1.02j - Polynomials, division and the factor theorem | OCR A-Level Mathematics A

Turn an awkward cubic into a product you can understand and solve. You will expand and factorise polynomials, divide by a linear expression and use the factor theorem to find roots and unknown coefficients.

Polynomial terms and operations

How can you find every solution of $x^3-2x^2-5x+6=0$ without guessing endlessly? Substituting $x=1$ gives $1-2-5+6=0$ . That small calculation will unlock the whole cubic. First, understand the expression you are handling.

You already know

Expand brackets by multiplying every term. Collect like terms. The zero-product rule says that a product is zero if and only if at least one factor is zero.

Powers tell you which terms belong together: $3x^2$ and $-7x^2$ combine, but $3x^2$ and $-7x$ do not.

Polynomial

A polynomial in $x$ is a sum of constant multiples of non-negative integer powers of $x$ . Its degree is the highest power with a non-zero coefficient. For example, $3x^3-2x+5$ is cubic, even with no $x^2$ term. A quadratic has degree $2$ , and its graph is a parabola. Neither $x^{-1}$ nor $\sqrt{x}$ is a polynomial in $x$ . A non-zero constant has degree $0$ ; the zero polynomial has no degree under this convention.

Keep powers in descending order. This gives every contribution a visible home.

Worked example 1: expand and collect

Question. Expand $(x-2)(x^2+3x+1)$ .

Plan. Distribute both terms, then collect equal powers.

\begin{aligned} (x-2)(x^2+3x+1)&=x^3+3x^2+x-2x^2-6x-2 && \text{distribute}\\ &=x^3+x^2-5x-2 && \text{collect like powers} \end{aligned}

Why this step? Both $3x^2$ and $-2x^2$ contribute to the same coefficient.

Answer. $x^3+x^2-5x-2$ .

Check. At $x=0$ , both forms give $-2$ . This checks the constant; distribution establishes the whole identity.

Change one feature: subtract another polynomial after expanding. The subtraction affects its whole bracket.

Worked example 2: subtract a whole polynomial

Question. Expand and simplify $(x-2)(x^2+3x+1)-(x^2-4x+2)$ .

Plan. Use the expansion above, then change every sign in the subtracted bracket.

\begin{aligned} &x^3+x^2-5x-2-x^2+4x-2 && \text{remove brackets}\\ &\qquad=x^3-x-4 && \text{collect like powers} \end{aligned}

Why this step? Subtracting $-4x$ adds $4x$ ; the minus sign belongs to the entire polynomial.

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

Check. At $x=1$ , the expression is $-5-(-1)=-4$ , matching $1-1-4$ .

Factorisation runs expansion backwards. First look for a common factor: $6x^3-12x^2=6x^2(x-2)$ . Grouping can reveal a common bracket:

x^3+2x^2-9x-18=x^2(x+2)-9(x+2)=(x+2)(x-3)(x+3).

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

Watch out

Wrong: $-(x^2-4x+2)=-x^2-4x+2$ . Right: $-(x^2-4x+2)=-x^2+4x-2$ . Distribute $-1$ to every term.

Try the same decision with a different linear factor.

Dividing by a linear polynomial

Division peels off the leading term, subtracts what it has accounted for and repeats with what remains.

Quotient and remainder

When a polynomial is divided by a non-zero linear polynomial, the quotient is a polynomial and the remainder is a constant. A zero remainder means the divisor is a factor. For example, $x^2+1=(x-1)(x+1)+2$ has quotient $x+1$ and remainder $2$ .

Writing division as an identity makes it easy to check by multiplying back.

Linear polynomial division

For $a\ne0$ , division gives

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

where $r$ is constant and $q(x)$ is the quotient.

Not in the formula sheet: learn it.

Choose each quotient term by dividing the highest remaining term by the leading term of the divisor.

Worked example 3: a monic divisor

Question. Divide $x^3-2x^2-5x+6$ by $x-1$ .

Plan. Remove $x^3$ , then $-x^2$ , then $-6x$ .

\begin{aligned} \mathrm f(x)-(x-1)x^2&=-x^2-5x+6 && \text{first term }x^2\\ (-x^2-5x+6)-(x-1)(-x)&=-6x+6 && \text{next term }-x\\ (-6x+6)-(x-1)(-6)&=0 && \text{last term }-6 \end{aligned}

Why this step? Each subtraction removes the highest remaining power, so the process finishes.

Answer. Quotient $x^2-x-6$ , remainder $0$ .

Check. $(x-1)(x^2-x-6)=x^3-2x^2-5x+6$ .

Change the divisor's leading coefficient to $2$ . The final quotient term can now be fractional.

Worked example 4: a non-monic divisor

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

Plan. Divide the leading terms at each stage and 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{constant remainder} \end{aligned}

Why this step? The final division is $-9x/(2x)=-9/2$ , not $-9$ .

Answer. Quotient $x^2-x-\frac92$ , remainder $\frac32$ .

Check. $(2x-1)(x^2-x-\frac92)+\frac32=2x^3-3x^2-8x+6$ .

Coefficient comparison is equally valid: expand $(ax-b)(Ax^2+Bx+C)+r$ and match powers. Missing powers still need a place: write $x^3+2=x^3+0x^2+0x+2$ in a division layout.

Watch out

Wrong: subtract only the leading term of a product. Right: subtract the entire product, including its signs. Otherwise the next remainder is already wrong.

Keep the non-monic divisor in this next division.

Why a zero gives a factor

The opening cubic left no remainder when divided by $x-1$ . That was already encoded in $\mathrm f(1)=0$ .

From $\mathrm f(x)\equiv(x-a)q(x)+r$ , substitute $x=a$ . The product vanishes, leaving $\mathrm f(a)=r$ . This proves both directions of the next result.

Factor theorem

For a polynomial $\mathrm f$ ,

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

More generally, for $a\ne0$ ,

\mathrm f\left(\frac ba\right)=0\quad\Longleftrightarrow\quad(ax-b)\text{ is a factor of }\mathrm f(x).

Not in the formula sheet: learn it.

A root, or zero, is an input at which the polynomial is zero. A failed trial rejects only that proposed root; it does not show there are none.

Worked example 5: show a factor

Question. Show that $x-2$ is a factor of $\mathrm f(x)=x^3+2x^2-5x-6$ , then factorise fully.

Plan. Evaluate at $2$ , divide by the established factor and factorise the quotient.

\begin{aligned} \mathrm f(2)&=8+8-10-6=0 && \text{test the factor}\\ \mathrm f(x)&=(x-2)(x^2+4x+3) && \text{linear division}\\ &=(x-2)(x+1)(x+3) && \text{factorise quotient} \end{aligned}

Why this step? The input $2$ makes the proposed factor zero, so the factor theorem establishes it before division.

Answer. $(x-2)(x+1)(x+3)$ .

Check. Expanding gives $x^3+2x^2-5x-6$ , matching every coefficient.

Change one feature: test a non-monic factor, whose zero is a fraction.

Worked example 6: a factor with leading coefficient 2

Question. Show that $2x-1$ is a factor of $\mathrm f(x)=2x^3+3x^2-8x+3$ , then factorise fully.

Plan. Evaluate at $\frac12$ and divide by $2x-1$ .

\begin{aligned} \mathrm f(\tfrac12)&=\tfrac14+\tfrac34-4+3=0 && \text{zero of }2x-1\\ \mathrm f(x)&=(2x-1)(x^2+2x-3) && \text{divide}\\ &=(2x-1)(x-1)(x+3) && \text{factorise quotient} \end{aligned}

Why this step? $2(1/2)-1=0$ ; substituting $1$ would test a different factor.

Answer. $(2x-1)(x-1)(x+3)$ .

Check. $(2x-1)(x^2+2x-3)=2x^3+3x^2-8x+3$ .

A supplied factor is something to establish, not an answer to copy. OCR's “Show that” needs the evaluation or division that makes the remainder visibly zero.

Exam tip

For a proposed factor $ax-b$ , write its zero $b/a$ , substitute with brackets, then state the factor-theorem conclusion. This protects you from sign and reciprocal errors.

Try a different non-unit coefficient.

Finding roots and unknown coefficients

A factor supplies a root and reduces the polynomial's degree so you can find the others.

Here is a useful surprise: a non-zero integer root of a polynomial with integer coefficients must divide its constant term. If $a$ is that root, moving all the other terms to the other side leaves a multiple of $a$ . This narrows integer trials. It does not rule out fractional or irrational roots.

Worked example 7: finish the opening cubic

Question. Solve $x^3-2x^2-5x+6=0$ .

Plan. Use the root $1$ , divide and factorise the quadratic.

\begin{aligned} \mathrm f(1)&=1-2-5+6=0 && \text{establish a factor}\\ \mathrm f(x)&=(x-1)(x^2-x-6) && \text{divide}\\ &=(x-1)(x-3)(x+2) && \text{factorise} \end{aligned}

Why this step? The zero-product rule accounts for the entire cubic, so no roots remain hidden.

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

Check. $\mathrm f(3)=27-18-15+6=0$ and $\mathrm f(-2)=-8-8+10+6=0$ .

Change one feature: a coefficient is unknown, but a factor is supplied. The theorem now creates an equation for that coefficient.

Worked example 8: recover a coefficient first

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

Plan. Impose $\mathrm g(-2)=0$ , then divide and factorise.

\begin{aligned} -8+4k+8-12&=0 && \text{substitute }-2\\ k&=3 && \text{solve for coefficient}\\ \mathrm g(x)&=(x+2)(x^2+x-6) && \text{divide with }k=3\\ &=(x+2)(x+3)(x-2) && \text{factorise} \end{aligned}

Why this step? The stated factor fixes $k$ ; trying arbitrary values wastes the information given.

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

Check. Expanding gives $x^3+3x^2-4x-12$ , including the required coefficient $3$ .

A repeated factor counts repeatedly in the degree but gives one distinct root: $(x-2)^2(x+1)=0$ has distinct roots $2$ and $-1$ . At this stage, factor-theorem searches normally use degree at most $3$ ; the next lesson develops equations through degree $4$ .

Watch out

Wrong: stop after finding one root of a cubic. Right: divide out its factor and solve the remaining quadratic, including irrational real roots.

Here the remaining quadratic does not factorise over the integers; use the quadratic formula you already know.

Quick Recap

Collect equal powers; subtraction changes every sign in a bracket.
Linear division gives $\mathrm f(x)\equiv(ax-b)q(x)+r$ , with constant remainder.
A zero at $b/a$ gives a factor $ax-b$ , and conversely.
A supplied factor can determine an unknown coefficient; divide and solve the remaining factor to finish.

The opening cubic needs no endless guessing: $(x-1)(x-3)(x+2)=0$ gives exactly $1,3,-2$ .

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1.02j - 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

Quick Recap

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