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For example, the leading term may be $x^2$."},{"text":"$$1$","correct":false,"explanation":"A highest power of $1$ defines a linear expression, not a quadratic expression. A quadratic has degree $2$, so its highest variable power is $2$."},{"text":"$$3$","correct":false,"explanation":"A highest power of $3$ defines a cubic expression. A quadratic expression has degree $2$, meaning its highest variable power is $2$."},{"text":"$$4$","correct":false,"explanation":"A highest power of $4$ defines a quartic expression rather than a quadratic. The highest variable power in a quadratic expression is $2$."}]},{"id":"1b7e4172-3c91-76f7-3083-010ecf4feb6d","cardType":"basic","points":1,"front":"Identify the constant term in the expression $t^2 + 8t + 15$.","back":"$$15$","mcqOptions":[{"text":"$$15$","correct":true,"explanation":"The constant term is the term containing no variable. In $t^2 + 8t + 15$, the only term without $t$ is $15$."},{"text":"$$8$","correct":false,"explanation":"The number $8$ multiplies $t$, so it is the coefficient of $t$, not the constant. The term without a variable is $15$."},{"text":"$$1$","correct":false,"explanation":"The implied coefficient of $t^2$ is $1$, but this is not the constant term. The term containing no variable is $15$."},{"text":"$$2$","correct":false,"explanation":"The number $2$ is the exponent in $t^2$, not a separate term in the expression. The constant term, which contains no variable, is $15$."}]},{"id":"a039f392-e054-2163-2ae0-4978ef924001","cardType":"basic","points":1,"front":"Identify the coefficient of $t$ in the expression $t^2 + 8t + 15$.","back":"$$8$","mcqOptions":[{"text":"$$8$","correct":true,"explanation":"A coefficient is the number multiplying a variable. Since the first-power term is $8t = 8 \\times t$, the coefficient of $t$ is $8$."},{"text":"$$15$","correct":false,"explanation":"The number $15$ is the constant term because it has no variable attached. The coefficient multiplying the first-power term $t$ is $8$."},{"text":"$$2$","correct":false,"explanation":"The number $2$ is the exponent in $t^2$, not the coefficient of the separate $t$ term. Since $8t = 8 \\times t$, the coefficient is $8$."},{"text":"$$1$","correct":false,"explanation":"The implied coefficient $1$ belongs to $t^2$, because $t^2 = 1t^2$. The coefficient of the first-power term $t$ is $8$."}]},{"id":"f338fad1-2ec6-6235-de04-d570f87b2d51","cardType":"basic","points":1,"front":"What is the primary goal when factorising a quadratic expression?","back":"To rewrite it as a product of brackets.","mcqOptions":[{"text":"To rewrite it as a product of brackets.","correct":true,"explanation":"Factorising reverses expansion by expressing a quadratic as brackets multiplied together. Expanding those brackets should reproduce every term of the original quadratic expression."},{"text":"To rewrite it as a sum of like terms.","correct":false,"explanation":"Collecting like terms simplifies an expression by combining matching terms, but it does not factorise it. Factorising rewrites the quadratic as a product of brackets."},{"text":"To rewrite it as a single linear term.","correct":false,"explanation":"A quadratic cannot generally be converted into one equivalent linear term because its squared term must remain represented. Factorising instead rewrites it as a product of brackets."},{"text":"To rewrite it as a quotient of brackets.","correct":false,"explanation":"Using a quotient means dividing brackets, which is not the goal of factorising a quadratic. Factorising expresses the original expression as a product of brackets."}]},{"id":"d34f013a-6c4e-5978-33f0-cfaa6539f519","cardType":"cloze","points":null,"front":"The mathematical process of [ ] is the reverse of expanding brackets.","back":"The mathematical process of factorising is the reverse of expanding brackets.","mcqOptions":[{"text":"factorising","correct":true,"explanation":"Factorising rewrites an expression as a product of factors, often brackets. Expanding multiplies those factors out, so factorising is the reverse process."},{"text":"expanding","correct":false,"explanation":"Expanding removes brackets by multiplying terms out, so it is the process being reversed. The reverse process, which creates a product of factors, is factorising."},{"text":"simplifying","correct":false,"explanation":"Simplifying is a broad process that can include collecting like terms and is not specifically the reverse of expanding brackets. The required reverse process is factorising."},{"text":"substituting","correct":false,"explanation":"Substituting replaces variables with given values rather than reversing an expansion. Factorising is the process that rewrites an expanded expression as a product of brackets."}]},{"id":"32059039-a82b-7bf5-7e6c-a78624747adb","cardType":"cloze","points":null,"front":"The mathematical process of factorising is the reverse of [ ] brackets.","back":"The mathematical process of factorising is the reverse of expanding brackets.","mcqOptions":[{"text":"expanding","correct":true,"explanation":"Expanding multiplies out the terms inside brackets to produce an expression. Factorising reverses this by rewriting the expression as a product of factors."},{"text":"collecting","correct":false,"explanation":"Collecting like terms combines terms with identical variable parts, but it is not the process reversed by factorising. Factorising reverses expanding brackets."},{"text":"substituting","correct":false,"explanation":"Substituting means replacing variables with values, so it is unrelated to forming or removing brackets. Factorising is specifically the reverse of expanding brackets."},{"text":"evaluating","correct":false,"explanation":"Evaluating finds the numerical value of an expression for given variable values. It is not reversed by factorising, which is the reverse of expanding brackets."}]},{"id":"276e4dd4-8cd4-2213-8524-ebb09b9209a7","cardType":"basic","points":1,"front":"When factorising $x^2 + bx + c$, what relationship must the two chosen numbers have with the constant term $c$?","back":"They must multiply to make $c$.","mcqOptions":[{"text":"They must multiply to make $c$.","correct":true,"explanation":"In $(x+m)(x+n)$, the constant term produced by expansion is $mn$. Therefore, the two chosen numbers must multiply to make $c$."},{"text":"They must divide to make $c$.","correct":false,"explanation":"Division is not used to match the constant term in this factorisation method. The two bracket constants must multiply to give $c$ when the brackets are expanded."},{"text":"They must add to make $c$.","correct":false,"explanation":"This confuses the two conditions: the chosen numbers add to make the coefficient $b$. They must multiply to make the constant term $c$."},{"text":"They must subtract to make $c$.","correct":false,"explanation":"Considering the difference may seem useful when signs differ, but it is not the constant-term condition. The two chosen numbers must multiply to make $c$."}]},{"id":"52de18be-a2f7-b261-edf5-741bc459fc2a","cardType":"basic","points":1,"front":"When factorising $x^2 + bx + c$, what relationship must the two chosen numbers have with the coefficient $b$?","back":"They must add to make $b$.","mcqOptions":[{"text":"They must add to make $b$.","correct":true,"explanation":"In $(x+m)(x+n)$, the middle terms are $mx$ and $nx$, which combine to $(m+n)x$. Therefore, the chosen numbers must add to make $b$."},{"text":"They must multiply to make $b$.","correct":false,"explanation":"This swaps the two factorisation conditions. The numbers multiply to make the constant term $c$, but they must add to make the coefficient $b$."},{"text":"They must subtract to make $b$.","correct":false,"explanation":"A difference may appear relevant when the numbers have opposite signs, but the signed numbers are still added. Their sum must equal the coefficient $b$."},{"text":"They must divide to make $b$.","correct":false,"explanation":"Division does not produce the middle coefficient when two brackets are expanded. The cross terms combine by addition, so the chosen numbers must add to make $b$."}]},{"id":"06a8e9d7-7962-d5e2-f6fe-c1c8b2545b5e","cardType":"basic","points":1,"front":"If the constant term $c$ is positive, what does this indicate about the signs in the factorised brackets?","back":"The signs are the same (both positive or both negative).","mcqOptions":[{"text":"The signs are the same (both positive or both negative).","correct":true,"explanation":"The constant term is the product of the two bracket constants. A positive product results when both constants have the same sign."},{"text":"The signs are different (one positive, one negative).","correct":false,"explanation":"Different signs produce a negative product, so they would make $c$ negative. When $c$ is positive, the signs must be the same."},{"text":"The signs are both positive in every case.","correct":false,"explanation":"A positive constant does not determine whether both signs are positive or negative. It only shows that the two signs are the same."},{"text":"The signs are both negative in every case.","correct":false,"explanation":"Two negative constants do produce a positive product, but two positive constants do as well. The middle coefficient determines which matching pair is needed."}]},{"id":"cb665460-7973-2f49-f080-4d938f1de33a","cardType":"cloze","points":null,"front":"In $x^2 + bx + c$, if $c$ is positive and $b$ is negative, then [ ] signs in the brackets must be negative.","back":"In $x^2 + bx + c$, if $c$ is positive and $b$ is negative, then both signs in the brackets must be negative.","mcqOptions":[{"text":"both","correct":true,"explanation":"A positive constant requires matching signs, while the negative middle coefficient requires their sum to be negative. Therefore, both bracket signs must be negative."},{"text":"no","correct":false,"explanation":"Using no negative signs would make both constants positive, giving a positive middle coefficient. The negative value of $b$ means both signs must be negative."},{"text":"opposite","correct":false,"explanation":"Opposite signs give a negative product, which would make $c$ negative. Since $c$ is positive and $b$ is negative, both signs must be negative."},{"text":"different","correct":false,"explanation":"Different signs would produce a negative constant term rather than a positive one. Matching negative signs give both the required positive product and negative sum."}]},{"id":"272c0c86-15af-bde8-ef7e-c146272a5d63","cardType":"cloze","points":null,"front":"In $x^2 + bx + c$, if $c$ is positive and $b$ is negative, then both signs in the brackets must be [ ].","back":"In $x^2 + bx + c$, if $c$ is positive and $b$ is negative, then both signs in the brackets must be negative.","mcqOptions":[{"text":"negative","correct":true,"explanation":"The bracket constants must have matching signs because $c$ is positive. Their sum must be negative because $b$ is negative, so both are negative."},{"text":"positive","correct":false,"explanation":"Two positive bracket constants would have a positive sum, making $b$ positive. As $b$ is negative, both matching signs must instead be negative."},{"text":"different","correct":false,"explanation":"Different signs produce a negative product, so $c$ would be negative. A positive $c$ requires matching signs, and a negative $b$ makes both negative."},{"text":"opposite","correct":false,"explanation":"Opposite signs are associated with a negative constant term because their product is negative. Here $c$ is positive, so both signs match and are negative."}]},{"id":"fff546ce-e126-c190-8740-69587ecd6745","cardType":"basic","points":1,"front":"If the constant term $c$ is negative, what does this indicate about the signs in the factorised brackets?","back":"The signs are different (one positive, one negative).","mcqOptions":[{"text":"The signs are different (one positive, one negative).","correct":true,"explanation":"The constant term is the product of the bracket constants. A negative product can only result when one constant is positive and the other is negative."},{"text":"The signs are the same (both positive or both negative).","correct":false,"explanation":"Constants with the same sign multiply to give a positive product. Since $c$ is negative, the bracket constants must instead have different signs."},{"text":"The signs are both positive in every case.","correct":false,"explanation":"Two positive constants have a positive product, so they would make $c$ positive. A negative constant requires one positive sign and one negative sign."},{"text":"The signs are both negative in every case.","correct":false,"explanation":"Multiplying two negative constants gives a positive result, not a negative one. A negative value of $c$ requires the two bracket signs to be different."}]},{"id":"62b3919e-5e23-d57a-563c-60b980f847fe","cardType":"basic","points":2,"front":"When factorising $y^2 + y - 30$, why is the larger factor ($6$) positive and the smaller factor ($5$) negative?","back":"Because the middle term ($+y$) is positive, so the factors must sum to $+1$.","mcqOptions":[{"text":"Because the middle term ($+y$) is positive, so the factors must sum to $+1$.","correct":true,"explanation":"The numbers must multiply to $-30$ and add to $+1$. The pair $+6$ and $-5$ satisfies both conditions, producing $(y+6)(y-5)$."},{"text":"Because the constant term ($-30$) is negative, so both factors must sum to $-1$.","correct":false,"explanation":"A negative constant determines that the signs differ, but not that the sum is negative. The coefficient of $y$ is $+1$, so the factors sum to $+1$."},{"text":"Because the middle term ($+y$) is positive, so the factors must multiply to $+1$.","correct":false,"explanation":"The middle coefficient comes from the sum of the bracket constants, not their product. The factors $+6$ and $-5$ multiply to $-30$ and sum to $+1$."},{"text":"Because the constant term ($-30$) is negative, so the larger factor must always be negative.","correct":false,"explanation":"A negative constant only requires opposite signs; it does not force the larger magnitude to be negative. Here the positive middle coefficient requires $+6-5=+1$."}]},{"id":"eaad3ed0-da06-f57c-b923-5c18cce96593","cardType":"cloze","points":null,"front":"To quickly check if a factorisation is correct, you should mentally [ ] the brackets.","back":"To quickly check if a factorisation is correct, you should mentally expand the brackets.","mcqOptions":[{"text":"expand","correct":true,"explanation":"Expanding reproduces the quadratic and checks the $x^2$ term, middle term and constant term. If all three match, the factorisation is correct."},{"text":"solve","correct":false,"explanation":"Solving finds the roots but does not directly verify every coefficient in the proposed factorisation. Mentally expanding the brackets confirms that the original quadratic is reproduced."},{"text":"simplify","correct":false,"explanation":"Simplifying alone does not specify the multiplication needed to recover the quadratic. The direct check is to expand the brackets and compare all resulting terms."},{"text":"differentiate","correct":false,"explanation":"Differentiation finds a gradient function and is unrelated to checking a factorisation. Expanding the brackets verifies whether they reproduce the original quadratic expression."}]},{"id":"2a26f3c1-ff4b-9133-48b4-912f127e4c2e","cardType":"basic","points":2,"front":"What does the zero product rule state regarding $(x+a)(x+b) = 0$?","back":"Either $x+a = 0$ or $x+b = 0$.","mcqOptions":[{"text":"Either $x+a = 0$ or $x+b = 0$.","correct":true,"explanation":"A product equals zero when at least one of its factors equals zero. Setting each bracket equal to zero therefore gives all possible solutions."},{"text":"Both $x+a = 0$ and $x+b = 0$.","correct":false,"explanation":"Requiring both factors to be zero is too restrictive; only one factor needs to be zero for the product to vanish. Each bracket must be considered separately."},{"text":"Either $x+a = 1$ or $x+b = 1$.","correct":false,"explanation":"The product is given as zero, so the zero product rule sets each possible factor equal to zero, not one. This produces the roots $x=-a$ and $x=-b$."},{"text":"The sum $x+a+x+b = 0$.","correct":false,"explanation":"The equation contains a product of factors, not their sum. The zero product rule requires $x+a=0$ or $x+b=0$, giving the two possible roots."}]},{"id":"e4faf512-a258-325f-23c6-82ba6ee6ea1f","cardType":"basic","points":1,"front":"What are the solutions or roots of a quadratic equation?","back":"The values of the variable that make the equation equal to zero.","mcqOptions":[{"text":"The values of the variable that make the equation equal to zero.","correct":true,"explanation":"A root is a variable value that makes the quadratic expression zero. Graphically, these values are the $x$-coordinates where the graph meets the $x$-axis."},{"text":"The values of the variable that make the equation equal to one.","correct":false,"explanation":"Roots are found by setting the quadratic equal to zero, not one. They identify where the graph meets the $x$-axis because the corresponding $y$-value is zero."},{"text":"The values of the variable that make the equation reach its maximum.","correct":false,"explanation":"A maximum is the turning point of a downward-opening quadratic and is not generally a root. Roots are the variable values that make the equation equal zero."},{"text":"The values of the variable that make the equation cross the y-axis.","correct":false,"explanation":"The graph crosses the $y$-axis when the input is zero, which gives the constant term. Roots instead occur where the output is zero, on the $x$-axis."}]},{"id":"307dcf5a-26d2-a2d3-6d40-0f13d3c05041","cardType":"basic","points":2,"front":"Solve the factorised equation $(x-4)(x+6)=0$.","back":"$$x = 4$ or $x = -6$","mcqOptions":[{"text":"$$x = 4$ or $x = -6$","correct":true,"explanation":"The zero-product rule means either bracket equals zero. Solving $x-4=0$ gives $x=4$, while solving $x+6=0$ gives $x=-6$."},{"text":"$$x = -4$ or $x = 6$","correct":false,"explanation":"This reverses both signs when solving the linear equations. From $x-4=0$ and $x+6=0$, the correct solutions are $x=4$ and $x=-6$."},{"text":"$$x = 4$ or $x = 6$","correct":false,"explanation":"The first root is correct, but $x+6=0$ requires subtracting $6$, giving $x=-6$. Therefore the solutions are $x=4$ or $x=-6$."},{"text":"$$x = -4$ or $x = -6$","correct":false,"explanation":"The second root is correct, but $x-4=0$ requires adding $4$, giving $x=4$. Therefore the solutions are $x=4$ or $x=-6$."}]},{"id":"c5663015-e1f1-3591-8659-ab4eb6317bc1","cardType":"cloze","points":null,"front":"Most quadratic equations solved by factorising will have [ ] distinct solutions.","back":"Most quadratic equations solved by factorising will have two distinct solutions.","mcqOptions":[{"text":"two","correct":true,"explanation":"A quadratic usually factorises into two linear brackets. Setting each bracket equal to zero normally produces two distinct solutions, although repeated roots are possible."},{"text":"one","correct":false,"explanation":"One solution occurs only when both factors produce the same repeated root. Most factorised quadratic equations have two linear factors giving two distinct solutions."},{"text":"three","correct":false,"explanation":"Three distinct solutions would generally require a cubic equation with three linear factors. A quadratic has degree two, so it normally has two distinct solutions."},{"text":"four","correct":false,"explanation":"Four distinct solutions would require an equation of at least degree four. A quadratic has only two linear factors and normally gives two distinct solutions."}]},{"id":"cdd51307-a853-ac50-920f-babe75f2d8b4","cardType":"basic","points":1,"front":"What must be done to the equation $x^2 + bx + c = k$ before it can be solved by factorising?","back":"It must be rearranged into the form $x^2 + bx + c = 0$.","mcqOptions":[{"text":"It must be rearranged into the form $x^2 + bx + c = 0$.","correct":true,"explanation":"The zero-product rule can be applied after one side has been made zero. Subtracting $k$ from both sides produces a quadratic expression equal to zero for factorising."},{"text":"It must be rearranged into the form $x^2 + bx + c = 1$.","correct":false,"explanation":"Factorising an expression equal to one does not allow the zero-product rule. The equation must be rearranged so the quadratic expression equals $0$."},{"text":"It must be rearranged into the form $x^2 + bx + c = kx$.","correct":false,"explanation":"Moving the constant to a term involving $x$ does not put the equation in the required zero form. All terms must be collected on one side, equal to $0$."},{"text":"It must be rearranged into the form $x^2 + bx = 0$.","correct":false,"explanation":"The constant term cannot simply be removed unless it cancels through valid rearrangement. The complete quadratic must be collected into the form $x^2+bx+c=0$."}]},{"id":"03c9cf48-2155-ea72-1c04-4630e94ee95b","cardType":"basic","points":3,"front":"Outline the steps to solve $r^2 - 5r - 14 = 0$.","back":"

Factorise: $(r-7)(r+2) = 0$
Set each bracket to zero: $r-7=0$ and $r+2=0$
Solve: $r=7, r=-2$

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

Factorise: $(r-7)(r+2) = 0$
Set each bracket to zero: $r-7=0$ and $r+2=0$
Solve: $r=7, r=-2$

","correct":true,"explanation":"The factors expand to $r^2-5r-14$. Applying the zero-product rule to each bracket gives the two correct roots, $r=7$ and $r=-2$."},{"text":"

Factorise: $(r-7)(r-2) = 0$
Set each bracket to zero: $r-7=0$ and $r-2=0$
Solve: $r=7, r=2$

","correct":false,"explanation":"The factors use two negative constants, so their product is $+14$ rather than $-14$. The correct factorisation is $(r-7)(r+2)=0$, giving $r=7,-2$."},{"text":"

Factorise: $(r+7)(r-2) = 0$
Set each bracket to zero: $r+7=0$ and $r-2=0$
Solve: $r=-7, r=2$

","correct":false,"explanation":"These factors expand to $r^2+5r-14$, so the middle term has the wrong sign. The correct factors are $(r-7)(r+2)$, giving $r=7,-2$."},{"text":"

Factorise: $(r-7)(r+2) = 0$
Set each bracket to zero: $r-7=0$ and $r+2=0$
Solve: $r=-7, r=2$

","correct":false,"explanation":"The factorisation is correct, but both linear equations have been solved with reversed signs. They give $r=7$ and $r=-2$, not $r=-7$ and $r=2$."}]},{"id":"1647fd9d-8bf4-4d8a-c281-7c16d4600365","cardType":"cloze","points":null,"front":"If the constant term in a quadratic is 18 and the middle term is $-9x$, the numbers in the brackets must be $\\text{[...]}$ and $-6$.","back":"If the constant term in a quadratic is 18 and the middle term is $-9x$, the numbers in the brackets must be $\\boxed{-3}$ and $-6$.","mcqOptions":[{"text":"$$-3$","correct":true,"explanation":"The bracket numbers must multiply to $18$ and add to $-9$. The pair $-3$ and $-6$ satisfies both conditions because their product is $18$ and sum is $-9$."},{"text":"$$3$","correct":false,"explanation":"Using $3$ with $-6$ gives a product of $-18$ and a sum of $-3$. The required number is $-3$, which gives product $18$ and sum $-9$."},{"text":"$$-12$","correct":false,"explanation":"Using $-12$ with $-6$ gives a product of $72$ and a sum of $-18$. The correct missing number is $-3$, producing product $18$ and sum $-9$."},{"text":"$$-2$","correct":false,"explanation":"Using $-2$ with $-6$ gives a product of $12$ and a sum of $-8$. The correct missing number is $-3$, producing product $18$ and sum $-9$."}]},{"id":"fdeb04ea-c9e4-0b61-caf1-1b2d75ade0d2","cardType":"cloze","points":null,"front":"If the constant term in a quadratic is 18 and the middle term is $-9x$, the numbers in the brackets must be $-3$ and $\\text{[...]}$.","back":"If the constant term in a quadratic is 18 and the middle term is $-9x$, the numbers in the brackets must be $-3$ and $\\boxed{-6}$.","mcqOptions":[{"text":"$$-6$","correct":true,"explanation":"The bracket numbers must multiply to $18$ and add to $-9$. The pair $-3$ and $-6$ satisfies both conditions because their product is $18$ and sum is $-9$."},{"text":"$$6$","correct":false,"explanation":"Using $6$ with $-3$ gives a product of $-18$ and a sum of $3$. The required number is $-6$, which gives product $18$ and sum $-9$."},{"text":"$$-3$","correct":false,"explanation":"Repeating $-3$ gives a product of $9$ and a sum of $-6$, so neither coefficient matches. The correct second number is $-6$."},{"text":"$$-9$","correct":false,"explanation":"Using $-9$ with $-3$ gives a product of $27$ and a sum of $-12$. The correct second number is $-6$, producing product $18$ and sum $-9$."}]},{"id":"51d4dd33-0e9b-a4dd-cd61-eea61dfa6742","cardType":"basic","points":2,"front":"Why is the pair $1$ and $8$ incorrect for factorising $x^2 + 6x + 8$?","back":"They multiply to $8$ but add to 9, not $6$.","mcqOptions":[{"text":"They multiply to $8$ but add to 9, not $6$.","correct":true,"explanation":"For $x^2+6x+8$, the bracket numbers must multiply to $8$ and add to $6$. Although $1\\times8=8$, their sum is $9$, so the pair fails."},{"text":"They multiply to $8$ but add to 8, not $6$.","correct":false,"explanation":"This makes an arithmetic error when adding the proposed numbers: $1+8=9$, not $8$. The pair is incorrect because it adds to $9$ rather than the required $6$."},{"text":"They add to $9$ but multiply to 9, not $8$.","correct":false,"explanation":"This confuses the sum with the product: $1\\times8=8$, so the product is correct. The pair fails because its sum is $9$ rather than $6$."},{"text":"They multiply to $6$ but add to 9, not $8$.","correct":false,"explanation":"The required product and sum have been interchanged and miscalculated. The numbers actually multiply to $8$ and add to $9$; they are wrong because the required sum is $6$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Component 1"},"heading":"Solving Quadratics","title":"Solving Quadratics Flashcards","topbar":{"backHref":"/gcse/maths/eduqas/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"56671f92-be0f-46f0-9806-cb344bf18453","walled":"$L38"}],"$L39","$L3a"]