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1.1 Roots of polynomial equations

Syllabus
9231–2028–2029
Topic
1.1
Level
AS

Learning objectives

1.1.1Polynomial roots• recall and use the relations between the roots and coefficients of polynomial equations e.g. to evaluate symmetric functions of the roots or to solve problems involving unknown coefficients in equations; restricted to equations of degree 2, 3 or 4 only.1.1.2Root transformations• use a substitution to obtain an equation whose roots are related in a simple way to those of the original equation Substitutions will not be given for the easiest cases, e.g. where the new roots are reciprocals or squares or a simple linear function of the old roots.

Read symmetric root information directly from a polynomial

For $A x^4+B x^3+C x^2+D x+E=0$ with roots $\alpha,\beta,\gamma,\delta$:\sum \alpha=-\frac{B}{A},\qquad \sum_{i<j}\alpha_i\alpha_j=\frac{C}{A},\qquad \sum_{i<j<k}\alpha_i\alpha_j\alpha_k=-\frac{D}{A},\qquad \alpha\beta\gamma\delta=\frac{E}{A}.Thesamealternatingpatterntruncatesfordegrees2and3.The same alternating pattern truncates for degrees 2 and 3.

These are Vieta's relations: expand A(x−α)(x−β)⋯A(x-\alpha)(x-\beta)\cdots and match coefficients. Each sigma means all distinct combinations of that size, not one selected pair or triple. Put the equation in descending powers first and divide every relation by the leading coefficient AA.

If $e_1=\sum\alpha$, $e_2=\sum_{i<j}\alpha_i\alpha_j$ and $e_3=\sum_{i<j<k}\alpha_i\alpha_j\alpha_k$, then\sum\alpha^2=e_1^2-2e_2,\qquad \sum\alpha^3=e_1^3-3e_1e_2+3e_3.

For 2x3−5x2−4x+3=02x^3-5x^2-4x+3=0, e1=5/2e_1=5/2, e2=−2e_2=-2 and e3=−3/2e_3=-3/2. Hence ∑α2=(5/2)2−2(−2)=41/4\sum\alpha^2=(5/2)^2-2(-2)=41/4. This obtains the symmetric value without solving the cubic; the same relations can instead form equations for unknown coefficients.

Do not lose the alternating signs: for a monic polynomial the coefficient pattern is 1,−e1,+e2,−e3,+e41,-e_1,+e_2,-e_3,+e_4. The syllabus restricts direct coefficient-root work here to equations of degree 2, 3 or 4.

Build a transformed-root equation by substituting backwards

For original roots x=αx=\alpha and required new roots y=g(α)y=g(\alpha): (1) rearrange to express xx in terms of yy; (2) substitute that expression for xx in f(x)=0f(x)=0; (3) clear denominators or eliminate radicals; (4) collect into a polynomial in yy and make the leading coefficient convenient; (5) check its degree and that each required transformed root satisfies it.

For reciprocal roots $y=1/x$, substitute $x=1/y$. Iff(x)=2x^3-3x^2+5x-7,thentheny^3f(1/y)=2-3y+5y^2-7y^3=0,soanequationforthereciprocalrootsisso an equation for the reciprocal roots is7y^3-5y^2+3y-2=0.

For a linear change y=mx+cy=m x+c with m≠0m\ne0, use x=(y−c)/mx=(y-c)/m. For example, if x2−3x+2=0x^2-3x+2=0 and y=2x−1y=2x-1, substituting x=(y+1)/2x=(y+1)/2 and multiplying by 44 gives (y+1)2−6(y+1)+8=0(y+1)^2-6(y+1)+8=0, hence y2−4y+3=0y^2-4y+3=0. Its roots 1,31,3 are exactly 2(1)−1,2(2)−12(1)-1,2(2)-1.

For powers such as y=x2y=x^2 or x3x^3, substitution may introduce y\sqrt{y} or y3\sqrt[3]{y}. Isolate the radical expression, raise to the necessary power, and simplify. Because raising powers can introduce extra solutions, verify the final roots against the original transformation and preserve the expected number of transformed roots, including multiplicity.

Transform roots, not coefficients. Multiplying the final equation by any non-zero constant changes no roots, but multiplying by a variable expression or clearing a denominator carelessly can add the forbidden value where that denominator was zero.