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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/as-level/maths/edexcel/as-pure-mathematics-7-algebraic-methods/flashcards","label":"Algebraic Methods","subjectCode":"as_level_maths","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"4e88eccf-d66b-4fa7-a18c-5e9d731f293d","title":"Algebraic Methods Flashcards","topicId":"987aba14-274c-4b8b-8dc2-774400179a63","topicSlug":"as-pure-mathematics-7-algebraic-methods","cards":[{"id":"bf77a449-da09-463e-996b-1c29891cdfa0","cardType":"basic","points":1,"front":"In the polynomial $f(x) = 3x^3 - 2x + 5$, what is the coefficient of $x^3$?","back":"3","mcqOptions":[{"text":"3","correct":true,"explanation":"The coefficient is the numerical multiplier of the term. Since the cubic term is $3x^3$, the coefficient of $x^3$ is 3."},{"text":"-2","correct":false,"explanation":"The value -2 is the coefficient of $x$, not $x^3$. The cubic term is $3x^3$, so the required coefficient is 3."},{"text":"5","correct":false,"explanation":"The value 5 is the constant term because it has no variable attached. The term containing $x^3$ is $3x^3$, whose coefficient is 3."},{"text":"1","correct":false,"explanation":"A coefficient of 1 would apply if the term were written simply as $x^3$. Here the term is $3x^3$, so its coefficient is 3."}]},{"id":"5d5e601a-fafc-419d-b32e-ec4c992f0162","cardType":"cloze","points":null,"front":"A root (or zero) of $f(x)$ is a value of $x$ that makes $\\text{[...]}$.","back":"A root (or zero) of $f(x)$ is a value of $x$ that makes $\\boxed{f(x) = 0}$.","mcqOptions":[{"text":"$$f(x) = 0$","correct":true,"explanation":"A root is an input at which the polynomial’s output equals zero. Therefore, a value of $x$ is a root precisely when $f(x) = 0$."},{"text":"$$f(x) = x$","correct":false,"explanation":"This condition describes an intersection with the line $y=x$, not necessarily a root. A root occurs where the graph meets the $x$-axis, so $f(x) = 0$."},{"text":"$$f(0) = x$","correct":false,"explanation":"This confuses the input with the output and evaluates the function only at zero. A root is a value of $x$ for which $f(x) = 0$."},{"text":"$$f(0) = 0$","correct":false,"explanation":"This only states that zero itself is a root, which need not be true. The general condition for any value $x$ to be a root is $f(x) = 0$."}]},{"id":"7c2db852-5137-435f-9cfc-3181ef2149ab","cardType":"basic","points":2,"front":"What happens to the sign of a negative value like $x = -3$ when it is squared vs cubed?","back":"

$(-3)^2$ becomes positive (9)
$(-3)^3$ stays negative (-27)

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

$(-3)^2$ becomes positive (9)
$(-3)^3$ stays negative (-27)

","correct":true,"explanation":"An even power pairs the negative factors, giving a positive result, while an odd power leaves one unpaired negative factor. Thus the square is 9 and the cube is -27."},{"text":"

$(-3)^2$ stays negative (-9)
$(-3)^3$ stays negative (-27)

","correct":false,"explanation":"The square is not negative because multiplying two negative factors gives a positive result. Therefore $(-3)^2=9$, while the odd power $(-3)^3=-27$."},{"text":"

$(-3)^2$ becomes positive (9)
$(-3)^3$ becomes positive (27)

","correct":false,"explanation":"The cube has three negative factors, so its result remains negative rather than becoming positive. The correct values are $(-3)^2=9$ and $(-3)^3=-27$."},{"text":"

$(-3)^2$ becomes positive (6)
$(-3)^3$ stays negative (-9)

","correct":false,"explanation":"This multiplies the base by the exponent instead of using repeated multiplication. Powers give $(-3)^2=(-3)(-3)=9$ and $(-3)^3=(-3)(-3)(-3)=-27$."}]},{"id":"ee636513-3474-4c83-99d7-c7c410b64d29","cardType":"cloze","points":null,"front":"According to the factor theorem, $(x - a)$ is a factor of $f(x)$ if and only if $\\text{[...]}$.","back":"According to the factor theorem, $(x - a)$ is a factor of $f(x)$ if and only if $\\boxed{f(a) = 0}$.","mcqOptions":[{"text":"$$f(a) = 0$","correct":true,"explanation":"Substituting $x=a$ makes the factor $(x-a)$ equal zero. The factor theorem therefore states that $(x-a)$ is a factor exactly when $f(a)=0$."},{"text":"$$f(-a) = 0$","correct":false,"explanation":"Using $-a$ reverses the sign and corresponds to testing the factor $(x+a)$. For the factor $(x-a)$, the factor theorem requires $f(a)=0$."},{"text":"$$f(a) = a$","correct":false,"explanation":"This condition concerns points where the graph meets the line $y=x$, not polynomial factors. The factor $(x-a)$ occurs when substituting $a$ gives $f(a)=0$."},{"text":"$$f(0) = a$","correct":false,"explanation":"Evaluating at zero gives the polynomial’s constant term and does not test $(x-a)$. The relevant substitution is $x=a$, and the required condition is $f(a)=0$."}]},{"id":"e5e5e739-742b-416f-b8ea-17e2ed7e2d47","cardType":"basic","points":1,"front":"To test if $(2x + 3)$ is a factor of a polynomial, what value of $x$ should be substituted?","back":"$$x = -\\frac{3}{2}$","mcqOptions":[{"text":"$$x = -\\frac{3}{2}$","correct":true,"explanation":"Set the proposed linear factor equal to zero: $2x+3=0$. Solving gives $2x=-3$, so the value required by the factor theorem is $x=-\\frac{3}{2}$."},{"text":"$$x = \\frac{3}{2}$","correct":false,"explanation":"This sign error comes from moving 3 across the equals sign without changing its sign. Solving $2x+3=0$ correctly gives $x=-\\frac{3}{2}$."},{"text":"$$x = -\\frac{2}{3}$","correct":false,"explanation":"This incorrectly takes the reciprocal after solving the linear equation. From $2x=-3$, division by 2 gives the required test value $x=-\\frac{3}{2}$."},{"text":"$$x = \\frac{2}{3}$","correct":false,"explanation":"This both reverses the fraction and loses the negative sign. Setting $2x+3=0$ and solving directly gives the correct substitution $x=-\\frac{3}{2}$."}]},{"id":"44a37ebd-7d3d-4131-8ed7-7b4952002ee8","cardType":"basic","points":1,"front":"If $f(x) = (x - 2)(Ax^2 + Bx + C)$, how are $A$, $B$, and $C$ typically found at AS level?","back":"Comparing coefficients after expanding the right-hand side.","mcqOptions":[{"text":"Comparing coefficients after expanding the right-hand side.","correct":true,"explanation":"After expansion, coefficients of matching powers of $x$ must be equal on both sides of the identity. Equating them produces simultaneous equations for $A$, $B$, and $C$."},{"text":"Completing the square after expanding the right-hand side.","correct":false,"explanation":"Completing the square rewrites a quadratic but does not directly identify all three unknown coefficients in this polynomial identity. Expanding and comparing coefficients gives equations for $A$, $B$, and $C$."},{"text":"Using the quadratic formula after expanding the right-hand side.","correct":false,"explanation":"The quadratic formula finds roots of a quadratic with known coefficients; it does not determine unknown coefficients in an identity. Instead, expand and compare coefficients of corresponding powers of $x$."},{"text":"Differentiating both sides after expanding the right-hand side.","correct":false,"explanation":"Differentiation changes the polynomial identity and is unnecessary for finding these constants. At AS level, the standard method is to expand, then compare coefficients of matching powers of $x$."}]},{"id":"88e98304-ca6a-4bd8-a266-7d2c33703094","cardType":"basic","points":1,"front":"Where should you look for possible rational roots of an integer cubic polynomial?","back":"Factors of the constant term divided by factors of the leading coefficient.","mcqOptions":[{"text":"Factors of the constant term divided by factors of the leading coefficient.","correct":true,"explanation":"The rational root theorem states that a rational root $\\frac{p}{q}$ in lowest terms has $p$ dividing the constant term and $q$ dividing the leading coefficient."},{"text":"Factors of the leading coefficient divided by factors of the constant term.","correct":false,"explanation":"This reverses the numerator and denominator in the rational root theorem. Possible roots have factors of the constant term on top and factors of the leading coefficient below."},{"text":"Factors of the linear coefficient divided by factors of the leading coefficient.","correct":false,"explanation":"The numerator comes from factors of the constant term, not the coefficient of $x$. The denominator comes from factors of the leading coefficient."},{"text":"Factors of the constant term divided by factors of the linear coefficient.","correct":false,"explanation":"The denominator must divide the leading coefficient, not the coefficient of $x$. Possible rational roots are factors of the constant term divided by factors of the leading coefficient."}]},{"id":"194a2eb2-abe7-4393-b6e1-dfe8f90b3355","cardType":"cloze","points":null,"front":"If a quadratic $ax^2 + bx + c$ has no real roots, its discriminant $\\text{[...]}$ must be $< 0$.","back":"If a quadratic $ax^2 + bx + c$ has no real roots, its discriminant $\\boxed{b^2 - 4ac}$ must be $< 0$.","mcqOptions":[{"text":"$$b^2 - 4ac$","correct":true,"explanation":"The discriminant of $ax^2+bx+c$ is $b^2-4ac$. A negative discriminant makes the square root in the quadratic formula non-real, so there are no real roots."},{"text":"$$b^2 + 4ac$","correct":false,"explanation":"The sign before $4ac$ is negative, as follows from the quadratic formula. The discriminant is $b^2-4ac$, and it must be negative for no real roots."},{"text":"$$b - 4ac$","correct":false,"explanation":"This omits the square on $b$, a common recall error. The discriminant is $b^2-4ac$, whose sign determines the number of real roots."},{"text":"$$b^2 - 2ac$","correct":false,"explanation":"The factor multiplying $ac$ is 4, not 2, because the quadratic formula contains $\\sqrt{b^2-4ac}$. No real roots require $b^2-4ac<0$."}]},{"id":"a3a7ab71-a5fd-4e89-94a7-cf2f32bce031","cardType":"cloze","points":null,"front":"If a quadratic $ax^2 + bx + c$ has no real roots, its discriminant $b^2 - 4ac$ must be $\\text{[...]}$.","back":"If a quadratic $ax^2 + bx + c$ has no real roots, its discriminant $b^2 - 4ac$ must be $\\boxed{< 0}$.","mcqOptions":[{"text":"$$< 0$","correct":true,"explanation":"A quadratic has no real roots when its graph does not meet the $x$-axis. This occurs precisely when the discriminant $b^2 - 4ac$ is strictly negative."},{"text":"$$= 0$","correct":false,"explanation":"A zero discriminant gives one repeated real root, where the graph touches the $x$-axis. No real roots require $b^2 - 4ac < 0$."},{"text":"$$> 0$","correct":false,"explanation":"A positive discriminant gives two distinct real roots, so the graph crosses the $x$-axis twice. No real roots require the discriminant to be less than zero."},{"text":"$$e 0$","correct":false,"explanation":"A non-negative discriminant allows at least one real root, including a repeated root when it equals zero. No real roots require the strict condition $b^2 - 4ac < 0$."}]},{"id":"40098186-8394-42ac-a5b5-f837431880c5","cardType":"basic","points":1,"front":"How does a repeated root appear on a cubic graph?","back":"The graph touches the $x$-axis and turns around (does not cross).","mcqOptions":[{"text":"The graph touches the $x$-axis and turns around (does not cross).","correct":true,"explanation":"A repeated root has even multiplicity, so the cubic reaches the $x$-axis and reverses direction instead of changing from positive to negative or vice versa."},{"text":"The graph crosses the $x$-axis and continues through it.","correct":false,"explanation":"Crossing the $x$-axis indicates a root of odd multiplicity, usually a simple root. A repeated root has even multiplicity, so the graph touches the axis and turns around."},{"text":"The graph touches the $x$-axis but continues without turning around.","correct":false,"explanation":"Touching without turning describes a stationary point of inflection more closely, not a repeated root. A repeated root makes the graph touch the axis and reverse direction."},{"text":"The graph has a vertical asymptote at the repeated root.","correct":false,"explanation":"Polynomials such as cubics have no vertical asymptotes because they are defined for every real $x$. A repeated root is shown by touching the $x$-axis and turning around."}]},{"id":"d3a1666a-f0e6-454c-afa3-fa8d94fe370a","cardType":"basic","points":2,"front":"How do you show a cubic equation has exactly one real root?","back":"

Show one linear factor $(x - a)$ gives $f(a) = 0$
Show the remaining quadratic factor has a negative discriminant

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

Show one linear factor $(x - a)$ gives $f(a) = 0$
Show the remaining quadratic factor has a negative discriminant

","correct":true,"explanation":"The linear factor supplies one real root. A negative discriminant means the remaining quadratic has no real roots, so no further real roots exist and the cubic has exactly one."},{"text":"

Show one linear factor $(x - a)$ gives $f(a) = 0$
Show the remaining quadratic factor has a positive discriminant

","correct":false,"explanation":"A positive quadratic discriminant gives two further real roots, producing three real roots altogether. The remaining quadratic must have a negative discriminant to contribute none."},{"text":"

Show one linear factor $(x - a)$ gives $f(a) = 0$
Show the remaining quadratic factor has a zero discriminant

","correct":false,"explanation":"A zero quadratic discriminant gives a repeated real root, so the cubic has another real root besides $a$. Exactly one real root requires a negative discriminant."},{"text":"

Show two linear factors $(x - a)(x - b)$ give $f(a) = f(b) = 0$
Show the remaining linear factor has a negative discriminant

","correct":false,"explanation":"A linear factor cannot have a discriminant, and two distinct linear factors already give two real roots. The valid method uses one linear factor and a remaining quadratic with negative discriminant."}]},{"id":"fd628800-b2f9-42cc-a486-71557c502eb4","cardType":"cloze","points":null,"front":"In algebraic proofs, an even integer is written as $\\text{[...]}$ and an odd integer as $2n + 1$.","back":"In algebraic proofs, an even integer is written as $\\boxed{2n}$ and an odd integer as $2n + 1$.","mcqOptions":[{"text":"$$2n$","correct":true,"explanation":"Every even integer is divisible by 2, so it can be written as $2n$ for some integer $n$. The corresponding odd form is $2n + 1$."},{"text":"$$2n + 1$","correct":false,"explanation":"This is the standard form for an odd integer, not an even one, because it is one more than a multiple of 2. An even integer is written as $2n$."},{"text":"$$2n + 2$","correct":false,"explanation":"Although $2n + 2$ is always even, it is not the standard general form used here and restricts the representation unnecessarily. The required form is $2n$."},{"text":"$$n^2$","correct":false,"explanation":"A square can be even or odd depending on whether $n$ is even or odd, so $n^2$ does not always represent an even integer. The general form is $2n$."}]},{"id":"d5333c93-cc18-4ba8-8d36-07ce07bc35b0","cardType":"cloze","points":null,"front":"In algebraic proofs, an even integer is written as $2n$ and an odd integer as $\\text{[...]}$.","back":"In algebraic proofs, an even integer is written as $2n$ and an odd integer as $\\boxed{2n + 1}$.","mcqOptions":[{"text":"$$2n + 1$","correct":true,"explanation":"An odd integer is one more than an even multiple of 2, so its general form is $2n + 1$ for an integer $n$. This includes every odd integer."},{"text":"$$2n$","correct":false,"explanation":"This expression is always even because it is divisible by 2. The general form for an odd integer is $2n + 1$, which is one greater than an even number."},{"text":"$$2n + 2$","correct":false,"explanation":"This expression is also always even, since both terms are divisible by 2. Odd integers must be represented by an even multiple of 2 plus or minus 1, such as $2n + 1$."},{"text":"$$n^2 + 1$","correct":false,"explanation":"Its parity varies with $n$: it is even when $n$ is odd and odd when $n$ is even. The expression that always represents an odd integer is $2n + 1$."}]},{"id":"18cefa81-b18f-4309-bb68-47714f555ee2","cardType":"basic","points":2,"front":"How do you disprove the claim: \"$n^2 - n + 3$ is prime for every positive integer $n$\"?","back":"Find a counterexample (e.g., if $n=3$, $3^2 - 3 + 3 = 9$, which is not prime).","mcqOptions":[{"text":"Find a counterexample (e.g., if $n=3$, $3^2 - 3 + 3 = 9$, which is not prime).","correct":true,"explanation":"A universal claim is disproved by one valid counterexample. Taking $n=3$ gives $9$, which is composite, so the expression is not prime for every positive integer."},{"text":"Check that $n^2 - n + 3$ is prime for several small positive integers.","correct":false,"explanation":"Testing several values can suggest a pattern but cannot prove a statement for every positive integer. One counterexample, such as $n=3$ giving $9$, is enough to disprove the claim."},{"text":"Show that $n^2 - n + 3$ is positive for every positive integer $n$.","correct":false,"explanation":"Being positive does not imply being prime; many positive integers are composite. For example, $n=3$ gives $9$, so a counterexample is needed to disprove the claim."},{"text":"Factorise $n^2 - n + 3$ and show that both factors are prime.","correct":false,"explanation":"This does not provide a valid disproof and the expression does not generally factorise into useful integer factors. The claim is disproved directly by the counterexample $n=3$, which gives $9$."}]},{"id":"a74139e2-22ea-48d0-b7ff-de2d9aec19dd","cardType":"basic","points":3,"front":"Prove that the sum of two neighbouring odd integers is a multiple of 4.","back":"

Let integers be $2n + 1$ and $2n + 3$
Sum: $(2n + 1) + (2n + 3) = 4n + 4$
Factorise: $4(n + 1)$, which is a multiple of 4

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

Let integers be $2n + 1$ and $2n + 3$
Sum: $(2n + 1) + (2n + 3) = 4n + 4$
Factorise: $4(n + 1)$, which is a multiple of 4

","correct":true,"explanation":"Neighbouring odd integers differ by 2, so they can be written as $2n + 1$ and $2n + 3$. Their sum is $4n + 4 = 4(n + 1)$, divisible by 4."},{"text":"

Let integers be $2n$ and $2n + 2$
Sum: $(2n) + (2n + 2) = 4n + 2$
Factorise: $2(2n + 1)$, which is a multiple of 4

","correct":false,"explanation":"These are neighbouring even integers, not odd integers, and their sum is $4n + 2$, which is not divisible by 4. Odd neighbours must be represented as $2n + 1$ and $2n + 3$."},{"text":"

Let integers be $2n + 1$ and $2n + 2$
Sum: $(2n + 1) + (2n + 2) = 4n + 3$
Factorise: $4(n + 3)$, which is a multiple of 4

","correct":false,"explanation":"The second expression is even, so the pair is not two neighbouring odd integers. Also, $4n + 3$ cannot equal $4(n + 3)$; the correct sum is $4n + 4 = 4(n + 1)$."},{"text":"

Let integers be $2n + 1$ and $2n + 5$
Sum: $(2n + 1) + (2n + 5) = 4n + 6$
Factorise: $2(2n + 3)$, which is a multiple of 4

","correct":false,"explanation":"These odd integers differ by 4, so they are not neighbouring odd integers. Their sum is $4n + 6$, which is divisible by 2 but not always by 4; the correct pair differs by 2."}]},{"id":"369834d4-679c-4f63-97dc-3b901fda102e","cardType":"basic","points":1,"front":"When proving a quadratic inequality is positive for all real $x$, what technique is most effective?","back":"Completing the square.","mcqOptions":[{"text":"Completing the square.","correct":true,"explanation":"Completing the square rewrites the quadratic in a form such as $(x - p)^2 + q$, making its minimum or maximum value clear and allowing positivity for all real $x$ to be established directly."},{"text":"Factorising.","correct":false,"explanation":"Factorising can identify roots and sign intervals, but it is less direct when proving positivity for every real $x$. Completing the square exposes the quadratic's minimum value immediately."},{"text":"Using the quadratic formula.","correct":false,"explanation":"The quadratic formula can locate roots, but it does not directly show the expression's minimum value. Completing the square is the most effective method for proving positivity for all real $x$."},{"text":"Substituting several positive values of $x$.","correct":false,"explanation":"Checking selected values cannot establish positivity for every real $x$, especially negative or non-integer values. Completing the square gives an algebraic argument covering the entire real domain."}]},{"id":"25198694-3e5e-4fe2-b893-56360f7bdf58","cardType":"cloze","points":null,"front":"To show $x^2 + 4x + 7 > 0$ for all $x$, complete the square to get $\\text{[...]}$.","back":"To show $x^2 + 4x + 7 > 0$ for all $x$, complete the square to get $\\boxed{(x + 2)^2 + 3}$.","mcqOptions":[{"text":"$$(x + 2)^2 + 3$","correct":true,"explanation":"Expanding $(x + 2)^2 + 3$ gives $x^2 + 4x + 4 + 3 = x^2 + 4x + 7$, so this is the correct completed-square form."},{"text":"$$(x + 2)^2 - 3$","correct":false,"explanation":"The constant is reduced instead of increased: expanding gives $x^2 + 4x + 1$. The correct form is $(x + 2)^2 + 3$, which expands to $x^2 + 4x + 7$."},{"text":"$$(x + 4)^2 + 7$","correct":false,"explanation":"The bracket would produce a linear term of $8x$, not $4x$. The correct form is $(x + 2)^2 + 3$, whose expansion matches the original expression."},{"text":"$$(x - 2)^2 + 3$","correct":false,"explanation":"Using $x - 2$ changes the linear term to $-4x$. The correct form is $(x + 2)^2 + 3$, which produces the required positive $4x$ term."}]},{"id":"a70296ef-ffd6-4475-874f-ac92bbc30fda","cardType":"basic","points":2,"front":"Why does $(x + 2)^2 + 3 > 0$ prove the expression is always positive?","back":"Because $(x + 2)^2 \\ge 0$ for all real $x$, so adding 3 makes the result strictly greater than zero.","mcqOptions":[{"text":"Because $(x + 2)^2 \\ge 0$ for all real $x$, so adding 3 makes the result strictly greater than zero.","correct":true,"explanation":"A square cannot be negative, so $(x + 2)^2$ is at least zero. Adding 3 therefore gives a minimum value of 3, which is strictly positive."},{"text":"Because $(x + 2)^2 > 0$ for all real $x$, so adding 3 makes the result strictly greater than zero.","correct":false,"explanation":"A square can equal zero when $x = -2$, so it is not always strictly positive. The correct reasoning uses $(x + 2)^2 ge 0$, then adding 3 ensures a positive result."},{"text":"Because $(x + 2)^2 ge 0$ for all real $x$, so subtracting 3 makes the result strictly greater than zero.","correct":false,"explanation":"Subtracting 3 could produce a negative value when the square is small. The expression adds 3 to a non-negative square, giving a minimum of 3 and proving it is strictly positive."},{"text":"Because $(x + 2)^2 le 0$ for all real $x$, so adding 3 makes the result strictly greater than zero.","correct":false,"explanation":"Squares are never less than or equal to zero; they are always greater than or equal to zero. That fact, followed by adding 3, proves the expression is strictly positive."}]},{"id":"9a90a7d2-3a6c-4eec-a1cc-4cba17e5b3d0","cardType":"basic","points":1,"front":"What is the danger of dividing an inequality by a variable expression?","back":"The sign of the expression might be negative, which would reverse the inequality symbol.","mcqOptions":[{"text":"The sign of the expression might be negative, which would reverse the inequality symbol.","correct":true,"explanation":"Dividing an inequality by a negative quantity reverses its direction, while dividing by a positive quantity does not. A variable expression may have either sign, so its sign must be known."},{"text":"The value of the expression might be zero, which would reverse the inequality symbol.","correct":false,"explanation":"Division by zero is undefined, but zero does not reverse an inequality because it cannot be used as a divisor. The inequality reverses specifically when the divisor is negative."},{"text":"The sign of the expression might be positive, which would reverse the inequality symbol.","correct":false,"explanation":"Dividing by a positive expression leaves the inequality direction unchanged. The symbol reverses only when the variable expression is negative, so its sign must be checked."},{"text":"The coefficient of the expression might be negative, which would reverse the inequality symbol.","correct":false,"explanation":"It is the sign of the entire divisor, not just a coefficient viewed separately, that determines whether the symbol reverses. A negative divisor reverses the inequality."}]},{"id":"f2b517f9-d3ad-4f95-83f3-c4e6032b5d93","cardType":"basic","points":1,"front":"If $n$ is an even integer ($n = 2k$), what is the algebraic form of $n^2 + 1$?","back":"$$4k^2 + 1$","mcqOptions":[{"text":"$$4k^2 + 1$","correct":true,"explanation":"Substituting $n = 2k$ gives $n^2 + 1 = (2k)^2 + 1 = 4k^2 + 1$, using the square of the coefficient correctly."},{"text":"$$2k^2 + 1$","correct":false,"explanation":"This squares $k$ but fails to square the coefficient 2. Since $(2k)^2 = 4k^2$, the correct form is $4k^2 + 1$."},{"text":"$$4k^2 + 2$","correct":false,"explanation":"The added 1 is unchanged when substituting for $n$; it does not become 2. Therefore $(2k)^2 + 1$ simplifies to $4k^2 + 1$."},{"text":"$$4k + 1$","correct":false,"explanation":"Squaring $2k$ requires squaring both the coefficient and $k$, giving $(2k)^2 = 4k^2$. The correct expression is therefore $4k^2 + 1$."}]},{"id":"9a462245-76e5-4613-9979-6c5e5a0296d6","cardType":"basic","points":1,"front":"How do you show that $4k^2 + 1$ is not a multiple of 4?","back":"It is one more than a multiple of 4 ($4(k^2) + 1$).","mcqOptions":[{"text":"It is one more than a multiple of 4 ($4(k^2) + 1$).","correct":true,"explanation":"The term $4k^2$ is divisible by 4, so it is a multiple of 4. Adding 1 leaves remainder 1, proving $4k^2 + 1$ is not divisible by 4."},{"text":"It is one less than a multiple of 4 ($4(k^2) + 1$).","correct":false,"explanation":"Adding 1 makes the number one more than the multiple $4k^2$, not one less. Its remainder on division by 4 is 1, so the correct description is one more."},{"text":"It is two more than a multiple of 4 ($4(k^2) + 1$).","correct":false,"explanation":"The final term is 1, not 2, so the expression differs from $4k^2$ by exactly one. Hence it is one more than a multiple of 4."},{"text":"It is equal to a multiple of 4 ($4(k^2) + 1$).","correct":false,"explanation":"The added 1 means the expression is not divisible by 4. It is one more than the multiple $4k^2$, leaving remainder 1 rather than remainder 0."}]},{"id":"a2feeb2a-a9a6-4386-acb1-bdd90a4ab83c","cardType":"cloze","points":null,"front":"A [ ] shows a statement is true for all cases, while a counterexample shows it is false for one case.","back":"A proof shows a statement is true for all cases, while a counterexample shows it is false for one case.","mcqOptions":[{"text":"proof","correct":true,"explanation":"A proof establishes that a mathematical statement follows logically in every permitted case, whereas a counterexample is enough to disprove a universal statement for one case."},{"text":"conjecture","correct":false,"explanation":"A conjecture is a statement believed to be true but not yet proved. The term that establishes truth for all cases is proof, while a counterexample disproves one."},{"text":"calculation","correct":false,"explanation":"A calculation may verify a particular example but does not establish a statement for every case. The correct term for a general justification is proof."},{"text":"pattern","correct":false,"explanation":"A pattern can suggest a rule from several examples, but it does not prove the rule holds universally. A proof is required to establish truth for all cases."}]},{"id":"c20d0d7e-104d-40c9-a519-cfdcf78db126","cardType":"cloze","points":null,"front":"A proof shows a statement is true for all cases, while a [ ] shows it is false for one case.","back":"A proof shows a statement is true for all cases, while a counterexample shows it is false for one case.","mcqOptions":[{"text":"counterexample","correct":true,"explanation":"A counterexample is one valid case for which a proposed statement fails. Therefore it can disprove a claim that is supposed to be true for all cases."},{"text":"proof","correct":false,"explanation":"A proof establishes a statement for all permitted cases rather than showing one failure. The single case that disproves a universal claim is called a counterexample."},{"text":"conjecture","correct":false,"explanation":"A conjecture is an unproved proposed statement, not a specific case showing that a statement is false. The required term is counterexample."},{"text":"assumption","correct":false,"explanation":"An assumption is something taken as true when starting an argument, whereas a counterexample is a particular case that contradicts a universal statement."}]}]}],"examCountdown":{"examDate":"2027-05-13","paperLabel":"Paper 1"},"heading":"Algebraic Methods","title":"Algebraic Methods Flashcards","topbar":{"backHref":"/as-level/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"987aba14-274c-4b8b-8dc2-774400179a63","walled":"$L39"}],"$L3a","$L3b"]