative middle term, what are the signs in the brackets?","back":"Both are negative.","mcqOptions":[{"text":"Both are negative.","correct":true,"explanation":"Two negative constants multiply to give a positive constant and add to give a negative middle coefficient. Therefore, both signs in the brackets must be negative."},{"text":"Both are positive.","correct":false,"explanation":"Two positive constants give a positive constant but also a positive middle coefficient. To produce a negative middle term while keeping the constant positive, both signs must be negative."},{"text":"The first is positive and the second is negative.","correct":false,"explanation":"Opposite signs make the bracket constants multiply to a negative value, contradicting the positive constant. Both signs must be negative to give a positive product and negative sum."},{"text":"The first is negative and the second is positive.","correct":false,"explanation":"Using one negative and one positive sign makes the constant term negative. A positive constant with a negative middle term requires two negative bracket constants."}]},{"id":"56fd9cee-942c-271b-cd51-4d560758292f","cardType":"basic","points":1,"front":"What are the bracket signs if the constant term of a quadratic is negative?","back":"One positive and one negative.","mcqOptions":[{"text":"One positive and one negative.","correct":true,"explanation":"A negative constant comes from multiplying numbers with opposite signs. Therefore, the two brackets must contain one positive term and one negative term."},{"text":"One positive and one positive.","correct":false,"explanation":"Two positive terms multiply to give a positive constant. A negative constant requires opposite signs, so there must be one positive sign and one negative sign."},{"text":"One negative and one negative.","correct":false,"explanation":"Two negative terms also multiply to give a positive constant. To produce a negative constant, one bracket must be positive and the other negative."},{"text":"The signs depend only on the linear term.","correct":false,"explanation":"The linear term helps determine which signed numbers are used, but the constant determines their sign relationship. A negative product requires one positive and one negative sign."}]},{"id":"3dc5fc6d-3be3-4d28-a789-c9d58ebef439","cardType":"basic","points":2,"front":"Factorise $n^2 - 4n - 21$.","back":"$$(n - 7)(n + 3)$","mcqOptions":[{"text":"$$(n - 7)(n + 3)$","correct":true,"explanation":"The numbers $-7$ and $3$ multiply to $-21$ and add to $-4$. Expanding the brackets therefore gives $n^2-4n-21$."},{"text":"$$(n - 3)(n + 7)$","correct":false,"explanation":"The numbers $-3$ and $7$ multiply to $-21$ but add to $4$, producing $n^2+4n-21$. The required pair is $-7$ and $3$."},{"text":"$$(n - 7)(n - 3)$","correct":false,"explanation":"Using two negative signs makes the constant positive because $(-7)(-3)=21$. The required factors have opposite signs: $(n-7)(n+3)$."},{"text":"$$(n + 7)(n - 3)$","correct":false,"explanation":"These constants multiply to $-21$, but they add to $4$, so the middle term would be $+4n$. They must instead be $-7$ and $3$."}]},{"id":"c177787b-db3f-74af-8e80-81848f0efcc0","cardType":"cloze","points":null,"front":"The difference of two squares formula is $\\text{[...]}$.","back":"The difference of two squares formula is $\\boxed{a^2 - b^2 = (a - b)(a + b)}$.","mcqOptions":[{"text":"$$a^2 - b^2 = (a - b)(a + b)$","correct":true,"explanation":"Expanding $(a-b)(a+b)$ gives $a^2+ab-ab-b^2$. The middle terms cancel, leaving precisely the difference $a^2-b^2$."},{"text":"$$a^2 + b^2 = (a - b)(a + b)$","correct":false,"explanation":"The product $(a-b)(a+b)$ equals $a^2-b^2$, not a sum of squares. Over the real numbers, $a^2+b^2$ does not factorise using this identity."},{"text":"$$a^2 - b^2 = (a - b)^2$","correct":false,"explanation":"Squaring $(a-b)$ produces $a^2-2ab+b^2$, including a middle term and a positive $b^2$. Difference of squares uses conjugate brackets instead."},{"text":"$$a^2 - b^2 = (a + b)^2$","correct":false,"explanation":"Squaring $(a+b)$ produces $a^2+2ab+b^2$, so neither the middle term nor the final sign matches. The correct factors have opposite signs."}]},{"id":"115b2074-16a0-51da-aaaf-325f75b82cee","cardType":"basic","points":1,"front":"Factorise $x^2 - 64$.","back":"$$(x - 8)(x + 8)$","mcqOptions":[{"text":"$$(x - 8)(x + 8)$","correct":true,"explanation":"Since $64=8^2$, the expression is $x^2-8^2$. Applying the difference of two squares identity gives $(x-8)(x+8)$."},{"text":"$$(x - 8)(x - 8)$","correct":false,"explanation":"Using two negative signs gives a positive constant and a middle term: $(x-8)^2=x^2-16x+64$. Difference of squares requires opposite bracket signs."},{"text":"$$(x + 8)(x + 8)$","correct":false,"explanation":"Using two positive signs gives $(x+8)^2=x^2+16x+64$, not $x^2-64$. The factors must have opposite signs so the middle terms cancel."},{"text":"$$(x - 4)(x + 4)$","correct":false,"explanation":"This confuses $64$ with $4^2$ rather than $8^2$. Expanding gives $x^2-16$, whereas the required factorisation is $(x-8)(x+8)$."}]},{"id":"b7b2c246-254e-f3d1-b61b-113a54a9f34f","cardType":"basic","points":1,"front":"Why is $x^2 + 64$ not a difference of two squares?","back":"It contains a plus sign instead of a subtraction sign.","mcqOptions":[{"text":"It contains a plus sign instead of a subtraction sign.","correct":true,"explanation":"The difference of two squares identity requires subtraction: $a^2-b^2$. Although both $x^2$ and $64$ are squares, they are joined by addition."},{"text":"It contains a subtraction sign instead of a plus sign.","correct":false,"explanation":"The expression actually contains addition, not subtraction. That plus sign is precisely why the difference of two squares identity cannot be applied."},{"text":"It contains terms that are not both perfect squares.","correct":false,"explanation":"Both terms are perfect squares because $x^2=(x)^2$ and $64=8^2$. The obstacle is the plus sign rather than either term’s form."},{"text":"It has a coefficient of one instead of two.","correct":false,"explanation":"A coefficient of $1$ makes $x^2$ a perfect square and does not prevent factorisation. The identity fails here because the squares are added rather than subtracted."}]},{"id":"ced319a4-b019-4cbd-5b9c-aee12592e22c","cardType":"basic","points":2,"front":"Factorise the perfect square $x^2 - 16x + 64$.","back":"$$(x - 8)^2$","mcqOptions":[{"text":"$$(x - 8)^2$","correct":true,"explanation":"Squaring $(x-8)$ gives $x^2-16x+64$: the middle term is $2(x)(-8)=-16x$, and the constant is $(-8)^2=64$."},{"text":"$$(x + 8)^2$","correct":false,"explanation":"The positive sign produces a positive middle term, since $(x+8)^2=x^2+16x+64$. The negative middle term requires $(x-8)^2$."},{"text":"$$(x - 4)^2$","correct":false,"explanation":"Halving $16$ correctly gives $8$, not $4$. Squaring $(x-4)$ produces $x^2-8x+16$, so both the middle and constant terms are wrong."},{"text":"$$(x - 8)(x + 8)$","correct":false,"explanation":"Opposite signs make the middle terms cancel, giving $x^2-64$. A perfect-square trinomial needs two identical negative factors, so the answer is $(x-8)^2$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Expanding and Factorising Quadratics","title":"Expanding and Factorising Quadratics Flashcards","topbar":{"backHref":"/gcse/maths/ocr/flashcards"},"topicSidebar":"$L91"}]}],"topicId":"fe86e79f-eeb0-41ed-a1b4-15885faaadbf","walled":"$L92"}],"$L93","$L94"]