Mf1, Mf2 - Polynomials, division and the factor theorem | OCR A-Level Mathematics B (MEI)

A polynomial equation becomes much friendlier when you can see its factors. Use Mf1 to expand and divide polynomials, then Mf2 to justify factors, recover unknown coefficients and find every real root of equations through degree four.

Polynomial terms and operations

Can a fourth-degree equation have four simple whole-number solutions? Try $x^4+x^3-7x^2-x+6=0$ : both $1$ and $-1$ work. We will turn those clues into a complete solution, rather than trusting a list of guesses.

You already know

Distribute every term when expanding brackets. The zero-product rule turns a complete factorisation into a complete solution: at least one factor must be zero.

Polynomials organise algebra by whole-number powers.

Polynomial

A polynomial in $x$ is a sum of constant multiples of non-negative integer powers of $x$ . The degree of a non-zero polynomial is its highest power with a non-zero coefficient. For example, $2x^3+5$ has degree $3$ , despite its missing powers. A quadratic has degree $2$ ; its graph is a parabola. The expressions $1/x$ and $\sqrt{x}$ are not polynomials in $x$ .

Only terms with the same power combine. Write them in descending order so that missing contributions stand out.

Worked example 1: multiply polynomials

Question. Expand $(x+2)(x^2-x+3)$ .

Plan. Multiply each term in the first bracket by the whole second bracket.

\begin{aligned} (x+2)(x^2-x+3)&=x^3-x^2+3x+2x^2-2x+6 && \text{distribute}\\ &=x^3+x^2+x+6 && \text{collect like terms} \end{aligned}

Why this step? Both brackets contribute to the middle powers; neither contribution can be skipped.

Answer. $x^3+x^2+x+6$ .

Check. At $x=1$ , the product is $3\times3=9$ , and the expansion is $1+1+1+6=9$ .

Change one feature: subtract a further polynomial from the product.

Worked example 2: subtraction after expansion

Question. Simplify $(x+2)(x^2-x+3)-(2x^2-3x+4)$ .

Plan. Expand first and distribute the subtraction across the final bracket.

\begin{aligned} &x^3+x^2+x+6-2x^2+3x-4 && \text{subtract all terms}\\ &\qquad=x^3-x^2+4x+2 && \text{collect powers} \end{aligned}

Why this step? The term $-3x$ is being subtracted, so it contributes $+3x$ .

Answer. $x^3-x^2+4x+2$ .

Check. At $x=1$ , the expression gives $9-3=6$ , matching $1-1+4+2$ .

To reverse expansion, take out a common factor or group terms. For example,

x^3-3x^2-4x+12=x^2(x-3)-4(x-3)=(x-3)(x-2)(x+2).

These rules are not in the formula sheet: learn them.

Watch out

Wrong: $x^3+x^2=2x^5$ . Right: unlike powers stay separate. Addition collects coefficients of equal powers; multiplication adds exponents.

Use a new multiplier but keep the subtraction decision.

Dividing by a linear polynomial

Polynomial division removes the highest remaining power at each step. Mf1 uses linear divisors; division by non-linear expressions is outside this requirement.

Quotient and remainder identity

For a linear divisor $ax+b$ , where $a\ne0$ ,

\mathrm f(x)\equiv(ax+b)q(x)+r.

The polynomial $q(x)$ is the quotient; $r$ is a constant remainder. The divisor is a factor exactly when $r=0$ .

Not in the formula sheet: learn it.

The identity is more than a layout: multiplication back checks every coefficient.

Worked example 3: linear division

Question. Divide $x^3+2x^2-5x-6$ by $x-2$ .

Plan. Select each quotient term to cancel the highest remaining power.

\begin{aligned} \mathrm f(x)-(x-2)x^2&=4x^2-5x-6 && \text{first term }x^2\\ (4x^2-5x-6)-(x-2)4x&=3x-6 && \text{next term }4x\\ (3x-6)-(x-2)3&=0 && \text{last term }3 \end{aligned}

Why this step? The remainder has smaller degree after each subtraction, so the division makes progress.

Answer. Quotient $x^2+4x+3$ , remainder $0$ .

Check. $(x-2)(x^2+4x+3)=x^3+2x^2-5x-6$ .

Change only the constant term of the dividend. All quotient terms stay the same, but the remainder changes.

Worked example 4: a non-zero remainder

Question. Divide $x^3+2x^2-5x-1$ by $x-2$ .

Plan. Repeat the division, carrying the new constant through each subtraction.

\begin{aligned} \mathrm f(x)-(x-2)x^2&=4x^2-5x-1 && \text{remove cubic term}\\ (4x^2-5x-1)-(x-2)4x&=3x-1 && \text{remove quadratic term}\\ (3x-1)-(x-2)3&=5 && \text{remove linear term} \end{aligned}

Why this step? A constant remainder is allowed. It need not be smaller than the divisor's coefficients.

Answer. Quotient $x^2+4x+3$ , remainder $5$ .

Check. Multiplying back gives $(x-2)(x^2+4x+3)+5=x^3+2x^2-5x-1$ .

A linear divisor need not be monic. Here is the same leading-term routine when its coefficient is $2$ .

Worked example 5: a non-monic linear divisor

Question. Divide $2x^3-3x^2-8x+6$ by $2x-1$ .

Plan. Divide each leading remainder term by $2x$ , then subtract the whole product.

\begin{aligned} \mathrm f(x)-(2x-1)x^2&=-2x^2-8x+6 && \text{first term }x^2\\ (-2x^2-8x+6)-(2x-1)(-x)&=-9x+6 && \text{next term }-x\\ (-9x+6)-(2x-1)(-\tfrac92)&=\tfrac32 && \text{last term }-9/2 \end{aligned}

Why this step? Dividing by $2x$ , rather than $x$ , makes the quotient's last term $-9/2$ . A fractional coefficient is perfectly valid.

Answer. Quotient $x^2-x-9/2$ , remainder $3/2$ .

Check. Multiplication gives $(2x-1)(x^2-x-9/2)+3/2=2x^3-3x^2-8x+6$ .

Any correct division method is valid, including coefficient comparison. If a power is missing, include its zero coefficient: $x^3+1=x^3+0x^2+0x+1$ .

Watch out

Wrong: call $5/(x-2)$ the remainder. Right: the remainder in the polynomial identity is $5$ ; the fraction appears only after dividing that identity by $x-2$ .

Change the constant again and keep track of it to the end.

Why a zero gives a factor

Substitute $x=a$ into $\mathrm f(x)\equiv(x-a)q(x)+r$ . The product vanishes and $\mathrm f(a)=r$ . This is why a single evaluation can certify a whole linear factor.

Factor theorem

For a polynomial $\mathrm f$ ,

\mathrm f(a)=0\quad\Longleftrightarrow\quad(x-a)\text{ is a factor of }\mathrm f(x).

Both directions hold. A zero of $\mathrm f$ is an input at which its value is zero.

Not in the formula sheet: learn it.

A rejected input rejects only its own factor. You still need an argument to say there are no other roots.

Worked example 6: establish a factor

Question. Show that $x+1$ is a factor of $\mathrm f(x)=x^3-2x^2-5x-2$ , then factorise completely.

Plan. Substitute $-1$ , divide and factorise the quotient.

\begin{aligned} \mathrm f(-1)&=-1-2+5-2=0 && \text{test the zero}\\ \mathrm f(x)&=(x+1)(x^2-3x-2) && \text{divide}\\ x^2-3x-2&=\left(x-\frac{3+\sqrt{17}}2\right)\left(x-\frac{3-\sqrt{17}}2\right) && \text{quadratic roots} \end{aligned}

Why this step? Complete factorisation may need irrational roots. Integer trials do not locate every real root.

Answer.

(x+1)\left(x-\frac{3+\sqrt{17}}2\right)\left(x-\frac{3-\sqrt{17}}2\right).

Check. The last two factors have root sum $3$ and product $-2$ , so their product is $x^2-3x-2$ ; multiplying by $x+1$ restores the cubic.

Change the proposed factor to $x-1$ . Its zero is now positive; the theorem itself is unchanged.

Worked example 7: a different proposed zero

Question. Show that $x-1$ is a factor of $\mathrm g(x)=x^3-3x^2-x+3$ , then factorise completely.

Plan. Evaluate at $1$ , divide and factorise the quadratic.

\begin{aligned} \mathrm g(1)&=1-3-1+3=0 && \text{factor theorem}\\ \mathrm g(x)&=(x-1)(x^2-2x-3) && \text{divide}\\ &=(x-1)(x-3)(x+1) && \text{factorise quotient} \end{aligned}

Why this step? A zero evaluation proves the first factor; division supplies the remaining factor exactly.

Answer. $(x-1)(x-3)(x+1)$ .

Check. Multiplying gives $x^3-3x^2-x+3$ .

For a factor such as $2x-3$ , find the zero first: it is $3/2$ , so test $\mathrm f(3/2)$ . This is the same theorem, since $2x-3=2(x-3/2)$ .

Exam tip

MEI's “Show that” requires evidence leading to the supplied result. Show the substituted terms and zero, then name the factor-theorem conclusion; a repeated assertion is not a proof.

Keep a positive proposed zero, with a new cubic.

Finding roots and unknown coefficients

A known factor can determine a missing coefficient before you solve the equation.

Worked example 8: a known cubic

Question. Solve $x^3-3x^2-4x+12=0$ .

Plan. Group terms to reveal a shared bracket, then use the zero-product rule.

\begin{aligned} x^3-3x^2-4x+12&=x^2(x-3)-4(x-3) && \text{group terms}\\ &=(x-3)(x^2-4) && \text{common bracket}\\ &=(x-3)(x-2)(x+2) && \text{difference of squares} \end{aligned}

Why this step? Grouping finds the factors directly; a trial-and-division routine is not compulsory.

Answer. $x=3,2,-2$ .

Check. Substitution gives respectively $27-27-12+12=0$ , $8-12-8+12=0$ and $-8-12+8+12=0$ .

Change one feature: the coefficient of $x^2$ is unknown, but the factor $x-3$ is given.

Worked example 9: a coefficient fixed by a factor

Question. $x-3$ is a factor of $p(x)=x^3+kx^2-4x+12$ . Find $k$ and solve $p(x)=0$ .

Plan. Impose $p(3)=0$ , then factorise the resulting cubic.

\begin{aligned} 27+9k-12+12&=0 && \text{factor theorem}\\ k&=-3 && \text{solve for }k\\ p(x)&=(x-3)(x-2)(x+2) && \text{factorise} \end{aligned}

Why this step? The factor gives an equation for $k$ , rather than another guess to make.

Answer. $k=-3$ , with roots $3,2,-2$ .

Check. The factor product expands to $x^3-3x^2-4x+12$ , confirming the coefficient.

For integer-coefficient polynomials, any non-zero integer root divides the constant term: if $p(a)=0$ , its other terms are all multiples of $a$ . Use that to choose sensible trials, but remember that roots need not be integers.

Watch out

Wrong: $x+2$ is a factor, so substitute $2$ . Right: substitute $-2$ , the input making that factor zero.

Now apply the same method with a different supplied factor.

Solving equations through degree four

Mf2 includes polynomial equations through degree $4$ . Some quartics are quadratics in disguise; others need successive linear factors. No general quartic formula is needed.

Worked example 10: a quadratic in a square

Question. Solve $x^4-5x^2+4=0$ .

Plan. Set $u=x^2$ , solve the quadratic and recover both square roots.

\begin{aligned} u^2-5u+4&=(u-1)(u-4)=0 && \text{set }u=x^2\\ x^2&=1\text{ or }4 && \text{return to }x\\ x&=\pm1\text{ or }\pm2 && \text{both signs} \end{aligned}

Why this step? Solving for $u$ is not solving for $x$ . Each positive square gives two real roots.

Answer. $x=-2,-1,1,2$ .

Check. For $x^2=1$ , $1-5+4=0$ ; for $x^2=4$ , $16-20+4=0$ .

Change one feature: include odd powers, so the substitution $u=x^2$ no longer captures the whole equation. Return to the opening problem and divide by its known linear factors.

Worked example 11: successive linear divisions

Question. Solve $x^4+x^3-7x^2-x+6=0$ .

Plan. Verify $1$ and $-1$ , divide them out and solve the remaining quadratic.

\begin{aligned} \mathrm f(1)&=1+1-7-1+6=0 && \text{first root}\\ \mathrm f(-1)&=1-1-7+1+6=0 && \text{second root}\\ \mathrm f(x)&=(x-1)(x^3+2x^2-5x-6) && \text{divide by }x-1\\ &=(x-1)(x+1)(x^2+x-6) && \text{divide by }x+1\\ &=(x-1)(x+1)(x+3)(x-2) && \text{factorise quadratic} \end{aligned}

Why this step? Each division is by a linear expression, within Mf1. The full product establishes that the root list is complete.

Answer. $x=1,-1,-3,2$ .

Check. Multiplying $(x^2-1)(x^2+x-6)$ restores the given quartic.

A quartic can have fewer than four distinct real roots: $x^4+x^2=0$ gives $x^2(x^2+1)=0$ , so only $x=0$ is real. Repeated factors count towards the degree without creating new values.

Watch out

Wrong: $x^2=4\Rightarrow x=4$ . Right: $x=\pm2$ . If the squared quantity is negative, it gives no real roots.

Keep the successive-factor route, with new coefficients and supplied starting roots.

Quick Recap

Expand and collect equal powers; factorisation reverses expansion.
Division by a linear expression leaves a constant remainder, checked by multiplication back.
$\mathrm f(a)=0$ if and only if $x-a$ is a factor; use this to find zeros or unknown coefficients.
A quartic may reduce through $u=x^2$ or successive linear factors; recover every real root.

Yes, the opening quartic has four whole-number solutions: its factorisation gives $1,-1,-3,2$ , and proves there are no others.

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Pure Mathematics

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Mechanics

Mf1, Mf2 - Polynomials, division and the factor theorem

Polynomial terms and operations

Dividing by a linear polynomial

Why a zero gives a factor

Finding roots and unknown coefficients

Solving equations through degree four

Quick Recap

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