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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/as-level/maths/edexcel/as-pure-mathematics-7-3-the-factor-theorem/flashcards","label":"7.3 The Factor Theorem","subjectCode":"as_level_maths","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"07e949eb-28da-4061-9b81-88a37199d15d","title":"7.3 The Factor Theorem Flashcards","topicId":"7e8b67b9-9739-460b-97d1-21c8d0a702fc","topicSlug":"as-pure-mathematics-7-3-the-factor-theorem","cards":[{"id":"5bb0a30e-d667-4121-a991-43c717c74e13","cardType":"basic","points":1,"front":"How is the degree of a polynomial determined?","back":"By the highest power of the variable.","mcqOptions":[{"text":"By the highest power of the variable.","correct":true,"explanation":"The degree of a polynomial is the greatest exponent of its variable among terms with non-zero coefficients. For example, a polynomial containing an $x^5$ term has degree $5$."},{"text":"By the lowest power of the variable.","correct":false,"explanation":"Confusing the degree with the smallest exponent gives the wrong classification. The degree is determined by the highest power of the variable with a non-zero coefficient."},{"text":"By the number of terms in the expression.","correct":false,"explanation":"Counting terms identifies whether an expression is a monomial, binomial or trinomial, not its degree. The degree is determined by the highest power of the variable."},{"text":"By the largest coefficient in the expression.","correct":false,"explanation":"Coefficients affect the size and shape of the polynomial but do not determine its degree. The degree is given by the highest power of the variable."}]},{"id":"75264034-5be3-487a-88e0-567efd672918","cardType":"basic","points":1,"front":"What is the relationship between a root ($x=a$) and a factor of a polynomial?","back":"If $x=a$ is a root, then $(x-a)$ is a factor.","mcqOptions":[{"text":"If $x=a$ is a root, then $(x-a)$ is a factor.","correct":true,"explanation":"A root means $f(a)=0$. By the Factor Theorem, this occurs exactly when division by $(x-a)$ has zero remainder, so $(x-a)$ is a factor."},{"text":"If $x=a$ is a root, then $(x+a)$ is a factor.","correct":false,"explanation":"This reverses the sign in the factor. A root at $x=a$ corresponds to $(x-a)$, because substituting $x=a$ makes that bracket equal to zero."},{"text":"If $x=a$ is a root, then $(a-x)$ is not a factor.","correct":false,"explanation":"Although the order is reversed, $(a-x)=-(x-a)$, so it is also a constant multiple of the corresponding factor. The standard factor is $(x-a)$."},{"text":"If $x=a$ is a root, then $(x-a)$ is a remainder.","correct":false,"explanation":"This confuses a divisor with the result of division. When $x=a$ is a root, division by $(x-a)$ gives remainder zero, making $(x-a)$ a factor."}]},{"id":"2ef41b40-b351-4bbe-9d1d-06d537aa7d9e","cardType":"cloze","points":null,"front":"If $x=-3$ is a root of a polynomial, the corresponding linear factor is $\\text{[...]}$.","back":"If $x=-3$ is a root of a polynomial, the corresponding linear factor is $\\boxed{x+3}$.","mcqOptions":[{"text":"$$x+3$","correct":true,"explanation":"A root $x=-3$ produces the factor $x-(-3)$, which simplifies to $x+3$. This factor equals zero when $x=-3$."},{"text":"$$x-3$","correct":false,"explanation":"This sign error would produce the root $x=3$, not $x=-3$. Using $x-a$ with $a=-3$ gives the correct factor $x+3$."},{"text":"$$3x+1$","correct":false,"explanation":"Setting this factor equal to zero gives $x=-\\frac{1}{3}$, so it corresponds to a different root. The root $-3$ gives the factor $x+3$."},{"text":"$$3x-1$","correct":false,"explanation":"Setting this factor equal to zero gives $x=\\frac{1}{3}$, reflecting both a reciprocal and sign error. The factor corresponding to $x=-3$ is $x+3$."}]},{"id":"20b1de7b-0707-490f-8b4f-69672e4c1916","cardType":"basic","points":1,"front":"If $x = \\frac{1}{2}$ is a root of $f(x)$, what is the equivalent linear factor with integer coefficients?","back":"$$2x-1$","mcqOptions":[{"text":"$$2x-1$","correct":true,"explanation":"The direct factor is $x-\\frac{1}{2}$. Multiplying the entire expression by $2$ clears the denominator without changing the root, giving the integer-coefficient factor $2x-1$."},{"text":"$$2x+1$","correct":false,"explanation":"This sign error gives the root $x=-\\frac{1}{2}$. For the root $x=\\frac{1}{2}$, clearing the denominator in $x-\\frac{1}{2}$ gives $2x-1$."},{"text":"$$x-2$","correct":false,"explanation":"This reciprocal error gives the root $x=2$ rather than $x=\\frac{1}{2}$. Clearing the denominator correctly produces the factor $2x-1$."},{"text":"$$x+2$","correct":false,"explanation":"This uses both the wrong sign and the reciprocal of the root, producing $x=-2$. The root $x=\\frac{1}{2}$ corresponds to the factor $2x-1$."}]},{"id":"b07bdfd9-abdd-4df3-8446-9f6f9566e094","cardType":"cloze","points":null,"front":"The Factor Theorem states that the bracket $(x-a)$ is a factor of $f(x)$ if and only if $\\text{[...]}$.","back":"The Factor Theorem states that the bracket $(x-a)$ is a factor of $f(x)$ if and only if $\\boxed{f(a)=0}$.","mcqOptions":[{"text":"$$f(a)=0$","correct":true,"explanation":"The Factor Theorem links the factor $(x-a)$ to the value obtained at $x=a$. It is a factor exactly when substitution gives zero, so $f(a)=0$."},{"text":"$$f(-a)=0$","correct":false,"explanation":"Substituting $-a$ tests the bracket $(x+a)$, not $(x-a)$. To test whether $(x-a)$ is a factor, the required condition is $f(a)=0$."},{"text":"$$f(a)=a$","correct":false,"explanation":"The theorem requires the remainder after division by $(x-a)$ to be zero, not equal to the root. Therefore, the correct condition is $f(a)=0$."},{"text":"$$f(0)=a$","correct":false,"explanation":"This evaluates the polynomial at zero and usually identifies its constant term; it does not test $(x-a)$. The factor condition is obtained by checking that $f(a)=0$."}]},{"id":"25cd0c79-bbc3-40b4-8376-d6eb7ee6a9c6","cardType":"basic","points":1,"front":"To use the Factor Theorem to show $(x+2)$ is a factor of $f(x)$, what specific value must be substituted for $x$?","back":"$$-2$","mcqOptions":[{"text":"$$-2$","correct":true,"explanation":"Write $(x+2)$ as $(x-(-2))$. The Factor Theorem therefore requires evaluating $f(-2)$; if the result is zero, $(x+2)$ is a factor."},{"text":"$$2$","correct":false,"explanation":"This misses the sign reversal between a factor and its root. Substituting $2$ tests $(x-2)$, whereas $(x+2)$ must be tested using $x=-2$."},{"text":"$$\\frac{1}{2}$","correct":false,"explanation":"This incorrectly takes the reciprocal of the constant. The value that makes $x+2$ equal to zero is $-2$, so the required substitution is $x=-2$."},{"text":"$$-\\frac{1}{2}$","correct":false,"explanation":"This combines an unnecessary reciprocal with the sign change. Solving $x+2=0$ gives $x=-2$, which is the value required by the Factor Theorem."}]},{"id":"4df49057-435e-4d61-82ed-edddf6848507","cardType":"basic","points":1,"front":"Where are the roots of a polynomial found on a graph of $y=f(x)$?","back":"The x-intercepts (where the curve crosses or touches the x-axis).","mcqOptions":[{"text":"The x-intercepts (where the curve crosses or touches the x-axis).","correct":true,"explanation":"Roots satisfy $f(x)=0$, so their graph points have $y=0$. These are the x-intercepts, including points where the curve crosses or merely touches the x-axis."},{"text":"The y-intercepts (where the curve crosses or touches the y-axis).","correct":false,"explanation":"The y-intercept is found by setting $x=0$, so it gives the polynomial’s constant term rather than all its roots. Roots occur at the x-intercepts where $y=0$."},{"text":"The turning points (where the curve changes its direction).","correct":false,"explanation":"Turning points describe where a graph changes from increasing to decreasing or vice versa, and they need not lie on an axis. Roots are the x-intercepts where $f(x)=0$."},{"text":"The stationary points (where the curve has a horizontal tangent).","correct":false,"explanation":"Stationary points occur where the gradient is zero, not necessarily where the polynomial itself is zero. Roots are found at the x-intercepts where the curve meets the x-axis."}]},{"id":"4e8c6833-2543-4e34-9355-98f8e64bd69e","cardType":"basic","points":1,"front":"In a cubic with integer coefficients, which term provides the candidate values for integer roots?","back":"The constant term (the number without an $x$).","mcqOptions":[{"text":"The constant term (the number without an $x$).","correct":true,"explanation":"For a polynomial with integer coefficients, every integer root must divide the constant term. Its positive and negative factors therefore provide the candidate integer roots to test."},{"text":"The leading coefficient (the number multiplying the highest power of $x$).","correct":false,"explanation":"The leading coefficient helps determine possible denominators of rational roots, but integer roots must divide the constant term. Its factors provide the candidate integer values."},{"text":"The linear term (the term containing only $x$).","correct":false,"explanation":"The coefficient of the linear term does not directly generate the possible integer roots. By the integer root result, candidates are factors of the polynomial’s constant term."},{"text":"The cubic term (the term containing $x^3$).","correct":false,"explanation":"The cubic term identifies the polynomial’s degree and leading coefficient, not its candidate integer roots. Those candidates come from the positive and negative factors of the constant term."}]},{"id":"581f4083-7fa5-4080-9cd8-7be5a679549f","cardType":"cloze","points":null,"front":"For the polynomial $f(x) = x^3 - 2x^2 + 5x + 6$, possible integer roots must be factors of [ ].","back":"For the polynomial $f(x) = x^3 - 2x^2 + 5x + 6$, possible integer roots must be factors of 6.","mcqOptions":[{"text":"6","correct":true,"explanation":"The integer root theorem says any integer root must divide the constant term. Here the constant term is 6, so its integer factors are the candidates."},{"text":"3","correct":false,"explanation":"The degree of the polynomial does not determine its possible integer roots. Integer roots must be factors of the constant term, which is 6 in this polynomial."},{"text":"5","correct":false,"explanation":"The coefficient of $x$ is not used to list possible integer roots. They must be factors of the constant term 6, not factors of 5."},{"text":"1","correct":false,"explanation":"The leading coefficient affects possible rational roots but is not the required value here. Possible integer roots must be factors of the constant term 6."}]},{"id":"5343ebb5-230a-4d12-a857-4cedbf999c0e","cardType":"basic","points":2,"front":"According to exam technique, what two things must be shown to prove $(x-a)$ is a factor?","back":"

The substitution $f(a)$ calculated to equal 0
A conclusion stating that $(x-a)$ is a factor

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

The substitution $f(a)$ calculated to equal 0
A conclusion stating that $(x-a)$ is a factor

","correct":true,"explanation":"Showing $f(a)=0$ applies the factor theorem, while the written conclusion explicitly links this result to $(x-a)$ being a factor. Both parts are needed for a complete proof."},{"text":"

The substitution $f(a)$ calculated to equal 1
A conclusion stating that $(x-a)$ is a factor

","correct":false,"explanation":"Obtaining a non-zero value shows that $(x-a)$ is not a factor, regardless of the conclusion stated. The substitution must give $f(a)=0$, followed by the factor conclusion."},{"text":"

The expression $(x-a)$ written in expanded form
A conclusion stating that $(x-a)$ is a factor

","correct":false,"explanation":"Expanding $(x-a)$ does not apply the factor theorem or establish divisibility. A valid proof calculates $f(a)=0$ and then concludes that $(x-a)$ is a factor."},{"text":"

The derivative $f'(a)$ calculated to equal 0
A remainder stating that division gives 1

","correct":false,"explanation":"Finding the derivative concerns gradients, not polynomial factors, and a remainder of 1 disproves divisibility. The proof requires $f(a)=0$ and an explicit factor conclusion."}]},{"id":"ddfbfa07-e2d8-429b-8f88-1eb43f64e8dc","cardType":"basic","points":1,"front":"What is the result of dividing a cubic polynomial by a linear factor?","back":"A quadratic expression.","mcqOptions":[{"text":"A quadratic expression.","correct":true,"explanation":"Dividing by a linear factor reduces the polynomial’s degree by one. Therefore, dividing a degree-three cubic by a degree-one factor produces a degree-two quadratic expression."},{"text":"A linear expression.","correct":false,"explanation":"A linear expression would result only if the degree were reduced by two. Division of a cubic by a linear factor reduces the degree from three to two, giving a quadratic."},{"text":"A cubic expression.","correct":false,"explanation":"Polynomial division by a non-constant factor reduces the degree rather than preserving it. A cubic divided by a linear factor therefore gives a quadratic expression."},{"text":"A quartic expression.","correct":false,"explanation":"Division cannot increase the degree from three to four. Subtracting the linear divisor’s degree from the cubic’s degree gives $3-1=2$, so the result is quadratic."}]},{"id":"00e01d94-18bf-4447-b694-a306d853e879","cardType":"basic","points":1,"front":"When factorising $f(x) = 2x^3 + ...$ into $(x-a)(Ax^2+Bx+C)$, how is the coefficient $A$ determined?","back":"By comparing the $x^3$ coefficients (in this case, $A=2$).","mcqOptions":[{"text":"By comparing the $x^3$ coefficients (in this case, $A=2$).","correct":true,"explanation":"The leading product is $x\\times Ax^2=Ax^3$. Matching this with the leading term $2x^3$ gives $A=2$ by comparing the cubic coefficients."},{"text":"By comparing the $x^2$ coefficients (in this case, $A=2$).","correct":false,"explanation":"The coefficient $A$ multiplies $x^2$, which combines with the leading $x$ to form the cubic term. Comparing quadratic coefficients does not directly determine $A$; $A=2$."},{"text":"By comparing the constant coefficients (in this case, $A=2$).","correct":false,"explanation":"Substituting $x=0$ compares constant terms and helps determine $C$, not the leading coefficient $A$. Comparing the $x^3$ coefficients gives $A=2$."},{"text":"By adding the factor coefficients (in this case, $A=2$).","correct":false,"explanation":"The leading coefficient is not found by adding the factor coefficients. Since $x\\times Ax^2=Ax^3$, comparison with $2x^3$ directly gives $A=2$."}]},{"id":"896f839a-e30c-45b7-a8a4-c494df0d25c4","cardType":"cloze","points":null,"front":"When substituting negative numbers into a polynomial, you should always use [ ] to avoid accuracy errors.","back":"When substituting negative numbers into a polynomial, you should always use brackets to avoid accuracy errors.","mcqOptions":[{"text":"brackets","correct":true,"explanation":"Brackets keep the negative value together before powers and products are evaluated. This prevents sign errors such as treating $(-2)^2$ as $-2^2$."},{"text":"absolute values","correct":false,"explanation":"Absolute values remove the sign and therefore change the number being substituted. Brackets should instead preserve the negative value while powers and products are evaluated correctly."},{"text":"positive signs","correct":false,"explanation":"A positive sign would change the negative input and produce a different polynomial value. Brackets should enclose the negative number so its sign is handled correctly."},{"text":"decimal approximations","correct":false,"explanation":"Decimal approximations do not control the order of operations and may introduce rounding errors. Brackets around negative substitutions preserve the sign and make exponentiation unambiguous."}]},{"id":"a31d2dd0-ff9c-4372-ab15-ac232cd7b3d4","cardType":"basic","points":1,"front":"If $f(x)$ contains two unknown constants $p$ and $q$, and you are given $f(1)=0$ and $f(2)=5$, what method is used to find $p$ and $q$?","back":"Simultaneous equations.","mcqOptions":[{"text":"Simultaneous equations.","correct":true,"explanation":"Substituting 1 and 2 into $f(x)$ produces two equations involving the two unknowns $p$ and $q$. Solving these together determines both constants."},{"text":"The quadratic formula.","correct":false,"explanation":"The factor theorem interprets $f(1)=0$ as a factor condition, but it does not alone determine both constants. The two given conditions must form simultaneous equations."},{"text":"Polynomial long division.","correct":false,"explanation":"Polynomial long division rewrites a polynomial after a factor or divisor is known; it does not directly find two unknown coefficients. Substitution creates simultaneous equations for $p$ and $q$."},{"text":"Completing the square.","correct":false,"explanation":"Completing the square is a technique for quadratic expressions, not a general method for determining two unknown polynomial constants. The two conditions give simultaneous equations."}]},{"id":"6bb79346-1ab9-4066-932d-45ea5f1e03f0","cardType":"basic","points":3,"front":"Outline the 3 standard steps to solve a cubic equation $f(x)=0$.","back":"

Find one root using the factor theorem
Divide by the matching linear factor to get a quadratic
Solve the quadratic (factorising or formula)

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

Find one root using the factor theorem
Divide by the matching linear factor to get a quadratic
Solve the quadratic (factorising or formula)

","correct":true,"explanation":"A known root identifies a linear factor by the factor theorem. Dividing removes that factor and leaves a quadratic, whose two roots can then be found by standard methods."},{"text":"

Find one gradient using differentiation
Divide by the matching linear factor to get a quadratic
Solve the quadratic (factorising or formula)

","correct":false,"explanation":"Differentiation finds stationary points, not necessarily roots, so it does not supply the required linear factor. First find a root using the factor theorem, then divide and solve the quadratic."},{"text":"

Find one root using the factor theorem
Multiply by the matching linear factor to get a quadratic
Solve the quadratic (factorising or formula)

","correct":false,"explanation":"Multiplying by a linear factor raises the degree and cannot reduce the cubic to a quadratic. After finding a root, divide by its matching factor and solve the resulting quadratic."},{"text":"

Find one root using the factor theorem
Divide by the matching linear factor to get a quadratic
Solve the linear expression (rearranging or formula)

","correct":false,"explanation":"The division step leaves a quadratic, not a linear equation, so solving only a linear expression misses the remaining two roots. The resulting quadratic must be solved."}]},{"id":"513ebce3-b40b-49b9-92e3-027ee8b65d2d","cardType":"cloze","points":null,"front":"If a quadratic factor cannot be factorised using integers, use the [ ] to find the remaining roots.","back":"If a quadratic factor cannot be factorised using integers, use the quadratic formula to find the remaining roots.","mcqOptions":[{"text":"quadratic formula","correct":true,"explanation":"The quadratic formula finds the roots of any quadratic with real or complex coefficients, including quadratics that cannot be factorised conveniently using integers."},{"text":"factor theorem","correct":false,"explanation":"The factor theorem tests whether a proposed value gives a root; it does not directly calculate both roots of an unfactorisable quadratic. The quadratic formula should be used."},{"text":"remainder theorem","correct":false,"explanation":"The remainder theorem finds the remainder after division by a linear expression, rather than directly solving a quadratic. The remaining roots should be found using the quadratic formula."},{"text":"binomial theorem","correct":false,"explanation":"The binomial theorem expands powers of binomial expressions and is not a root-solving method. An integer-unfactorisable quadratic should instead be solved using the quadratic formula."}]},{"id":"8a70ab9e-c460-41eb-a15a-a5120ed61952","cardType":"basic","points":1,"front":"When asked to \"factorise completely\", what should you do if the remaining quadratic factor cannot be factorised into integers?","back":"Leave the answer as (linear factor)(quadratic factor).","mcqOptions":[{"text":"Leave the answer as (linear factor)(quadratic factor).","correct":true,"explanation":"Once an integer linear factor has been removed, an irreducible quadratic cannot be split into integer linear factors, so this is the required complete factorisation."},{"text":"Expand the answer into a cubic polynomial in descending powers.","correct":false,"explanation":"Expanding would reverse the factorisation process rather than complete it. The correct response is to leave the expression as a product of its linear factor and irreducible quadratic factor."},{"text":"Rewrite the answer as three linear factors with integer coefficients.","correct":false,"explanation":"A quadratic with no suitable integer factor pair cannot be written as two integer linear factors. It should remain as the quadratic factor alongside the known linear factor."},{"text":"Leave the answer as the linear factor on its own.","correct":false,"explanation":"Discarding the quadratic would change the polynomial and lose two of its roots. The complete answer must retain both the linear factor and the irreducible quadratic factor."}]},{"id":"03438fc2-904c-401f-8f8c-2827a4ebe71d","cardType":"basic","points":1,"front":"If $f(2) = -12$, what does the factor theorem tell us about the expression $(x-2)$?","back":"$$(x-2)$ is not a factor of $f(x)$.","mcqOptions":[{"text":"$$(x-2)$ is not a factor of $f(x)$.","correct":true,"explanation":"The factor theorem requires $f(2)=0$ for $(x-2)$ to be a factor. Since $f(2)=-12$, division by $(x-2)$ leaves a non-zero remainder."},{"text":"$$(x-2)$ is definitely a factor of $f(x)$.","correct":false,"explanation":"Confusing any evaluated value with zero causes this error. The factor theorem says $(x-2)$ is a factor only when $f(2)=0$, but here $f(2)=-12$."},{"text":"$$(x-2)$ is twice a factor of $f(x)$.","correct":false,"explanation":"A repeated factor would still require $f(2)=0$, together with an additional derivative condition. Since $f(2)=-12$, $(x-2)$ is not even a single factor."},{"text":"$$(x-2)$ is possibly a factor of $f(x)$.","correct":false,"explanation":"The value $-12$ is the remainder when $f(x)$ is divided by $(x-2)$, not evidence of an unknown factor status. Its non-zero value proves it is not a factor."}]},{"id":"375bec91-f77c-48af-995f-5fd1e10d6abb","cardType":"cloze","points":null,"front":"For a cubic $f(x)$, if $x=a$ is the only root that results in $f(x)=0$, the curve will cross the x-axis [ ].","back":"For a cubic $f(x)$, if $x=a$ is the only root that results in $f(x)=0$, the curve will cross the x-axis once.","mcqOptions":[{"text":"once","correct":true,"explanation":"An x-axis crossing occurs at a real root. If $x=a$ is the cubic’s only root, there is only one x-intercept, so the curve crosses the axis once."},{"text":"twice","correct":false,"explanation":"Crossing twice would require two distinct real roots and therefore two x-values satisfying $f(x)=0$. The premise gives only one such root, so the curve crosses once."},{"text":"three times","correct":false,"explanation":"A cubic can cross the x-axis three times only when it has three distinct real roots. Here $x=a$ is the only root, so there is one crossing."},{"text":"nowhere","correct":false,"explanation":"An odd-degree polynomial must have at least one real root, and the given root $x=a$ produces an x-intercept. Therefore, the curve crosses the x-axis once, not nowhere."}]},{"id":"51b0b11d-bc37-4600-bf97-fa2947514220","cardType":"basic","points":1,"front":"What is the degree of a cubic polynomial?","back":"Degree 3.","mcqOptions":[{"text":"Degree 3.","correct":true,"explanation":"The degree of a polynomial is its highest power of $x$. A cubic polynomial has highest exponent $3$, so its degree is 3."},{"text":"Degree 2.","correct":false,"explanation":"Degree 2 describes a quadratic polynomial, not a cubic. The term “cubic” means the greatest power of the variable is 3."},{"text":"Degree 4.","correct":false,"explanation":"Degree 4 describes a quartic polynomial. A cubic is named after its highest exponent, which is 3, so its degree is 3."},{"text":"Degree 6.","correct":false,"explanation":"Counting a cubic’s possible terms can lead to this mistake, but degree is determined only by the highest exponent. For a cubic, that exponent is 3."}]}]}],"examCountdown":{"examDate":"2027-05-13","paperLabel":"Paper 1"},"heading":"7.3 The Factor Theorem","title":"7.3 The Factor Theorem Flashcards","topbar":{"backHref":"/as-level/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"7e8b67b9-9739-460b-97d1-21c8d0a702fc","walled":"$L39"}],"$L3a","$L3b"]