Roots of Polynomials (Edexcel A Level Further Maths: Core Pure): Revision Note

Roots of quadratics

How are the roots of a quadratic linked to its coefficients?

• Because a quadratic equation $a{x}^2+bx+c=0$  (where $a\ne 0$) has roots $\alpha$ and $\beta$, you can write this equation instead in the form $a(x-\alpha )(x-\beta )=0$

  • Note that $(x-\alpha )(x-\beta )=x^2-(\alpha +\beta )x+\alpha \beta$

    • It is possible that the roots are repeated, i.e. that $\alpha =\beta$

  • You can then equate the two forms:

    • $a{x}^2+bx+c=a(x-\alpha )(x-\beta )$

  • Then (because $a\ne 0$) you can divide both sides of that by a and expand the brackets:

    • $x^2+\frac{b}{a}x+\frac{c}{a}=x^2-(\alpha +\beta )x+\alpha \beta$

  • Finally, compare the coefficients

    • Coefficients of x: $-\frac{b}{a}=(\alpha +\beta )$

    • Constant terms: $\frac{c}{a}=\alpha \beta$

• Therefore for a quadratic equation $a{x}^2+bx+c=0$  :

  • The sum of the roots $\alpha +\beta$ is equal to $-\frac{b}{a}$

  • The product of the roots $\alpha \beta$ is equal to $\frac{c}{a}$

  • Unless an exam question specifically asks you to prove these results, you can always use them without proof to answer questions about quadratics

Related Roots

• You may be asked to consider two quadratic equations, with the roots of the second quadratic linked to the roots of the first quadratic in some way

  • You are usually required to find the sum or product of the roots of the second equation

• The strategy is to use identities which contain $\alpha \beta$ and $\alpha +\beta$(where $\alpha$ and $\beta$ are the roots of the first quadratic)

  • If you know the values of $\alpha$ and $\beta$ from the first quadratic, you can use them to help find the sum or product of the new roots

  • If the second quadratic has roots ${\alpha }^3$ and ${\beta }^3$ , then use the identities:

    • $(\alpha +\beta {)}^3={\alpha }^3+{\beta }^3+3\alpha \beta (\alpha +\beta )$

    • $(\alpha \beta {)}^3={\alpha }^3{\beta }^3$

  • If the second quadratic has roots $\frac{1}{\alpha }$ and $\frac{1}{\beta }$, then use the identities:

    • $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }$

    • $\frac{1}{\alpha }\times \frac{1}{\beta }=\frac{1}{\alpha \beta }$

• You can then form a new equation for a quadratic with the new roots

  • This is done by recalling that a quadratic with a given pair of roots can be written in the form x² – (sum of the roots)x + (product of the roots) = 0

  • Be aware that this will not give a unique answer

    • This is because multiplying an entire quadratic by a constant does not change its roots

    • You can use this fact, for example, to find a quadratic that has a particular pair of roots AND has all integer coefficients

• See the worked example below for an example of how to do some of this!

Roots of cubics

How are the roots of a cubic linked to its coefficients?

• Because a cubic equation $a{x}^3+b{x}^2+cx+d=0$ (where $a\ne 0$) has roots $\alpha$, $\beta$ and $\gamma$, you can write this equation instead in the form $a(x-\alpha )(x-\beta )(x-\gamma )=0$

  • Note that $(x-\alpha )(x-\beta )(x-\gamma )=x^3-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +\alpha \gamma )x-\alpha \beta \gamma$

  • It is possible that some of the roots are repeated, i.e. that some or all of them are equal to each other
    

  • You can then equate the two forms:

    • $a{x}^3+b{x}^2+cx+d=a(x-\alpha )(x-\beta )(x-\gamma )$

  • Then (because $a\ne 0$) you can divide both sides of that by a and expand the brackets:

    • $x^3+\frac{b}{a}{x}^2+\frac{c}{a}x+\frac{d}{a}=x^3-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +\alpha \gamma )x-\alpha \beta \gamma$

  • Finally, compare the coefficients

    • Coefficients of x²: $-\frac{b}{a}=\alpha +\beta +\gamma$

    • Coefficients of x: $\frac{c}{a}=\alpha \beta +\beta \gamma +\alpha \gamma$

    • Constant terms: $-\frac{d}{a}=\alpha \beta \gamma$

• Therefore for a cubic equation $a{x}^3+b{x}^2+cx+d=0$ :

  • The sum of the roots $\alpha +\beta +\gamma$ is equal to $-\frac{b}{a}$

    • The sum of roots $\alpha +\beta +\gamma$ can also be denoted by 

  • The sum of the product pairs of roots $\alpha \beta +\beta \gamma +\alpha \gamma$ is equal to $\frac{c}{a}$

    • This ‘sum of pairs’ $\alpha \beta +\beta \gamma +\alpha \gamma$ can also be denoted by 

  • The product of the roots $\alpha \beta \gamma$ is equal to $-\frac{d}{a}$

    • The product of roots $\alpha \beta \gamma$can also be denoted by 

    • See quartic equations where using this ‘sum of triples’ notation makes more sense!

  • Unless an exam question specifically asks you to prove these results, you can always use them without proof to answer questions about cubics

Related Roots

• You may be asked to consider two cubic equations, with the roots of the second cubic linked to the roots of the first cubic in some way

  • You are usually required to find the sum or product of the roots of the second equation

• The strategy is to use identities which contain , , and   (where $\alpha$, $\beta$ and $\gamma$ are the roots of the first cubic)

  • If you know the values of $\alpha$, $\beta$, and $\gamma$ from the first cubic, you can use them to help find the sum or product of the new roots

  • If the second cubic has roots ${\alpha }^2$, ${\beta }^2$, and ${\gamma }^2$, then use the identities:

    • $(\alpha +\beta +\gamma {)}^2={\alpha }^2+{\beta }^2+{\gamma }^2+2(\alpha \beta +\beta \gamma +\alpha \gamma )$

    • i.e., $(\alpha +\beta +\gamma {)}^2={\alpha }^2+{\beta }^2+{\gamma }^2+2\sum \alpha \beta$

    • $(\alpha \beta \gamma {)}^2={\alpha }^2{\beta }^2{\gamma }^2$

  • If the second cubic has roots ${\alpha }^3,{\beta }^3,{\gamma }^3$, then use the identities:

    • $(\alpha +\beta +\gamma {)}^3={\alpha }^3+{\beta }^3+{\gamma }^3+3(\alpha +\beta +\gamma \left)\right(\alpha \beta +\beta \gamma +\alpha \gamma )-3\alpha \beta \gamma$

    • i.e., $(\alpha +\beta +\gamma {)}^3={\alpha }^3+{\beta }^3+{\gamma }^3+3\sum \alpha \sum \alpha \beta -3\sum \alpha \beta \gamma$

    • $(\alpha \beta \gamma {)}^3={\alpha }^3{\beta }^3{\gamma }^3$

  • If the second cubic has roots $\frac{1}{\alpha }$, $\frac{1}{\beta }$, and $\frac{1}{\gamma }$, then use the identities:

    • $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }$

    • $\frac{1}{\alpha }\times \frac{1}{\beta }\times \frac{1}{\gamma }=\frac{1}{\alpha \beta \gamma }$

• You can then form a new equation for a cubic with the new roots

  • This is done by recalling that a cubic with three given roots can be written in the form $x^3-\left(\text{sum of roots}\right)x^2+\left(\text{sum of product pairs}\right)x-\left(\text{product of roots}\right)=0$

  • Be aware that this will not give a unique answer

    • This is because multiplying an entire cubic by a constant does not change its roots

    • You can use this fact, for example, to find a cubic that has a particular pair of roots AND has all integer coefficients

Roots of quartics

How are the roots of a quartic linked to its coefficients?

• Because a quartic equation $a{x}^4+b{x}^3+c{x}^2+dx+e=0$ (where $a\ne 0$) has roots $\alpha$, $\beta$, $\gamma$ and $\delta$, you can write this equation instead in the form $a(x-\alpha )(x-\beta )(x-\gamma )(x-\delta )=0$

• Note that $(x-\alpha )(x-\beta )(x-\gamma )(x-\delta )$ expands to $x^4-(\alpha +\beta +\gamma +\delta )x^3+(\alpha \beta +\beta \gamma +\alpha \gamma +\gamma \delta +\alpha \delta +\beta \delta )x^2-(\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta )x+\alpha \beta \gamma \delta$

  • It is possible that some of the roots are repeated, i.e. that some or all of them are equal to each other
    

  • You can then equate the two forms:

    • $a{x}^4+b{x}^3+c{x}^2+dx+e=a(x-\alpha )(x-\beta )(x-\gamma )(x-\delta )$

  • Then (because $a\ne 0$) you can divide both sides of that by a and expand the brackets:

    • $x^3+\frac{b}{a}{x}^2+\frac{c}{a}x+\frac{d}{a}=$ $x^4-(\alpha +\beta +\gamma +\delta )x^3+(\alpha \beta +\beta \gamma +\alpha \gamma +\gamma \delta +\alpha \delta +\beta \delta )x^2-(\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta )x+\alpha \beta \gamma \delta$

  • Finally, compare the coefficients

    • Coefficients of x³: $-\frac{b}{a}=\alpha +\beta +\gamma +\delta$

    • Coefficients of x²: $\frac{c}{a}=\alpha \beta +\beta \gamma +\alpha \gamma +\gamma \delta +\alpha \delta +\beta \delta$

    • Coefficients of x: $-\frac{d}{a}=\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta$

    • Constant terms: $\frac{e}{a}=\alpha \beta \gamma \delta$

• Therefore for a quartic equation $a{x}^4+b{x}^3+c{x}^2+dx+e=0$ :

  • The sum of the roots $\alpha +\beta +\gamma +\delta$ is equal to $-\frac{b}{a}$

    • The sum of roots $\alpha +\beta +\gamma +\delta$ can also be denoted by $\sum \alpha$

  • The sum of the product pairs of roots $\alpha \beta +\beta \gamma +\alpha \gamma +\gamma \delta +\alpha \delta +\beta \delta$ is equal to $\frac{c}{a}$

    • This ‘sum of pairs’ $\alpha \beta +\beta \gamma +\alpha \gamma +\gamma \delta +\alpha \delta +\beta \delta$ can also be denoted by $\sum \alpha \beta$

  • The sum of the product triples of roots $\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta$ is equal to $-\frac{d}{a}$

    • This ‘sum of triples’ can also be denoted by $\sum \alpha \beta \gamma$

  • The product of the roots $\alpha \beta \gamma \delta$ is equal to $\frac{e}{a}$

    • The product of roots (or ‘product of fours’) $\alpha \beta \gamma \delta$ can also be denoted by $\sum \alpha \beta \gamma \delta$

  • Unless an exam question specifically asks you to prove these results, you can always use them without proof to answer questions about quartics

Related Roots

• You may be asked to consider two quartic equations, with the roots of the second quartic linked to the roots of the first quartic in some way

  • You are usually required to find the sum or product of the roots of the second equation

• The strategy is to use identities which contain $\sum \alpha$, $\sum \alpha \beta$, $\sum \alpha \beta \gamma$, and $\sum \alpha \beta \gamma \delta$ (where $\alpha$, $\beta$,$\gamma$ and $\delta$ are the roots of the first quartic)

  • If you know the values of $\alpha$, $\beta$, $\gamma$, and $\delta$ from the first quartic, you can use them to help find the sum or product of the new roots

  • If the second quartic has roots ${\alpha }^2$, ${\beta }^2$, ${\gamma }^2$ and ${\delta }^2$, then use the identities:

    • $(\alpha +\beta +\gamma +\delta {)}^2={\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2+2(\alpha \beta +\beta \gamma +\alpha \gamma +\gamma \delta +\alpha \delta +\beta \delta )$

    • i.e., $(\alpha +\beta +\gamma +\delta {)}^2={\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2+2\sum \alpha \beta$

    • $(\alpha \beta \gamma \delta {)}^2={\alpha }^2{\beta }^2{\gamma }^2{\delta }^2$

  • (Note that you will not be asked about a quartic with roots ${\alpha }^3,{\beta }^3,{\gamma }^3$ and ${\delta }^3$)

  • If the second quartic has roots $\frac{1}{\alpha }$, $\frac{1}{\beta }$, $\frac{1}{\gamma }$ and $\frac{1}{\delta }$, then use the identities:

    • $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }+\frac{1}{\delta }=\frac{\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta }{\alpha \beta \gamma \delta }$

    • $\frac{1}{\alpha }\times \frac{1}{\beta }\times \frac{1}{\gamma }\times \frac{1}{\delta }=\frac{1}{\alpha \beta \gamma \delta }$

• You can then form a new equation for a quartic with the new roots

  • This is done by recalling that a quartic with four given roots can be written in the form $x^4-\left(\text{sum of roots}\right)x^3+\left(\text{sum of product pairs}\right)x^2-\left(\text{sum of product triples}\right)x+\left(\text{product of roots}\right)=0$

  • Be aware that this will not give a unique answer

    • This is because multiplying an entire quartic by a constant does not change its roots

    • You can use this fact, for example, to find a quartic that has a particular pair of roots AND has all integer coefficients

Roots of polynomials

What is the general pattern linking the roots to the coefficients of a polynomial?

• By looking at the links between the coefficients and the roots of quadratics, cubics, and quartics, you can see that a pattern emerges, which also holds true for higher order polynomials

• It is useful to use sigma notation to keep expressions for sums of roots concise

  • For a quartic with roots $\alpha ,\beta ,\gamma ,\delta$, for example:

    • The sum of the roots $\alpha +\beta +\gamma +\delta$ is denoted by $\sum \alpha$

    • The sum of the pairs of roots $\alpha \beta +\beta \gamma +\alpha \gamma +\gamma \delta +\alpha \delta +\beta \delta$is denoted by $\sum \alpha \beta$

    • The sum of the triples of roots $\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta$ is denoted by $\sum \alpha \beta \gamma$

    • The sum of the sets of fours (in this case just one term) $\alpha \beta \gamma \delta$ is denoted by $\sum \alpha \beta \gamma \delta$

• The table below summarises the relationships between the coefficients and roots of quadratics, cubics, and quartics:

How can I find sums and products of related roots?

• You may be asked to consider a second equation, that has roots linked to the roots of the first equation in some way

  • You are usually required to find the sum or product of the roots of the second equation

• The strategy is to use identities containing $\sum \alpha ,\sum \alpha \beta ,\sum \alpha \beta \gamma$, and/or $\sum \alpha \beta \gamma \delta$ (depending on the question and the degree of the polynomial)

  • If you know the value of the roots from the first equation, these identities can help you find the sum or product of the roots of the second equation

• The table below shows useful identities for finding a new quadratic equation whose roots are related to the roots $\alpha$ and $\beta$ of the original quadratic equation

  • In each case the sum or the product of the ‘new roots’ can be linked back to $\alpha \beta$ or $\alpha +\beta$ for the original equation

• Similar identities that could be useful for cubics and quartics are listed earlier in this revision note in the cubics and quartics sections

• A good place to start if the new roots are squared, is by considering $(\sum \alpha {)}^2$

  • or if the new roots are cubed, then start by considering $(\sum \alpha {)}^3$

  • or if the new roots are reciprocals (i.e., $\frac{1}{\alpha },\frac{1}{\beta }$, etc.), then start by adding the new roots together to form a single algebraic fraction