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How to Factorise Cubic Equations (with Examples)

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A cubic equation refers to any equation containing an x^3 term. In order to factorize a cubic equation we must follow the following steps:

Steps to Factorise a Cubic Equation:

  1. Let p(x) denote the cubic equation.
  2. Try different values to substitute in place of x for which the resulting value is 0. Suppose you find an a such that p(a)=0.
  3. By the factor theorem we can conclude that x-a is a factor of p(x).
  4. Perform long division to divide p(x) by x-a. Let q(x) denote the quotient.
  5. We have that p(x) = (x-a)q(x). Now the quotient q(x) is a quadratic which can be factorised using the usual method of splitting the middle term.

Example 1: Factorise the cubic polynomial p(x) = x^3 - 2x^2 - 9x + 18

Solution: We try various values of x. For example, x=1 gives p(1)=8 which is nonzero and therefore does not work.

Let us now try x=2. We have that p(2)=2^3 - 2(2)^2 - 9(2) + 18 = 0. So the required value for ‘a’ is 2 and we conclude that x-2 is a factor of p(x). We now perform long division as follows,

Factorising cubic equations using long division example 1

So we have that p(x)=x^3 - 2x^2 - 9x + 18 = (x-2)(x^2-9)x^3 - 2x^2 - 9x + 18 = (x-2)(x^2-3^2)x^3 - 2x^2 - 9x + 18 = (x-2)(x-3)(x+3)

Example 2: Factorise the cubic polynomial p(x) =  2x^3 + 5x^2 - 28x - 15

Solution: As a guess we substitute x=-5 in the above cubic equation. We have that p(-5)=2(-5)^3 + 5(-5)^2 - 28(-5) - 15 = 0. So the required value for ‘a’ is -5 and we conclude that x+5 is a factor of p(x). We now perform long division as follows,

Factorising cubic equations using long division example 2

So we have that p(x)=2x^3 + 5x^2 - 28x - 15 = (x + 5) (2x^2 - 5x - 3)2x^3 + 5x^2 - 28x - 15= (x + 5) (2x^2 - 6x + x - 3) 2x^3 + 5x^2 - 28x - 15= (x + 5) [2x(x - 3) + 1(x - 3)]2x^3 + 5x^2 - 28x - 15 = (x + 5) (2x + 1) (x - 3)

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