Hard Further Maths Roots of polynomials Questions

Challenging, exam-style Further Maths Roots of polynomials questions with worked solutions. Stretch yourself on the hardest roots-of-polynomials, symmetric-functions, quadratic, cubic problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
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
Question 2
9 markschallenging
The cubic equation x3−6x2+11x−6=0x^3-6x^2+11x-6=0 has roots α\alpha, β\beta and γ\gamma. Which of the following is an equation whose roots are α2\alpha^2, β2\beta^2 and γ2\gamma^2?
Show worked solution

Worked solution

  1. Identify the coefficients of the cubic

    a=1,b=−6,c=11,d=−6a=1,\quad b=-6,\quad c=11,\quad d=-6

    Read the coefficients directly from the given equation.

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

    α+β+γ=−ba=6\alpha+\beta+\gamma=-\frac{b}{a}=6

    The first elementary symmetric function of the roots.

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

    αβ+βγ+γα=ca=11\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}=11

    The second elementary symmetric function of the roots.

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

    αβγ=−da=6\alpha\beta\gamma=-\frac{d}{a}=6

    The third elementary symmetric function of the roots.

  5. Find σ1′\sigma_{1}' for the new roots

    σ1′=σ12−2σ2=14\sigma_{1}'=\sigma_{1}^{2} - 2 \sigma_{2}=14

    The first symmetric function of the squared roots.

  6. Find σ2′\sigma_{2}' for the new roots

    σ2′=−2σ1σ3+σ22=49\sigma_{2}'=- 2 \sigma_{1} \sigma_{3} + \sigma_{2}^{2}=49

    The second symmetric function of the squared roots.

  7. Find σ3′\sigma_{3}' for the new roots

    σ3′=σ32=36\sigma_{3}'=\sigma_{3}^{2}=36

    The third symmetric function of the squared roots.

  8. Build the monic equation from the new symmetric functions

    y3−14y2+49y−36=0y^{3}-14y^{2}+49y-36=0

    Use yn−σ1′yn−1+σ2′yn−2−⋯=0y^n-\sigma_1'y^{n-1}+\sigma_2'y^{n-2}-\cdots=0.

  9. Multiply through to obtain integer coefficients

    y3−14y2+49y−36=0y^3-14y^2+49y-36=0

    Clear the fractions and make the leading coefficient positive.

  10. Record the numerical symmetric functions

    σ1=6,σ2=11,σ3=6\sigma_{1}=6,\quad \sigma_{2}=11,\quad \sigma_{3}=6

    These values drive every symmetric calculation for this polynomial.

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

    p2=σ12−2σ2=14p_2=\sigma_1^{2}-2\sigma_2=14

    The second power sum follows from Newton's identity.

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

    p3=36p_3=36

    Newton's identities extend the calculation to cubes.

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

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

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

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

    ∏i(1−αi)=p(1)a=0\prod_i\left(1-\alpha_i\right)=\frac{p(1)}{a}=0

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

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

    p(−1)=−24p(-1)=-24

    The alternating sum of the coefficients.

  16. Select the correct equation

    y3−14y2+49y−36=0y^3-14y^2+49y-36=0

    This equation has exactly the required roots.

Answer
y3−14y2+49y−36=0y^3-14y^2+49y-36=0
Question 3
9 markschallenging
The quadratic equation x2−2x−5=0x^2-2x-5=0 has roots α\alpha and β\beta. Which of the following is the value of α4+β4\alpha^4+\beta^4?
Show worked solution

Worked solution

  1. Identify the coefficients of the quadratic

    a=1,b=−2,c=−5a=1,\quad b=-2,\quad c=-5

    Read the coefficients directly from the given equation.

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

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

    The first elementary symmetric function of the roots.

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

    αβ=ca=−5\alpha\beta=\frac{c}{a}=-5

    The second elementary symmetric function of the roots.

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

    α4+β4=σ14−4σ12σ2+2σ22\alpha^4+\beta^4=\sigma_{1}^{4} - 4 \sigma_{1}^{2} \sigma_{2} + 2 \sigma_{2}^{2}

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

  5. Substitute the values of the symmetric functions

    α4+β4=(2)4−4(2)2(−5)+2(−5)2\alpha^4+\beta^4=\left(2\right)^{4} - 4 \left(2\right)^{2} \left(-5\right) + 2 \left(-5\right)^{2}

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

  6. Record the numerical symmetric functions

    σ1=2,σ2=−5\sigma_{1}=2,\quad \sigma_{2}=-5

    These values drive every symmetric calculation for this polynomial.

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

    p2=σ12−2σ2=14p_2=\sigma_1^{2}-2\sigma_2=14

    The second power sum follows from Newton's identity.

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

    p3=38p_3=38

    Newton's identities extend the calculation to cubes.

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

    p(1)=−6p(1)=-6

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

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

    ∏i(1−αi)=p(1)a=−6\prod_i\left(1-\alpha_i\right)=\frac{p(1)}{a}=-6

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

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

    p(−1)=−2p(-1)=-2

    The alternating sum of the coefficients.

  12. Confirm the leading coefficient is used as the divisor

    a=1a=1

    Every symmetric function is divided by the leading coefficient.

  13. 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.

  14. 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.

  15. Recall the root-coefficient relations for a quartic

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

    For ax4+bx3+cx2+dx+e=0ax^4+bx^3+cx^2+dx+e=0 the four symmetric functions alternate in sign.

  16. Select the correct value

    α4+β4=146\alpha^4+\beta^4=146

    This agrees with the value found from the symmetric functions.

Answer
α4+β4=146\alpha^4+\beta^4=146
Question 4
9 markschallenging
The quadratic equation x2−4x+1=0x^2-4x+1=0 has roots α\alpha and β\beta. Which of the following is an equation whose roots are 2α−32\alpha-3 and 2β−32\beta-3?
Show worked solution

Worked solution

  1. Identify the coefficients of the quadratic

    a=1,b=−4,c=1a=1,\quad b=-4,\quad c=1

    Read the coefficients directly from the given equation.

  2. Write the substitution that produces the new roots

    y=2x−3 ⇒ x=y+32y=2x-3\ \Rightarrow\ x=\frac{y+3}{2}

    Invert the transformation so the original equation can be used.

  3. Substitute x=y+32x=\frac{y+3}{2} into the original equation

    (y+32)2−4(y+32)+1=0\left(\frac{y+3}{2}\right)^{2}-4\left(\frac{y+3}{2}\right)+1=0

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

  4. Multiply through by 222^{2} to clear fractions

    22[(y+32)2−4(y+32)+1]=02^{2}\left[\left(\frac{y+3}{2}\right)^{2}-4\left(\frac{y+3}{2}\right)+1\right]=0

    This removes every denominator and leaves integer coefficients.

  5. Expand and collect like terms

    y2−2y−11=0y^2-2y-11=0

    Gather the powers of yy.

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

    σ1′=∑2α−3=2\sigma_1'=\sum 2\alpha-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=4,σ2=1\sigma_{1}=4,\quad \sigma_{2}=1

    These values drive every symmetric calculation for this polynomial.

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

    p2=σ12−2σ2=14p_2=\sigma_1^{2}-2\sigma_2=14

    The second power sum follows from Newton's identity.

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

    p3=52p_3=52

    Newton's identities extend the calculation to cubes.

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

    p(1)=−2p(1)=-2

    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=−2\prod_i\left(1-\alpha_i\right)=\frac{p(1)}{a}=-2

    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)=6p(-1)=6

    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

    y2−2y−11=0y^2-2y-11=0

    This equation has exactly the required roots.

Answer
y2−2y−11=0y^2-2y-11=0
Question 5
9 markschallenging
The cubic equation x3−2x2−x+2=0x^3-2x^2-x+2=0 has roots α\alpha, β\beta and γ\gamma. Which of the following is the value of α4+β4+γ4\alpha^4+\beta^4+\gamma^4?
Show worked solution

Worked solution

  1. Identify the coefficients of the cubic

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

    Read the coefficients directly from the given equation.

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

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

    The first elementary symmetric function of the roots.

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

    αβ+βγ+γα=ca=−1\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}=-1

    The second elementary symmetric function of the roots.

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

    αβγ=−da=−2\alpha\beta\gamma=-\frac{d}{a}=-2

    The third elementary symmetric function of the roots.

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

    α4+β4+γ4=σ14−4σ12σ2+4σ1σ3+2σ22\alpha^4+\beta^4+\gamma^4=\sigma_{1}^{4} - 4 \sigma_{1}^{2} \sigma_{2} + 4 \sigma_{1} \sigma_{3} + 2 \sigma_{2}^{2}

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

  6. Substitute the values of the symmetric functions

    α4+β4+γ4=(2)4−4(2)2(−1)+4(2)(−2)+2(−1)2\alpha^4+\beta^4+\gamma^4=\left(2\right)^{4} - 4 \left(2\right)^{2} \left(-1\right) + 4 \left(2\right) \left(-2\right) + 2 \left(-1\right)^{2}

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

  7. Record the numerical symmetric functions

    σ1=2,σ2=−1,σ3=−2\sigma_{1}=2,\quad \sigma_{2}=-1,\quad \sigma_{3}=-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=8p_3=8

    Newton's identities extend the calculation to cubes.

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

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

    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=0\prod_i\left(1-\alpha_i\right)=\frac{p(1)}{a}=0

    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)=0p(-1)=0

    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 value

    α4+β4+γ4=18\alpha^4+\beta^4+\gamma^4=18

    This agrees with the value found from the symmetric functions.

Answer
α4+β4+γ4=18\alpha^4+\beta^4+\gamma^4=18

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