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The cubic equation is expressed by a formula known as the cubic equation formula. An equation that has three degrees of freedom is known as a cubic equation. In every cubic equation, the roots can take on one of two possible forms: either one real root and two imaginary roots or three real roots. The polynomials are referred to as cubic polynomials if they have a degree value of three.
Read more: Discriminant
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KeyTerms: Cubic Equation, Cubic Root, Polynomial, Synthetic Division, Zeros
What is Cubic Equation Formula?
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The formula for a cubic equation to figure out the curve of a cubic equation. Using a cubic equation formula to write down a cubic equation is a great way to find the roots of a cubic equation. A polynomial with degree n has n zeros, also called roots. The equation for a cube looks like this:
| ax3 + bx2 + cx + d = 0 |
Two methods to solve the cubic equation include:
- Trial - Error and Synthetic Division
- Factorization.
The video below explains this:
Polynomials Detailed Video Explanation:
Cubic roots of unity
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Roots of the equation x3 −1=0 are the cubic roots of unity.
the roots are x=1, ( −1+i√ 3)/2 , (−1−i√3)/2
Properties of cubic roots of unity are
- One root is real and the other two are complex; one of the complex roots is the conjugate of the other.
- Ω = (−1+i √3)/2 and ω 2 = (−1−i√3)/2
- 1+ω+ω 2 =0
- ω 3 =1
- ω3n = 1
ω3n+1 = ω
ω 3n+2 =ω 2
Read more:
| Relevant Concepts | ||
|---|---|---|
| Roots of polynomials | Zeroes of polynomials | Complex numbers |
| Imaginary numbers | Factor theorem | Factorial |
Depressing the Cubic Equation
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Substitute,
x = y - b/3a
In the above cubic equation, then we get,
a(y - b/3a)3 + b(y - b/3a)2 + c(y - b/3a) + d = 0
Further simplification gives us the following depressed cubic equation:
Ay3 + (c - b2/3a)y + (d + 2b3/27a2 + bc/3a) = 0
It is imperative that it have the term in x3; otherwise, it would not be cubic (and, hence, a would not be zero), but b, c, or d might all be zero.
Things to Remember
- There are n zeros for a polynomial of degree n.
- A cubic equation is a polynomial equation of degree 3 in one variable.
- There are only three roots or zeros of a quadratic equation which may be either real or imaginary.
- The roots of a cubic equation are the values of x satisfying the cubic equation.
- General formula for cubic equation is, ax3+bx2+cx+d=0
- Sum of products = α+β+γ=−b/a, where b is the coefficient of x2 and a is the coefficient of x3.
Solved Examples
Ques. What Is Cubic Equation Formula? (2 marks)
Ans. The formula for the cubic equation can also be utilized to derive the curve of a cubic equation. Representing a cubic equation using a cubic equation formula is extremely useful for locating the cubic equation's roots. A polynomial with degree n will include n zeros or roots. This is the form of the cubic equation: ax3+bx2+cx+d=0
Ques. Mention two solve cubic equations. (2 marks)
Ans. Two methods to solve the cubic equation are follows:
- Trial - Error and Synthetic Division
- Factorization.
Ques. How to Solve Cubic Polynomials Using Cubic Polynomial Formula? (2 marks)
Ans. The most prevalent method for solving a cubic problem is
- Reduce a cubic polynomial to a quadratic equation.
- Using the quadratic formula, solve the quadratic equation
Ques. What Is the Equation for Cubic Polynomials Formula? (1 mark)
Ans. Cubic equations are third-degree algebraic equations of the form ax3 + bx2 + cx + d = 0, where a, b, and c are coefficients and d is the constant.
Ques. Solve x3 – 6x2 + 11x – 6 = 0 (2 marks)
Ans. You can factorize this equation to get the answer 0 which is
(x-1)(x-2)(x-3)=0.
The solutions to this equation are x = 1, x = 2, and x = 3, which are all distinct from one another.
Ques. Solve the cubic equation x3 – 23x2 + 142x – 120. (5 marks)
Ans. Factorize the polynomial to get;
x3 – 23x2 + 142x – 120 = (x – 1) (x2 – 22x + 120)
But, x2 – 22x + 120 = x2 – 12x – 10x + 120
= x (x – 12) – 10(x – 12)
= (x – 12) (x – 10)
Therefore, x3 – 23x2 + 142x – 120 = (x – 1) (x – 10) (x – 12)
Equate each factor to zero to get;
x = 1
x = 10
x = 12
Hence, The roots of the equation are x = 1, 10 and 12.
Ques. Select the cubic polynomials from the following: (2 marks)
(a) p(x): 5x2 + 6x + 1
(b) p(x): 2x + 3
(c) q(z): z2 − 1
(d) r(z): z2 + (√2)9
(e) r(z): √5z2
(f) s(x): 10x
(g) p(y): y3 − 6y2 + 11y − 6
(h) q(y): 81y3 − 1
(i) r(z): z + 3
Ans. The following are the polynomials that are cubic:
- p(y): y3 − 6y2 + 11y − 6
- q(y): 81y3 − 1
- r(z): z2 + (√2)9
Ques. Find the roots of the following cubic equation 2x3 + 3x2 – 11x – 6 = 0 (5 marks)
Ans. To locate the roots of a given equation.
This problem cannot be solved using the approach of factorization; thus, we will apply the method of trial and error to identify a root.
Typically, we begin with the value "1."
f (1) = 2 + 3 – 11 – 6 ≠ 0
f (–1) = –2 + 3 + 11 – 6 ≠ 0
f (1) = 2 + 3 – 11 – 6 ≠ 0
The value "2" causes the L.H.S to equal "0." Two is therefore one of the three roots.
We will now utilize the Synthetic Division Method to locate the remaining two roots.
The remainder of dividing our equation by (x-2) will give us the other two roots. The remainder of dividing our equation by (x-2) will give us the other two roots.
Quotient: (2x2 + 7x + 3)
This fraction can be factored as (2x+1) (x+3)
From this, we obtain the values
x = -1/2 and x = -3.
Ques. Using the cubic equation formula, solve the cubic equation x3 – 2x2 – x + 2. (5 marks)
Ans. To locate the roots of the equation above
We shall first determine whether the cubic equation can be factored, and if it cannot be, we will employ the synthetic division approach. In this instance, however, examination reveals that this problem can be solved using factorization.
x3 – 2x2 – x + 2.
= x2(x – 2) – (x – 2)
= (x2 – 1) (x – 2)
= (x + 1) (x – 1) (x – 2)
We can conclude that,
x = -1, x = 1 and x = 2.
So, the three roots of the cubic equation are x = -1, x = 1 and x = 2.
Ques. Find the roots of equation f(x) = 3x3 − 16x2 + 23x − 6 = 0. (5 marks)
Ans. Given expression: f(x) = 3x3 − 16x2 + 23x − 6 = 0.
Obtain roots by factoring the polynomial first.
Since the constant is +6 the possible factors are 1, 2, 3, 6.
- f(1) = 3 – 16 + 23 – 6 ≠ 0
- f(2) = 24 – 64 + 46 – 6 = 0
- f(3) = 81 – 144 + 69 – 6 = 0
- f(6) = 648 – 576 + 138 – 6 ≠ 0
We know that, if f(a) = 0, then (x-a) is a factor of f(x).
So, (x – 2) and (x – 3) are factors of f(x). To determine the remaining components, utilize synthetic division.
(x – 2)(x – 3) = (x2 – 5x + 6)
So, (3x- 1) is another factor of f(x).
So,
The roots of the given equation are 1/3, 2, and 3.
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