Further Maths Roots of polynomials Practice Questions

Free Further Maths Roots of polynomials practice questions with full step-by-step worked solutions. Covers roots-of-polynomials, vieta, quadratic, cubic. Practise exam-style problems and check your method.

roots-of-polynomialsvietaquadraticcubicquarticgeneral-form
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The quadratic equation 2x2−7x+3=02x^2-7x+3=0 has roots α\alpha and β\beta. Find the value of α+β\alpha+\beta.
Show worked solution

Worked solution

  1. Identify the coefficients of the quadratic

    a=2,b=−7,c=3a=2,\quad b=-7,\quad c=3

    Read the coefficients directly from the given equation.

  2. Recall the required root-coefficient relation

    α+β=−ba\alpha+\beta=-\frac{b}{a}

    The relation follows from comparing coefficients with the factorised form.

  3. Substitute the coefficients

    α+β=−−72=72\alpha+\beta=-\frac{-7}{2}=\frac{7}{2}

    Put the numerical coefficients into the relation.

  4. State the value

    α+β=72\alpha+\beta=\frac{7}{2}

    This is the required symmetric function of the roots.

Answer
α+β=72\alpha+\beta=\frac{7}{2}
Question 2
2 markseasy
The quadratic equation ax2+bx+c=0ax^2+bx+c=0 has roots α\alpha and β\beta. Which expression is equal to αβ\alpha\beta?
Show worked solution

Worked solution

  1. Write the general quadratic equation

    ax2+bx+c=0ax^2+bx+c=0

    The coefficients are a, b, ca,\ b,\ c.

  2. Write the same polynomial in factorised form

    a(x−α)(x−β)=0a\left(x-\alpha\right)\left(x-\beta\right)=0

    A polynomial of degree 22 factorises over its 22 roots.

  3. Compare the coefficient of x0x^{0}

    αβ=ca\alpha\beta=\frac{c}{a}

    Expanding the factorised form and comparing coefficients gives the relation.

  4. Select the correct expression

    ca\frac{c}{a}

    This is the root-coefficient relation required.

Answer
ca\frac{c}{a}
Question 3
4 marksintermediate
The quartic equation ax4+bx3+cx2+dx+e=0ax^4+bx^3+cx^2+dx+e=0 has roots α\alpha, β\beta, γ\gamma and δ\delta. Which expression is equal to αβγδ\alpha\beta\gamma\delta?
Show worked solution

Worked solution

  1. Write the general quartic equation

    ax4+bx3+cx2+dx+e=0ax^4+bx^3+cx^2+dx+e=0

    The coefficients are a, b, c, d, ea,\ b,\ c,\ d,\ e.

  2. Write the same polynomial in factorised form

    a(x−α)(x−β)(x−γ)(x−δ)=0a\left(x-\alpha\right)\left(x-\beta\right)\left(x-\gamma\right)\left(x-\delta\right)=0

    A polynomial of degree 44 factorises over its 44 roots.

  3. Compare the coefficient of x3x^{3}

    α+β+γ+δ=−ba\alpha+\beta+\gamma+\delta=-\frac{b}{a}

    The 11-fold products of the roots appear with sign (−1)1(-1)^{1}.

  4. Compare the coefficient of x2x^{2}

    αβ+αγ+αδ+βγ+βδ+γδ=ca\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=\frac{c}{a}

    The 22-fold products of the roots appear with sign (−1)2(-1)^{2}.

  5. Compare the coefficient of x1x^{1}

    αβγ+αβδ+αγδ+βγδ=−da\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=-\frac{d}{a}

    The 33-fold products of the roots appear with sign (−1)3(-1)^{3}.

  6. Compare the coefficient of x0x^{0}

    αβγδ=ea\alpha\beta\gamma\delta=\frac{e}{a}

    The 44-fold products of the roots appear with sign (−1)4(-1)^{4}.

  7. Select the correct expression

    ea\frac{e}{a}

    This is the root-coefficient relation required.

Answer
ea\frac{e}{a}
Question 4
6 markshard
The quartic equation x4+2x3−3x2+x−1=0x^4+2x^3-3x^2+x-1=0 has roots α\alpha, β\beta, γ\gamma and δ\delta. Which of the following is the value of α2+β2+γ2+δ2\alpha^2+\beta^2+\gamma^2+\delta^2?
Show worked solution

Worked solution

  1. Identify the coefficients of the quartic

    a=1,b=2,c=−3,d=1,e=−1a=1,\quad b=2,\quad c=-3,\quad d=1,\quad e=-1

    Read the coefficients directly from the given equation.

  2. Apply the relation for σ1\sigma_{1}

    α+β+γ+δ=−ba=−2\alpha+\beta+\gamma+\delta=-\frac{b}{a}=-2

    The first elementary symmetric function of the roots.

  3. Apply the relation for σ2\sigma_{2}

    αβ+αγ+αδ+βγ+βδ+γδ=ca=−3\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=\frac{c}{a}=-3

    The second elementary symmetric function of the roots.

  4. Apply the relation for σ3\sigma_{3}

    αβγ+αβδ+αγδ+βγδ=−da=−1\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=-\frac{d}{a}=-1

    The third elementary symmetric function of the roots.

  5. Apply the relation for σ4\sigma_{4}

    αβγδ=ea=−1\alpha\beta\gamma\delta=\frac{e}{a}=-1

    The fourth elementary symmetric function of the roots.

  6. Write the target expression in terms of the symmetric functions

    α2+β2+γ2+δ2=σ12−2σ2\alpha^2+\beta^2+\gamma^2+\delta^2=\sigma_{1}^{2} - 2 \sigma_{2}

    Every symmetric function of the roots reduces to the σk\sigma_k.

  7. Substitute the values of the symmetric functions

    α2+β2+γ2+δ2=(−2)2−2(−3)\alpha^2+\beta^2+\gamma^2+\delta^2=\left(-2\right)^{2} - 2 \left(-3\right)

    Replace each σk\sigma_k by the value found from the coefficients.

  8. Record the numerical symmetric functions

    σ1=−2,σ2=−3,σ3=−1,σ4=−1\sigma_{1}=-2,\quad \sigma_{2}=-3,\quad \sigma_{3}=-1,\quad \sigma_{4}=-1

    These values drive every symmetric calculation for this polynomial.

  9. Compute the power sum p2=∑α2p_2=\sum\alpha^2

    p2=σ12−2σ2=10p_2=\sigma_1^{2}-2\sigma_2=10

    The second power sum follows from Newton's identity.

  10. Compute the power sum p3=∑α3p_3=\sum\alpha^3

    p3=−29p_3=-29

    Newton's identities extend the calculation to cubes.

  11. Evaluate the polynomial at x=1x=1

    p(1)=0p(1)=0

    The value at x=1x=1 is the sum of the coefficients.

  12. Select the correct value

    α2+β2+γ2+δ2=10\alpha^2+\beta^2+\gamma^2+\delta^2=10

    This agrees with the value found from the symmetric functions.

Answer
α2+β2+γ2+δ2=10\alpha^2+\beta^2+\gamma^2+\delta^2=10
Question 5
9 markschallenging
The quartic equation x4+2x3−x2+3x−2=0x^4+2x^3-x^2+3x-2=0 has roots α\alpha, β\beta, γ\gamma and δ\delta. Which of the following is an equation whose roots are 1α\frac{1}{\alpha}, 1β\frac{1}{\beta}, 1γ\frac{1}{\gamma} and 1δ\frac{1}{\delta}?
Show worked solution

Worked solution

  1. Identify the coefficients of the quartic

    a=1,b=2,c=−1,d=3,e=−2a=1,\quad b=2,\quad c=-1,\quad d=3,\quad e=-2

    Read the coefficients directly from the given equation.

  2. Write the substitution that produces the new roots

    y=1x ⇒ x=1yy=\frac{1}{x}\ \Rightarrow\ x=\frac{1}{y}

    Invert the transformation so the original equation can be used.

  3. Substitute x=1yx=\frac{1}{y} into the original equation

    (1y)4+2(1y)3−(1y)2+3(1y)−2=0\left(\frac{1}{y}\right)^{4}+2\left(\frac{1}{y}\right)^{3}-\left(\frac{1}{y}\right)^{2}+3\left(\frac{1}{y}\right)-2=0

    Every root α\alpha of the original gives a root yy of the new equation.

  4. Multiply through by y4y^{4} to clear fractions

    y4[(1y)4+2(1y)3−(1y)2+3(1y)−2]=0y^{4}\left[\left(\frac{1}{y}\right)^{4}+2\left(\frac{1}{y}\right)^{3}-\left(\frac{1}{y}\right)^{2}+3\left(\frac{1}{y}\right)-2\right]=0

    This removes every denominator and leaves integer coefficients.

  5. Expand and collect like terms

    −2y4+3y3−y2+2y+1=0-2y^4+3y^3-y^2+2y+1=0

    Gather the powers of yy.

  6. Check the sum of the new roots against the new coefficients

    σ1′=∑1α=32\sigma_1'=\sum \frac{1}{\alpha}=\frac{3}{2}

    The sum of the new roots must equal −b′/a′-b'/a' for the new equation.

  7. Record the numerical symmetric functions

    σ1=−2,σ2=−1,σ3=−3,σ4=−2\sigma_{1}=-2,\quad \sigma_{2}=-1,\quad \sigma_{3}=-3,\quad \sigma_{4}=-2

    These values drive every symmetric calculation for this polynomial.

  8. Compute the power sum p2=∑α2p_2=\sum\alpha^2

    p2=σ12−2σ2=6p_2=\sigma_1^{2}-2\sigma_2=6

    The second power sum follows from Newton's identity.

  9. Compute the power sum p3=∑α3p_3=\sum\alpha^3

    p3=−23p_3=-23

    Newton's identities extend the calculation to cubes.

  10. Evaluate the polynomial at x=1x=1

    p(1)=3p(1)=3

    The value at x=1x=1 is the sum of the coefficients.

  11. Relate p(1)p(1) to the roots

    ∏i(1−αi)=p(1)a=3\prod_i\left(1-\alpha_i\right)=\frac{p(1)}{a}=3

    Because p(x)=a∏i(x−αi)p(x)=a\prod_i(x-\alpha_i).

  12. Evaluate the polynomial at x=−1x=-1

    p(−1)=−7p(-1)=-7

    The alternating sum of the coefficients.

  13. Confirm the leading coefficient is used as the divisor

    a=1a=1

    Every symmetric function is divided by the leading coefficient.

  14. Recall the root-coefficient relations for a quadratic

    α+β=−ba,αβ=ca\alpha+\beta=-\frac{b}{a},\quad \alpha\beta=\frac{c}{a}

    For ax2+bx+c=0ax^2+bx+c=0 the sum and product of the roots come straight from the coefficients.

  15. Recall the root-coefficient relations for a cubic

    ∑α=−ba,∑αβ=ca,αβγ=−da\sum\alpha=-\frac{b}{a},\quad \sum\alpha\beta=\frac{c}{a},\quad \alpha\beta\gamma=-\frac{d}{a}

    For ax3+bx2+cx+d=0ax^3+bx^2+cx+d=0 the three symmetric functions alternate in sign.

  16. Select the correct equation

    2y4−3y3+y2−2y−1=02y^4-3y^3+y^2-2y-1=0

    This equation has exactly the required roots.

Answer
2y4−3y3+y2−2y−1=02y^4-3y^3+y^2-2y-1=0

Unlock 65 more Roots of polynomials questions

Create a free account to work through every Further Maths Roots of polynomials question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Roots of polynomials practice

Hard QuestionsWorked Solutions

Related Pure Maths topics

Argand diagramsComplex numbers: arithmeticModelling with differential equationsComplex numbers: de MoivreFirst-order differential equationsFurther calculus