Problems Based on Solving Cubic Equations

Problem 1: Find the roots of f(x) = x³ – 4x² -3x + 6 = 0.

Solution: 

Given expression:  f(x) = x³ – 4x² -3x + 6 = 0.

First, factorize the polynomial to get roots.

Since the constant is +6 the possible factors are 1, 2, 3, 6.

f(1) = 1 – 4 – 3 + 6 = 7 – 7 = 0

f(2) = 8 – 16 – 6 + 6 ≠ 0

f(3) = 27 – 36 – 9 + 6 ≠ 0

f(6) = 216 – 144 -18 + 6 = -48 ≠ 0

So,  according to Factor Theorem (x – 1) is a factor of the given equation. Now to find the remaining factors use the long division method.

According to Division Algorithm we can write,

So, f(x) = x³ – 4x² -3x + 6 = (x – 1) (x² – 3x – 6) = 0

⇒ (x – 1) = 0 or (x² – 3x – 6) = 0

We know that the roots of a quadratic equation ax² + bx + c = 0 are,

x = [-b ± √(b²-4ac)]/2a

Hence, for (x² – 3x – 6) = 0

x = [3 ± √(3² – 4(1)(-6)]/2(1)

x = (3 ± √33)/2

Hence, the roots of the given cubic equation are 1, (3+√33)/2, and (3–√33)/2.

Problem 2: Find the roots of equation f(x) = 4x³ – 10x² + 4x = 0.

Solution:

Given expression:  f(x) = 4x³ – 10x² + 4x = 0

⇒ x (4x² – 10x + 4) = 0

⇒ x (4x² – 8x – 2x + 4) = 0

⇒ x(4x(x – 2) – 2(x – 2)) = 0

⇒ x (4x – 2) (x – 2) = 0

⇒ x = 0 or 4x – 2 = 0, x – 2 = 0

⇒ x = 0 or x = 1/2 or x = 2

Hence, the roots of the given equation are 0, 1/2 and 2.

Problem 3: Find the roots of equation f(x) = x³ + 3x² + x + 3 = 0.

Solution:

Given expression:  f(x) = x³ + 3x² + x + 3 = 0.

⇒ x² (x + 3) + 1(x + 3) = 0

⇒ (x + 3) (x² + 1) = 0

⇒ x + 3 = 0 or x² + 1 = 0

⇒ x = -3, ±i

So, the given equation has a real root, i.e., -3, and two imaginary roots, i.e., ±i.

Problem 4: Find the roots of equation f(x) = x³ – 7x² – x + 7 = 0.

Solution:

Given expressions,

f(x) = x³ – 3x² – 5x + 7 = 0

First, factorize the equation, f(x): x³ – 3x² – 5x + 7= 0

It can be factored into (x-7)(x+1)(x-1) = 0

After factoring the polynomial, we can find the roots by equating each factor to zero. For example:

• x – 7 = 0, so x = 7
• x + 1 = 0, so x = -1
• x – 1 = 0, so x = 1

So the roots of the equation f(x): x³ – 3x² – 5x + 7 = 0 are

• x = 7
• x = -1
• x = 1

Problem 5: Find the roots of equation f(x) = x³ − 6x² + 11x − 6 = 0, using the graphical method.

Solution:

Given expression: f(x) = x³ − 6x² + 11x − 6 = 0.

Now, simply substitute random values for x in the graph for the given function:

x	1	2	3	4	5
f(x)	0	0	0	6	24

We can see that the graph has cut the X-axis at 3 points, therefore, there are 3 real solutions.

From the graph, the solutions are: x = 1, x = 2, and x = 3.

Hence, the roots of the given equation are 1, 2, and 3.

Cubic Equation is a mathematical equation in which a polynomial of degree 3 is equated to a constant or another polynomial of maximum degree 2. The standard representation of the cubic equation is ax³+bx²+cx+d = 0 where a, b, c, and d are real numbers. Some examples of cubic equation are x³ – 4x² + 15x – 9 = 0, 2x³ – 4x² = 0 etc.

For learning How to Solve Cubic Equations we must first learn about polynomials, the degree of the polynomial, and others. In this article, we will learn about, Polynomials, Polynomial Equations, Solving Cubic Equations Or how to solve cubic equations, and others in detail.