Cubic roots and coefficients

The roots in polynomials are closely linked to the coefficients of that polynomial.

All polynomials in this chapter will be written as follows - the letters are important here:

ax^3+bx^2+cx+d

Sum of roots

If p, q and r are roots of ax^3+bx^2+cx+d=0, then p+q+r=-\frac ba.

This means that the sum of roots of a cubic is equal to -\frac ba.

Find the sum of the roots of x^3-6x^2+11x-6=0

• The sum of roots is -\frac ba:
• =-\frac{-6}1
• =6
• Answer: sum of roots = 6

Find the sum of the roots of 2x^3+3x^2-4x+5=0

• The sum of roots is -\frac ba:
• =-\frac32
• Answer: sum of roots = -\frac32

The sum of the roots of the polynomial x^3+kx^2+4x-8=0 is 2. Find the value of k.

• Sum of roots = 2 which represents -\frac ba.
• 2=-\frac k1
• k=-2
• Answer: k=-2

The sum of the roots of the polynomial 3x^3+mx^2-9x+12=0 is -4. Find the value of m.

• Sum of roots = -4 which represents -\frac ba.
• -4=-\frac m3
• m=12
• Answer: m=12

Sum of products of roots taken two at a time (sum of product ‘pairs’)

If p, q and r are roots of ax^3+bx^2+cx+d=0, then pq+pr+qr=\frac ca.

In other words, if we multiply together each possible pair of roots and then add those products, we get the same result as \frac ca.

Find the sum of product pairs for x^3-6x^2+11x-6=0

• The sum of product pairs is \frac ca:
• =\frac{11}1
• =11
• Answer: sum of product pairs = 11

Find the sum of product pairs for 2x^3+3x^2-4x+5=0

• The sum of product pairs is \frac ca:
• =\frac{-4}2
• =-2
• Answer: sum of product pairs = -2

The actual roots of this polynomial (+ some others here) are complex numbers. This shows us that actually, the answer above isn’t really an answer! If we found the actual roots and found the sum of their product pairs, we wouldn’t get a real number.

Product of roots

If p, q and r are roots of ax^3+bx^2+cx+d=0, then p\times q\times r=-\frac da.

In words, the product of roots of a cubic is equal to -\frac da.

Find the product of the roots of x^3-6x^2+11x-6=0

• The product of roots is -\frac da:
• =-\frac{-6}1
• =6
• Answer: product of roots = 6

Find the product of the roots of x^3-3x^2+3x-1=0

• The product of roots is -\frac da:
• =-\frac{-1}1
• =1
• Answer: product of roots = 1

The equation x^3+4x^2-x+2k=0 has product of roots 6. Find the value of k.

• The product of roots is 6, which represents -\frac da.
• 6=-\frac{2k}1
• 2k=-6
• k=-3
• Answer: k=-3

The equation x^3-3x^2+4x+7=0 has roots p, q and r. Find the value of \frac1p+\frac1q+\frac1r.

• \frac1p+\frac1q+\frac1r
  • =\frac{pq}{pqr}+\frac{pr}{pqr}+\frac{qr}{pqr}
  • =\frac{pq+pr+qr}{pqr}
• Sum of product pairs = \frac ca:
  • =\frac41=4
  • So pq+pr+qr=4.
• Product of roots = -\frac da:
  • =-\frac71=-7
  • So pqr=-7.
• Therefore, \frac1p+\frac1q+\frac1r
  • =\frac{pq+pr+qr}{pqr}
  • =\frac4{-7}
  • =-\frac47
• Answer: \frac1p+\frac1q+\frac1r=-\frac47

Summary of finding roots and coefficients

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