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Polynomials are algebraic expressions made up of variables, constants, and exponents that are whole numbers. The degree of a polynomial means the highest power of the variable in the expression. A cubic polynomial is a type of polynomial where the degree is 3. This means the highest exponent of the variable is 3. In other words, the variable can go up to the power of 3, but not higher. An example of a cubic polynomial is:
f(x) = x^3 + 2x^2 − 5x + 6
Here, the term x^3 shows that the degree is 3.
In this maths article, we shall read about cubic polynomials, its general equation, roots, and factorization. We shall also solve some examples of cubic polynomials for better understanding of the concept.
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Polynomials are grouped into four types based on their degree, which means the highest power of the variable. These types are: zero polynomial, linear polynomial, quadratic polynomial, and cubic polynomial. A cubic polynomial is a polynomial where the highest power of the variable is 3. It is written in the general form:
p(x) = ax³ + bx² + cx + d
Here, a ≠ 0 and a, b, c, d are real numbers. The values a, b, and c are called the coefficients of x³, x², and x, and d is called the constant term.
Also, a cubic equation is an equation that involves a cubic polynomial. Some of the examples of cubic polynomial are:
\(p\left(y\right)=4y^3+4y^2-y-1\), and \(r\left(z\right)=\pi z^3+(\sqrt{2})^{10}.\).
A cubic polynomial is a type of algebraic expression that has the highest power of 3. It is written in this general form: ax³ + bx² + cx + d,
where a, b, c, and d are numbers (constants), and a ≠ 0.
When we write it as an equation and set it equal to 0, it becomes a cubic equation:
ax³ + bx² + cx + d = 0
The values of x that make this equation true are called the roots or zeros of the cubic polynomial. These are the solutions we try to find.
We have already discussed that a cubic polynomial can be expressed in the form \(y=ax^3+bx^2+cx+d\). Here, \(a\ne0\) and a, b, c, and d are real numbers, out of which a, b, and c are coefficients of \(x^3,\ x^2,\ and\ x\) respectively and d is the constant term.
We can find the solution for a cubic polynomial graphically. When we plot the graph of a cubic polynomial on the xy plane, the points at which the graph crosses the x-axis are the solutions or the roots of the polynomial.
The two important things to be considered before plotting a graph of a cubic polynomial are:
A cubic polynomial can be graphically expressed as:
The solution of a cubic polynomial are called the roots of a cubic polynomial or zeroes of a cubic polynomial. As the degree of the polynomial is three, the number of roots of a cubic polynomial is three.
We get these roots of the polynomial by either plotting the polynomial on graph paper or by solving the equation with a formula. Let p, q, and r be three roots of the polynomial\(ax^3+bx^2+cx+d\), then the relation between these roots can be given as:
Expressing a cubic polynomial as the product of its factors is termed as factorization of cubic polynomials. Different methods can be adopted to express the polynomial as the product of its factors. These are long division of the polynomial and algebraic identities, and grouping.
Learn about Zeros of a Cubic Polynomial.
Factorizing polynomial can be done using different methods. Here are the steps that need to be followed:
Let us understand this using an example:
Example: Express as the product of factors: \(f(x)=x^3−5x^2+4x−20\)
Solution:
The given polynomial is \(f(x)=x^3−5x^2+4x−20\)
As 5 is the root of the polynomial, so, (x-5) becomes one of the factors.
Dividing the cubic polynomial by (x-5) to get the quadratic polynomial as quotient.
We get, \((x^2+4)\) as quotients.
So, the two factors for the given polynomial are (x-5) and \((x^2+4)\).
Or we can write :
\(x^3−5x^2+4x−20= (x-5)(x^2+4)\)
Learn about x axis and y axis
Example 1: Check whether 2y + 1 is a factor of the polynomial \(4y^3+4y^2-y-1\)or not.
Solution: Given factor is 2y-1=0
So, 2y = 1, and y = 1/2.
Replacing the value of y = 1/2 in the given polynomial.
\(4\left(\frac{1}{2}\right)^3+4\left(\frac{1}{2}\right)^2-\frac{1}{2}-1\)
\(4\left(\frac{1}{8}\right)+4\left(\frac{1}{4}\right)-\frac{1}{2}-1\)
\(\left(\frac{1}{2}\right)+\left(1\right)-\frac{1}{2}-1\)
=0
As y = 1/2 satisfies the given polynomial. So 2y-1 is the factor of \(4y^3+4y^2-y-1\).
Example 2: Classify the given polynomials as cubic polynomials:
p(x):\(5x^2+6x+1\)
q(z): \(z^2−1\)
p(y): \(y^3−6y^2+11y−6\)
q(y): \(81y^3−1\)
Solution: Out of the given polynomials p(y) and q(y) are cubic polynomials as the degree of these polynomials is 3.
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