Zeroes and Coefficients of a Cubic

The idea from the last section extends beautifully to cubics. For a cubic polynomial

p(x)=ax3+bx2+cx+d,a≠0p(x) = ax^3 + bx^2 + cx + d, \quad a \neq 0

with zeroes α\alpha, β\beta, γ\gamma (gamma), the three relationships are:

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

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

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

Key Point: Sum of zeroes =−b/a= -b/a, sum of products taken two at a time =+c/a= +c/a, product of all three =−d/a= -d/a. Watch the signs — they alternate −,+,−-, +, -.

[Board Important] The middle relation (sum of products in pairs) is the one students forget. Write all three down before substituting.

Where the Cubic Relations Come From

Just like the quadratic case, write the cubic in factored form:

ax3+bx2+cx+d=a(x−α)(x−β)(x−γ)ax^3 + bx^2 + cx + d = a(x - \alpha)(x - \beta)(x - \gamma)

Expanding (x−α)(x−β)(x−γ)(x-\alpha)(x-\beta)(x-\gamma) gives

x3−(α+β+γ)x2+(αβ+βγ+γα)x−αβγ.x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma.

Multiplying by aa and comparing with ax3+bx2+cx+dax^3 + bx^2 + cx + d:

  • x2x^2 coefficient: −a(α+β+γ)=b⇒α+β+γ=−ba-a(\alpha+\beta+\gamma) = b \Rightarrow \alpha+\beta+\gamma = -\dfrac{b}{a}.
  • xx coefficient: a(αβ+βγ+γα)=c⇒αβ+βγ+γα=caa(\alpha\beta+\beta\gamma+\gamma\alpha) = c \Rightarrow \alpha\beta+\beta\gamma+\gamma\alpha = \dfrac{c}{a}.
  • constant: −aαβγ=d⇒αβγ=−da-a\alpha\beta\gamma = d \Rightarrow \alpha\beta\gamma = -\dfrac{d}{a}.

Key Point: Same method as the quadratic — expand the factored form, match coefficients. The signs alternate because of the (x−zero)(x - \text{zero}) factors.

[Board Important] Knowing the derivation helps you reconstruct the signs if you forget them in the exam.

Forming a Quadratic from Its Zeroes

This is the reverse problem, and a guaranteed exam favourite. If a quadratic has sum of zeroes SS and product of zeroes PP, then one such polynomial is:

p(x)=x2−Sx+Pp(x) = x^2 - Sx + P

or more generally k(x2−Sx+P)k(x^2 - Sx + P) for any non-zero constant kk.

Worked outline

Find a quadratic whose zeroes have sum −3-3 and product 22.

  • p(x)=x2−(sum)x+(product)=x2−(−3)x+2=x2+3x+2p(x) = x^2 - (\text{sum})x + (\text{product}) = x^2 - (-3)x + 2 = x^2 + 3x + 2.

Key Point: The template is x2−(sum)x+(product)x^2 - (\text{sum})x + (\text{product}). Be careful: the sum comes in with a minus sign in front.

[Board Important] If the zeroes are given directly (say α\alpha and β\beta), first compute S=α+βS = \alpha + \beta and P=αβP = \alpha\beta, then plug into the template.

Forming a Cubic from Its Zeroes

The same pattern extends to cubics. If a cubic has zeroes with:

  • sum =S1=α+β+γ= S_1 = \alpha + \beta + \gamma,
  • sum of products in pairs =S2=αβ+βγ+γα= S_2 = \alpha\beta + \beta\gamma + \gamma\alpha,
  • product =S3=αβγ= S_3 = \alpha\beta\gamma,

then a cubic with these zeroes is:

p(x)=x3−S1x2+S2x−S3p(x) = x^3 - S_1 x^2 + S_2 x - S_3

Worked outline

Form a cubic with zeroes 1, 2, 3.

  • S1=1+2+3=6S_1 = 1+2+3 = 6; S2=(1)(2)+(2)(3)+(3)(1)=2+6+3=11S_2 = (1)(2)+(2)(3)+(3)(1) = 2+6+3 = 11; S3=1⋅2⋅3=6S_3 = 1\cdot2\cdot3 = 6.
  • p(x)=x3−6x2+11x−6p(x) = x^3 - 6x^2 + 11x - 6.

Key Point: The cubic template is x3−S1x2+S2x−S3x^3 - S_1 x^2 + S_2 x - S_3 — signs alternate −,+,−-, +, -.

[Board Important] Compute S1S_1, S2S_2, S3S_3 carefully and substitute with correct signs. A sign slip here loses easy marks.

Solved Examples

Example 1: Form a quadratic from sum and product

Find a quadratic polynomial whose zeroes have sum 14\dfrac{1}{4} and product −1-1.

Solution:

  1. Use p(x)=x2−(sum)x+(product)p(x) = x^2 - (\text{sum})x + (\text{product}).
  2. p(x)=x2−14x+(−1)=x2−14x−1p(x) = x^2 - \dfrac{1}{4}x + (-1) = x^2 - \dfrac{1}{4}x - 1.
  3. Multiply by 4 to clear fractions: 4x2−x−44x^2 - x - 4.

Final Answer: 4x2−x−44x^2 - x - 4 (or x2−14x−1x^2 - \dfrac{1}{4}x - 1).

Takeaway: Use the template, then scale to clear fractions if you wish.

Example 2: Quadratic from given zeroes

Find a quadratic polynomial whose zeroes are 3 and −2-2.

Solution:

  1. Sum S=3+(−2)=1S = 3 + (-2) = 1; Product P=3×(−2)=−6P = 3 \times (-2) = -6.
  2. p(x)=x2−Sx+P=x2−x−6p(x) = x^2 - Sx + P = x^2 - x - 6.

Final Answer: x2−x−6x^2 - x - 6.

Takeaway: Compute sum and product first, then apply the template.

Example 3: Quadratic with irrational zeroes

Form a quadratic whose zeroes are 2+32 + \sqrt{3} and 2−32 - \sqrt{3}.

Solution:

  1. Sum =(2+3)+(2−3)=4= (2+\sqrt{3}) + (2-\sqrt{3}) = 4.
  2. Product =(2+3)(2−3)=4−3=1= (2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1.
  3. p(x)=x2−4x+1p(x) = x^2 - 4x + 1.

Final Answer: x2−4x+1x^2 - 4x + 1.

Takeaway: Conjugate surds add to a rational and multiply using a2−b2a^2 - b^2.

Example 4: Verify cubic relations

Verify the relations between zeroes and coefficients for p(x)=x3−6x2+11x−6p(x) = x^3 - 6x^2 + 11x - 6, given its zeroes are 1, 2, 3.

Solution:

  1. Here a=1,b=−6,c=11,d=−6a=1, b=-6, c=11, d=-6.
  2. α+β+γ=1+2+3=6=−−61=−ba\alpha+\beta+\gamma = 1+2+3 = 6 = -\dfrac{-6}{1} = -\dfrac{b}{a}. ✓
  3. αβ+βγ+γα=2+6+3=11=111=ca\alpha\beta+\beta\gamma+\gamma\alpha = 2+6+3 = 11 = \dfrac{11}{1} = \dfrac{c}{a}. ✓
  4. αβγ=6=−−61=−da\alpha\beta\gamma = 6 = -\dfrac{-6}{1} = -\dfrac{d}{a}. ✓

Final Answer: All three relations verified.

Takeaway: Match each computed value to the corresponding −b/a-b/a, c/ac/a, −d/a-d/a.

Example 5: Form a cubic from its zeroes

Find a cubic polynomial whose zeroes are −2-2, 1 and 3.

Solution:

  1. S1=−2+1+3=2S_1 = -2 + 1 + 3 = 2.
  2. S2=(−2)(1)+(1)(3)+(3)(−2)=−2+3−6=−5S_2 = (-2)(1) + (1)(3) + (3)(-2) = -2 + 3 - 6 = -5.
  3. S3=(−2)(1)(3)=−6S_3 = (-2)(1)(3) = -6.
  4. p(x)=x3−S1x2+S2x−S3=x3−2x2−5x+6p(x) = x^3 - S_1 x^2 + S_2 x - S_3 = x^3 - 2x^2 - 5x + 6.

Final Answer: x3−2x2−5x+6x^3 - 2x^2 - 5x + 6.

Takeaway: Compute S1,S2,S3S_1, S_2, S_3 then substitute into x3−S1x2+S2x−S3x^3 - S_1x^2 + S_2x - S_3.

Example 6: Find the third zero of a cubic

Two zeroes of x3−4x2+x+6x^3 - 4x^2 + x + 6 are −1-1 and 2. Find the third zero.

Solution:

  1. Sum of zeroes =−ba=−−41=4= -\dfrac{b}{a} = -\dfrac{-4}{1} = 4.
  2. If two zeroes are −1-1 and 2, their sum is 1.
  3. Third zero =4−1=3= 4 - 1 = 3.

Final Answer: The third zero is 3.

Takeaway: Use the sum-of-zeroes relation to recover a missing zero.

Example 7: Find product of all zeroes of a cubic

Find the product of the zeroes of 2x3−5x2−14x+82x^3 - 5x^2 - 14x + 8.

Solution:

  1. Here a=2a = 2, d=8d = 8.
  2. Product of zeroes =−da=−82=−4= -\dfrac{d}{a} = -\dfrac{8}{2} = -4.

Final Answer: Product =−4= -4.

Takeaway: Product of all three zeroes of a cubic =−d/a= -d/a.

Example 8: Sum of products in pairs

For the cubic x3+2x2−7x−4x^3 + 2x^2 - 7x - 4, find αβ+βγ+γα\alpha\beta + \beta\gamma + \gamma\alpha.

Solution:

  1. Here a=1a = 1, c=−7c = -7.
  2. αβ+βγ+γα=ca=−71=−7\alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{c}{a} = \dfrac{-7}{1} = -7.

Final Answer: −7-7.

Takeaway: Sum of products in pairs =c/a= c/a.

Example 9: Quadratic from reciprocal zeroes

If α,β\alpha, \beta are zeroes of x2−5x+6x^2 - 5x + 6, form a quadratic whose zeroes are 1α\dfrac{1}{\alpha} and 1β\dfrac{1}{\beta}.

Solution:

  1. For original: sum =5= 5, product =6= 6.
  2. New sum =1α+1β=α+βαβ=56= \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = \dfrac{5}{6}.
  3. New product =1α⋅1β=1αβ=16= \dfrac{1}{\alpha}\cdot\dfrac{1}{\beta} = \dfrac{1}{\alpha\beta} = \dfrac{1}{6}.
  4. Polynomial: x2−56x+16x^2 - \dfrac{5}{6}x + \dfrac{1}{6}, or 6x2−5x+16x^2 - 5x + 1.

Final Answer: 6x2−5x+16x^2 - 5x + 1.

Takeaway: Find the new sum and product, then build the polynomial.

Example 10: Form a cubic from sums S1,S2,S3S_1, S_2, S_3

Find a cubic polynomial with α+β+γ=2\alpha+\beta+\gamma = 2, αβ+βγ+γα=−7\alpha\beta+\beta\gamma+\gamma\alpha = -7, and αβγ=−14\alpha\beta\gamma = -14.

Solution:

  1. Use p(x)=x3−S1x2+S2x−S3p(x) = x^3 - S_1 x^2 + S_2 x - S_3.
  2. p(x)=x3−2x2+(−7)x−(−14)p(x) = x^3 - 2x^2 + (-7)x - (-14).
  3. =x3−2x2−7x+14= x^3 - 2x^2 - 7x + 14.

Final Answer: x3−2x2−7x+14x^3 - 2x^2 - 7x + 14.

Takeaway: Substitute the three symmetric sums with alternating signs.