Cubic Polynomials
The polynomial in the form of ax3 + bx2 + cx + d is known as cubic or cubical polynomial where a ≠ 0.
Today we are going to learn how can we factorise the cubic polynomial.
General Form of a Cubic Polynomial: ax3 + bx2 + cx + d
Mind It that
Standard Form Of Cubic Polynomial
When a is one ( a=1) then the cubic polynomial is known as standard form i.e. x3 + bx2 + cx + d
Initial Step:
* Check a + b + c + d = 0, If sum of all coefficients = 0 then directly we say that +1 is the solution of the Cubic Polynomial or (x –1) is a factor of this.
* Check a + c = b + d , If sum of odd terms = sum of even terms then we can say that –1 is the solution of the Cubic Polynomial or (x+1) is a factor of this.
3. Check
α + β + y = b or
α × β × y = d and
αβ + βy + yα = c
Now apply Paravartya Sutra rule to get a quadratic Equation and apply usual Combo rule of Adyamadyena and Adyamadyena for solving quadratic equation.
We can divided the equation into two parts
01. Totally Positive and
(a) x3 + bx2 + cx + d
02. Partially Negative
(i) x3 + bx2 + cx + d
(ii) x3 – bx2 + cx + d
(iii) x3 + bx2 – cx + d
(iv) x3 + bx2 + cx – d
(v) x3 – bx2 – cx – d
(vi) x3 – bx2 – cx + d
Examples: (1)
Solve x3 + 6x2 + 11x +6
1. Sum of odd terms = sum of even terms or a + c = b + d
1 + 11 = 6 + 6 = > 12 = 12
So (x+1) is one factor
2. Using Paravatya rule divide (X3 + 6x2 + 11x +6) by (x+1) it gives x2 + 5x + 6 as a Quotient
3. Solve quadratic equation and get (x+2)(x+3).
4. Ans = (x+1)(x+2)(x+3).
If none of the Initial steps are satisfied then we apply following approach.
Examples: (2)
Solve x3 – 6x2 + 11x –6
1. Sum of all coefficients = 0 so (x–1) is 1 factor.
2. Using Paravatya rule (x3 – 6x2 + 11x –6)/(x–1) gives x2 – 5x + 6
3. Solve quadratic equation using Combo Rule to get (x–2)(x–3)
4. Ans. = (x–1)(x–2)(x–3).
Examples: (3)
Solve x3 + 12x2 + 44x + 48
1. None of the above Initial Condition is satisfied.
2. Last term (d) = 48.
Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
3. By trial and error method find any three numbers which satisfied the Cubic Polynomial.
or
Check
α + β + y = b or
α × β × y = d and
αβ + βy + yα = c
We need to find 3 numbers of these factors such that their sum = b (12) and product = d (44).
4. We observe 2+ 4 + 6 = 12 and 2 x 4 x 6 = 48
5. So the factors are (x+2)(x+4)(x+6)
Examples: (4)
Solve x3 -2x2 -23x + 60
1. None of the above Initial Condition is satisfied.
2. Last term(d) = 60.
Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
3. We need to find 3 numbers of these factors such that their sum = b(–2) and product = d(60).
4. By trial and error method we observe –3 –4 + 5 = –2 and –3 x –4 x 5 = 60.
5. So the factors are (x–3)(x–4)(x+5).
Exercise 01
01. x3 + 6x2 + 11x +6
02. x3 + 12x2 + 44x + 48
03. x3 + 9x2 + 23x + 15
04. x3 + 7x2 + 14x + 8
05. x3 + 8x2 + 17x + 10
06. x3 + 10x2 + 27x + 18
07. x3 + 9x2 + 24x + 16
08. x3 + 13x2 + 32x + 20
Ans.
01. [ (x+1)(x+2)(x+3)]
02. [ (x+2)(x+4)(x+6)]
03. [ (x+1)(x+3)(x+5)]
04. [ (x+1)(x+2)(x+4)]
05. [ (x+1)(x+2)(x+5)]
06. [ (x+1)(x+3)(x+6)]
07. [ (x+ )(x+ )(x+ )]
08. [ (x+1)(x+2)(x+3)]
Exercise 02
01. x3 + 2x2 – x – 2
02. x3 + 7x2 – 21x – 27
03. x3 + 3x2 –19x – 24
04. x3 + x2 – 17x + 15
05. x3 + 6x2 – 37x + 30
06. x3 – 6x2 + 11x – 6
07. x3 – 5x2 – 2x + 24
08. x3 – 7x2 + 4x + 12
09. x3 – 7x2 +11x – 5
10. x3 + x2 – 16x – 16
11. x3 + x2 –3x – 9
12. x3 – 3x2 – 9x – 5
13. x3 + x2 – x – 1
14. x3 –23x2 +142x – 120
Ans.
01. [ (x+1)(x–1)(x–2)]
02. [ (x+2)(x+4)(x+6)]
03. [ (x–2)(x–4)(x+3)]
04. [ (x–1)(x–3)(x+5)]
05. [ (x–1)(x–3)(x+10)]
06. [ (x–1)(x–3)(x–3)]
07. [ (x+2 )(x–3)(x–4)]
08. [ (x+1)(x+2)(x+6)]
09. [ (x+1)(x+1)(x+5)]
10. [ (x+1)(x+4)(x–4)]
11. [ (x+3)(x±√2)]
12. [ (x+1)(x+1)(x–5)]
13. [ (x+1)(x+1)(x–1)]
14. [ (x–1)(x–10)(x–12)]
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