Pa1 - Roots and coefficients of polynomials | OCR A-Level Further Mathematics B (MEI)

A polynomial reveals useful facts about its roots even when you cannot find the roots individually. Pa1 connects the coefficients of quadratics, cubics and quartics to sums and products of their roots. You will use those connections to evaluate expressions and solve questions in which the roots satisfy an extra condition.

Quadratic roots and coefficient signs

Can the squares of three roots add to a negative number? The equation $2x^3-6x^2+10x-8=0$ will give us exactly that surprise. We will find the sum without solving the cubic; complex roots make it possible.

Begin with the simpler link between a quadratic and its two roots. If the roots are $\alpha$ and $\beta$ , its factorisation is

\begin{aligned} A(x-\alpha)(x-\beta) &=A\bigl(x^2-(\alpha+\beta)x+\alpha\beta\bigr) && \text{expand the factors} \end{aligned}

Comparing coefficients with $Ax^2+Bx+C$ shows where both the ratios and the minus sign come from.

You already know

A root $\alpha$ corresponds to a factor $x-\alpha$ . Roots can be real or complex, and a repeated root counts as many times as its factor occurs. For example, $(x-2)^2$ has roots $2$ and $2$ , not just one place in the root list.

A symmetric expression has the same value when the roots exchange labels. Their sum and product are symmetric: it does not matter which root you call $\alpha$ .

Roots of a quadratic

If $Ax^2+Bx+C=0$ , where $A\ne0$ , has roots $\alpha$ and $\beta$ , counting multiplicity, then

\alpha+\beta=-\frac BA,\qquad \alpha\beta=\frac CA

These relations hold for real and complex roots alike. Divide by the leading coefficient; do not assume it is $1$ .

Not in the formula booklet: learn it.

You are reading the combined behaviour of the roots, rather than using the quadratic formula to discover each one.

Worked example 1: a quadratic that is not monic

Question. The equation $2x^2-10x+7=0$ has roots $\alpha$ and $\beta$ . Find $\alpha+\beta$ and $\alpha\beta$ .

Plan. Identify the signed coefficients and divide each relevant coefficient by the leading coefficient.

\begin{aligned} A&=2,\quad B=-10,\quad C=7 && \text{include coefficient signs} \\ \alpha+\beta&=-\frac{-10}{2}=5 && \text{negative coefficient ratio} \\ \alpha\beta&=\frac72 && \text{constant ratio} \end{aligned}

Why this step? The expansion gives the negative sum in the coefficient of $x$ , but the positive product in the constant term.

Answer. $\alpha+\beta=5$ and $\alpha\beta=\frac72$ .

Check. $2\bigl(x^2-5x+\frac72\bigr)=2x^2-10x+7$ , recovering the given coefficients.

Worked example 2 changes only the constant term from $7$ to $0$ . The sum stays fixed, while the product becomes zero.

Worked example 2: a zero root

Question. The equation $2x^2-10x=0$ has roots $\gamma$ and $\delta$ . Find their sum and product.

Plan. Treat the missing constant as zero and apply the same two relations.

\begin{aligned} \gamma+\delta&=-\frac{-10}{2}=5 && \text{same linear coefficient} \\ \gamma\delta&=\frac02=0 && \text{zero constant} \end{aligned}

Why this step? A missing term has coefficient zero. It still contributes a definite value to the relationship.

Answer. $\gamma+\delta=5$ and $\gamma\delta=0$ .

Check. $2x(x-5)=0$ has roots $0$ and $5$ , whose sum and product agree.

The sign must be applied to the coefficient as it actually appears.

Watch out

Wrong: for $2x^2-10x+7=0$ , take the sum as $-5$ . Here $B=-10$ , so $-B/A=-(-10)/2=5$ . Keep a negative coefficient in brackets when substituting.

Change the constant once more. The sign of the product now changes, but the sum does not.

Three roots and three symmetric sums

A cubic connects its three remaining coefficients to the sum of its roots, the sum of their pairwise products and their full product.

Expand three factors in stages:

\begin{aligned} (x-\alpha)(x-\beta)(x-\gamma) &=x^3-(\alpha+\beta+\gamma)x^2 \\ &\quad +(\alpha\beta+\alpha\gamma+\beta\gamma)x-\alpha\beta\gamma \end{aligned}

The signs alternate. To obtain the coefficient of $x$ , choose two negative roots from the three brackets and $x$ from the remaining bracket. There are three distinct choices, so there are three pairwise products.

Pairwise products

For three roots $\alpha$ , $\beta$ , $\gamma$ , the sum of their pairwise products is $\alpha\beta+\alpha\gamma+\beta\gamma$ . Each unordered pair appears once. It is not $\alpha\beta\gamma$ , and it does not include squares such as $\alpha^2$ .

Multiplying the expansion by the leading coefficient gives the relations for a cubic that is not monic.

Roots of a cubic

If $Ax^3+Bx^2+Cx+D=0$ , where $A\ne0$ , has roots $\alpha$ , $\beta$ , $\gamma$ , counting multiplicity, then

\begin{aligned} \alpha+\beta+\gamma&=-\frac BA \\ \alpha\beta+\alpha\gamma+\beta\gamma&=\frac CA \\ \alpha\beta\gamma&=-\frac DA \end{aligned}

Not in the formula booklet: learn it.

The constant term now has the opposite sign to the product: three negative root selections make a negative term.

Worked example 3: reading all three relationships

Question. The equation $2x^3-6x^2+10x-8=0$ has roots $\alpha$ , $\beta$ , $\gamma$ . Find their sum, the sum of their pairwise products and their product.

Plan. Read the coefficient ratios in the order sum, pairs, product, using the sign pattern negative, positive, negative.

\begin{aligned} \alpha+\beta+\gamma&=-\frac{-6}{2}=3 && \text{coefficient of }x^2 \\ \alpha\beta+\alpha\gamma+\beta\gamma&=\frac{10}{2}=5 && \text{coefficient of }x \\ \alpha\beta\gamma&=-\frac{-8}{2}=4 && \text{constant term} \end{aligned}

Why this step? The coefficient of $x$ contains pairs of roots, while the constant contains all three roots. Keep those objects separate.

Answer. The required values are $3$ , $5$ and $4$ , respectively.

Check. The corresponding monic cubic is $x^3-3x^2+5x-4$ ; multiplying by $2$ gives the stated equation.

Worked example 4 changes only the coefficient of $x^2$ to zero. A zero sum of roots does not mean every root is zero.

Worked example 4: a missing squared term

Question. The equation $2x^3+10x-8=0$ has roots $p$ , $q$ , $r$ . Find their sum, the sum of their pairwise products and their product.

Plan. Insert the missing coefficient $B=0$ , then use the same relations.

\begin{aligned} p+q+r&=-\frac02=0 && \text{no squared term} \\ pq+pr+qr&=\frac{10}{2}=5 && \text{unchanged linear term} \\ pqr&=-\frac{-8}{2}=4 && \text{unchanged constant} \end{aligned}

Why this step? Only the relationship attached to the changed coefficient changes; the other two ratios stay the same.

Answer. The values are $0$ , $5$ and $4$ .

Check. $2(x^3-0x^2+5x-4)=2x^3+10x-8$ . Since the product is $4$ , none of the roots is zero despite their zero sum.

Signs matter especially when the degree changes.

Watch out

Wrong: the product of roots of $Ax^3+Bx^2+Cx+D=0$ is $D/A$ . Right: it is $-D/A$ . For a quadratic the constant comes from two negative factors; for a cubic it comes from three.

The following cubic has a different leading coefficient and a positive constant. Keep both changes in the ratios.

Four roots and cancellation in pairs

A quartic has six pairwise products, four products of three roots and one product of all four roots.

In $(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$ , choosing one root, two roots, three roots or four roots produces alternating signs. Writing each combination once prevents missed or duplicated terms.

Roots of a quartic

If $Ax^4+Bx^3+Cx^2+Dx+E=0$ , where $A\ne0$ , has roots $\alpha$ , $\beta$ , $\gamma$ , $\delta$ , counting multiplicity, then

\begin{aligned} \alpha+\beta+\gamma+\delta&=-\frac BA \\ \alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta&=\frac CA \\ \alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta&=-\frac DA \\ \alpha\beta\gamma\delta&=\frac EA \end{aligned}

Not in the formula booklet: learn it.

For example, $3x^4-6x^3+15x^2+12x-9=0$ gives the four symmetric sums $2$ , $5$ , $-4$ , $-3$ , in that order. With repeated roots, keep the repeated places: $(x^2+1)^2=0$ has roots $i$ , $i$ , $-i$ , $-i$ . Its six pairwise products sum to $-1+1+1+1+1-1=2$ , matching the coefficient of $x^2$ .

Questions often arrange the roots so that some terms cancel. The pair $\beta$ , $-\beta$ has zero sum and product $-\beta^2$ ; use that structure before expanding many terms.

Worked example 5: a quartic with two cancelling roots

Question. A quartic $x^4-3x^3+cx^2+dx+8=0$ has roots $\alpha$ , $\frac2\alpha$ , $\beta$ , $-\beta$ . Determine all four roots and the values of $c$ and $d$ .

Plan. Use the sum to find the first pair, the full product to find the second pair, then group the pairwise and triple products.

\begin{aligned} \alpha+\frac2\alpha+\beta-\beta&=3 && \text{sum of roots} \\ \alpha^2-3\alpha+2&=0 && \text{multiply by non-zero }\alpha \\ (\alpha-1)(\alpha-2)&=0 && \text{factorise} \\ \left\{\alpha,\frac2\alpha\right\}&=\{1,2\} && \text{either labelling gives the same pair} \\ \alpha\frac2\alpha\beta(-\beta)&=8 && \text{product of four roots} \\ -2\beta^2&=8\quad\Longrightarrow\quad\beta=\pm2i && \text{solve }\beta^2=-4 \end{aligned}

For the remaining coefficients, the cross-pair terms cancel:

\begin{aligned} c&=2+\left(\alpha+\frac2\alpha\right)(\beta-\beta)-\beta^2 \\ &=2-(-4)=6 && \text{sum of six pairwise products} \\ -d&=2(\beta-\beta)-\left(\alpha+\frac2\alpha\right)\beta^2 \\ &=-3(-4)=12 && \text{sum of four triple products} \end{aligned}

Why this step? Grouping the cancelling roots avoids six separate pair calculations and keeps the minus sign before $d$ visible.

Answer. The roots are $1$ , $2$ , $2i$ , $-2i$ , with $c=6$ and $d=-12$ .

Check. $(x-1)(x-2)(x^2+4)=(x^2-3x+2)(x^2+4)=x^4-3x^3+6x^2-12x+8$ .

Worked example 6 changes only the constant term from $8$ to $18$ . The first pair of roots stays fixed, while the imaginary pair moves.

Worked example 6: changing the full root product

Question. A quartic $x^4-3x^3+cx^2+dx+18=0$ has roots $\alpha$ , $\frac2\alpha$ , $\beta$ , $-\beta$ . Determine the roots and the coefficients $c$ , $d$ .

Plan. Use the unchanged sum for the first pair, then recalculate the full product and the remaining symmetric sums.

\begin{aligned} \alpha+\frac2\alpha&=3 && \text{the other roots cancel} \\ \alpha^2-3\alpha+2&=0\quad\Longrightarrow\quad \left\{\alpha,\frac2\alpha\right\}=\{1,2\} && \text{same quadratic} \\ -2\beta^2&=18\quad\Longrightarrow\quad\beta=\pm3i && \text{new full product} \\ c&=2-\beta^2=2-(-9)=11 && \text{pairwise products} \\ -d&=-3\beta^2=27 && \text{triple products} \\ d&=-27 && \text{coefficient has the opposite sign} \end{aligned}

Why this step? The constant determines the product of all four roots. Changing it does not alter the sum equation for $\alpha$ .

Answer. The roots are $1$ , $2$ , $3i$ , $-3i$ , with $c=11$ , $d=-27$ .

Check. $(x^2-3x+2)(x^2+9)=x^4-3x^3+11x^2-27x+18$ .

The signs of the pair products and triple products are different. A correct numerical sum of triples can still lead to the wrong coefficient if its leading minus sign is dropped.

Exam tip

For a monic quartic, write “ $-d=$ sum of triple products”, then solve for $d$ . Under “Determine” or detailed reasoning, show the equations for the root sum and product, and a method of solving the resulting quadratic; a final list of roots is insufficient.

The next constant changes the imaginary pair again. Use the same cancellation, but show enough work to justify each result.

Watch out

Wrong: the second quartic has $d=27$ because its triple products sum to $27$ . In a monic quartic that sum is $-d$ . The correct line is $-d=27$ , hence $d=-27$ .

Keep the alternating coefficient signs visible when using the next root relations.

Expressions without individual roots

A sum of squares or reciprocals can often be rewritten using the coefficient relationships you already know.

Expand the square of a sum: every pairwise product appears twice. For three roots,

(\alpha+\beta+\gamma)^2 =\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\alpha\gamma+\beta\gamma)

For reciprocals, a common denominator turns the numerator into the pairwise-product sum. This requires every root to be non-zero.

Squares and reciprocals of three roots

Write $S_1=\alpha+\beta+\gamma$ , $S_2=\alpha\beta+\alpha\gamma+\beta\gamma$ and $S_3=\alpha\beta\gamma$ . Then

\alpha^2+\beta^2+\gamma^2=S_1^2-2S_2

If $S_3\ne0$ , then

\frac1\alpha+\frac1\beta+\frac1\gamma =\frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma} =\frac{S_2}{S_3}

For a quadratic or quartic, the same square-expansion idea subtracts twice the sum of distinct pairwise products from the square of the root sum.

Not in the formula booklet: learn it.

Choose the identity before doing any arithmetic. There is usually no benefit in solving the polynomial first.

Worked example 7: a negative sum of squares

Question. The roots of $2x^3-6x^2+10x-8=0$ are $\alpha$ , $\beta$ , $\gamma$ . Determine $\alpha^2+\beta^2+\gamma^2$ and $\frac1\alpha+\frac1\beta+\frac1\gamma$ .

Plan. Read $S_1$ , $S_2$ , $S_3$ , then use the two identities instead of finding the roots.

\begin{aligned} S_1&=3,\quad S_2=5,\quad S_3=4 && \text{coefficient relationships} \\ \alpha^2+\beta^2+\gamma^2&=3^2-2(5)=-1 && \text{expand the squared sum} \\ \frac1\alpha+\frac1\beta+\frac1\gamma&=\frac54 && \text{common denominator} \end{aligned}

Why this step? The roots need not be real. A negative sum of their squares is possible because squaring a non-real number need not give a non-negative real number.

Answer. The sum of squares is $-1$ and the reciprocal sum is $\frac54$ .

Check. The square identity balances: $-1+2(5)=9=3^2$ . The reciprocal denominator is valid because the full product $4$ is non-zero.

Worked example 8 changes only the constant from $-8$ to $-4$ . The first two symmetric sums stay fixed, so only the reciprocal sum changes.

Worked example 8: the constant changes reciprocals

Question. The roots of $2x^3-6x^2+10x-4=0$ are $p$ , $q$ , $r$ . Determine their sum of squares and their reciprocal sum.

Plan. Recalculate the full product, retaining the unchanged sum and pairwise-product sum.

\begin{aligned} p+q+r&=3,\quad pq+pr+qr=5,\quad pqr=2 && \text{new constant ratio} \\ p^2+q^2+r^2&=3^2-2(5)=-1 && \text{unchanged first two sums} \\ \frac1p+\frac1q+\frac1r&=\frac52 && \text{new full product} \end{aligned}

Why this step? The squares depend on the first two sums, while the reciprocal sum also depends on the full product.

Answer. The sum of squares is $-1$ and the reciprocal sum is $\frac52$ .

Check. Multiplying the reciprocal sum by $pqr=2$ gives $5$ , exactly the numerator $pq+pr+qr$ .

The opening puzzle is answered: the squares of the roots of the given cubic sum to $-1$ . That result also proves that not all its roots are real, without finding any root individually.

Watch out

Wrong: $\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2$ . Right: subtract $2(\alpha\beta+\alpha\gamma+\beta\gamma)$ . The cross terms do not disappear merely because the individual roots are unknown.

The next polynomial changes the symmetric sums. Its non-zero constant ensures that the reciprocals exist.

Using a condition linking the roots

When one root is expressed using the others, substitute that expression into the root sum first; it often reduces the problem to a quadratic.

Suppose the three roots are $\alpha$ , $\beta$ and $\alpha+\beta$ . Their sum is $2(\alpha+\beta)$ , and the sum of their pairwise products is

\alpha\beta+\alpha(\alpha+\beta)+\beta(\alpha+\beta) =\alpha\beta+(\alpha+\beta)^2

Once $\alpha+\beta$ and $\alpha\beta$ are known, the two roots $\alpha$ , $\beta$ satisfy $u^2-(\alpha+\beta)u+\alpha\beta=0$ . This is the quadratic relationship used backwards.

Problem-solving

For linked roots, find their sum before trying to find each root. Then use the pairwise-product sum to recover a two-root product, and form a quadratic for the remaining pair.

The next examples use that same structure. One changed coefficient decides whether the final pair is real or non-real.

Worked example 9: a linked pair of complex roots

Question. The equation $x^3-6x^2+13x+c=0$ has roots $\alpha$ , $\beta$ , $\alpha+\beta$ . Determine all three roots and $c$ .

Plan. Use the sum to find $\alpha+\beta$ , use the pairwise sum to find $\alpha\beta$ , then solve their quadratic and use the full product.

\begin{aligned} 2(\alpha+\beta)&=6\quad\Longrightarrow\quad\alpha+\beta=3 && \text{root sum} \\ \alpha\beta+(\alpha+\beta)^2&=13 && \text{pairwise products} \\ \alpha\beta&=13-9=4 && \text{use the known sum} \\ u^2-3u+4&=0 && \text{quadratic with roots }\alpha,\beta \\ u&=\frac{3\pm\sqrt{9-16}}2=\frac32\pm\frac{\sqrt7}{2}i && \text{quadratic formula} \\ -c&=\alpha\beta(\alpha+\beta)=4(3)=12 && \text{product of three roots} \end{aligned}

Why this step? The third root is the sum of the first two; it is not an additional independent unknown.

Answer. The roots are $3$ , $\frac32+\frac{\sqrt7}{2}i$ , $\frac32-\frac{\sqrt7}{2}i$ , and $c=-12$ .

Check. $(x-3)(x^2-3x+4)=x^3-6x^2+13x-12$ .

Worked example 10 changes only the coefficient of $x$ from $13$ to $11$ . The sum is unchanged, but the two-root product is smaller.

Worked example 10: the linked pair becomes real

Question. The equation $x^3-6x^2+11x+c=0$ has roots $\alpha$ , $\beta$ , $\alpha+\beta$ . Determine its roots and $c$ .

Plan. Keep the same root sum and use the changed pairwise-product equation.

\begin{aligned} \alpha+\beta&=3 && \text{unchanged root sum} \\ \alpha\beta+3^2&=11\quad\Longrightarrow\quad\alpha\beta=2 && \text{new pairwise sum} \\ u^2-3u+2&=(u-1)(u-2)=0 && \text{quadratic for the pair} \\ \{\alpha,\beta\}&=\{1,2\},\qquad\alpha+\beta=3 && \text{retain the third root} \\ -c&=2(3)=6 && \text{full product} \end{aligned}

Why this step? Changing the pairwise-product sum can change the nature of the pair without changing the third root.

Answer. The roots are $1$ , $2$ , $3$ , and $c=-6$ .

Check. $(x-1)(x-2)(x-3)=x^3-6x^2+11x-6$ .

These particular derived expressions are not in the formula booklet: rebuild them from the root relationships. A careful final sign check on the constant is especially useful.

Watch out

Wrong: from $\alpha\beta(\alpha+\beta)=6$ , conclude $c=6$ . For the monic cubic, the root product equals $-c$ , so $c=-6$ .

The final cubic has a different root sum and a non-real linked pair. State all three roots as well as the coefficient.

Quick Recap

Expand factors $x-\alpha$ to recover the alternating signs in the coefficient relationships; divide every coefficient by the leading coefficient.
Count each combination once: a cubic has three pairwise products; a quartic has six pairwise products and four triple products. The coefficients give their sums.
Count repeated roots with their multiplicities, and use cancelling pairs before expanding long expressions.
Rewrite a sum of squares or reciprocals using symmetric sums instead of solving for each root.
A condition linking roots can reveal their sum and product, leaving only a familiar quadratic.

The opening cubic has root sum $3$ and pairwise-product sum $5$ , so its roots' squares sum to $3^2-2(5)=-1$ .

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Pa1 - Roots and coefficients of polynomials

Quadratic roots and coefficient signs

Three roots and three symmetric sums

Four roots and cancellation in pairs

Expressions without individual roots

Using a condition linking the roots

Quick Recap

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