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Polynomial equations establish a significant pattern between algebraic terms and exponents. The graph of a polynomial equation can also be plotted using intercepts, turning points, and the Intermediate Value Theorem. A polynomial equation is an equation made up of multiple terms consisting of variables as well as numbers. This equation can have different types of exponents. The highest exponent can be found out by the degree of the polynomial equation. The degree of a polynomial equation also tells us the roots of that particular equation.
Also Read: Zeroes of Polynomial
Key Terms: Polynomial equations, Polynomial functions, variables, number, exponent, constants, coefficients, linear, quadratic
Polynomial Equations
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Polynomial equations are equations that have multiple components including constants, variables, coefficients, and exponents. The equation is built on a single independent variable and the terms can be of different exponential degrees. A polynomial equation can be solved by factoring them on the basis of their degrees and variables in the given equation.
Also Read: Degree of Polynomial
How to Solve Polynomial equations?
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Polynomial equations can be easily solved if one has a good grasp of the concepts of basic algebra. These equations can be solved using general algebraic and factorization rules. The first and foremost step in solving a polynomial equation is to somehow produce or obtain a zero on the right-hand side of the polynomial equation.
Read More About: Quadratic equations formula
Examples of Polynomial Equations
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These are few examples of polynomial equations i.e. their expressions-
- 4x3 - 2x2 + 8x - 21 = 0
- 3x2 - 4x - 12 = 0
- 2y2 + 3y + 4 = f(x)
- 9x3 + 5x2 – 4x – 2 = 0
- 2x3 - 13x2 + 17x + 12 = g(x)
- 3x4 - 4x3 - 3x - 1 = p(x)
Also Read: Complex Numbers and Quadratic equations
Degrees of Polynomial equations
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The degree of a polynomial is the highest exponential term of the polynomial equation. Based upon the degree of a polynomial, there are five types for polynomial equations, namely,
- Constant Polynomial - A polynomial which has a degree of 0 is known as a constant polynomial.
Example - f(x) = 2, g(x) = -14, h(y) =5/2 etc are constant polynomials. The zero polynomial is also known as the constant polynomial 0 or f(x) =o.
- Linear Polynomial - A polynomial which has a degree 1 is known as a constant polynomial.
Example- f(x) = x-12 , g(x) = 12x, h(x) = -7x + 8 are linear polynomials.
- Quadratic Polynomial - A polynomial which has a degree 2 is known as a quadratic polynomial.
Example - f(x) = 2x2 - 3x + 15, g(x) = (3/2) y2 - 4y + 11/3 etc are quadratic polynomials.
- Cubic Polynomial- A polynomial that has a degree 3 is known as a cubic polynomial.
Example - f(x) = 12x3 - 4x2 + 7x - 6, g(x) = 7x3 + 4x - 12 are cubic polynomials.
- Bi-quadratic Polynomial-A polynomial with degree 4 is known as a biquadratic polynomial.
Example - f(x) = 12x - 7x3 + 8x² - 12c - 20 is biquadratic polynomial.
Also Read: Polynomial Notations
Polynomial Equations Formula
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The Polynomial Equation Formula is generally denoted in the form of anxn, where a is considered as a coefficient, x is a variable, and n is considered as an exponent.
Polynomial Equation Formula in expanded form can be written in the following manner:
f(x) = aoxn + an-1xn-1 + an-2xn-2 + …….. + a1x +a0
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Things to Remember
- A polynomial equation is an equation made up of multiple terms consisting of variables as well as numbers.
- Polynomial Equation Formula in expanded form can be written in the following manner,
- f(x) = anxn + an-1xn-1 + an-2xn-2 + …….. + a1x +a0
- where a is considered as a coefficient, x is a variable, and n is considered as an exponent.
- There are five types of polynomial equations, namely,
- Constant Polynomial
- Linear Polynomial
- Quadratic Polynomial
- Cubic Polynomial
- Bi-quadratic Polynomial.
- A polynomial equation can be solved by factoring them on the basis of their degrees and variables in the given equation.
- A few examples of polynomial equations i.e. their expressions-
- 4x3 - 2x2 + 8x - 21 = 0
- 3x2 - 4x - 12 = 0
Also Read: Polynomials in one variable
Sample Questions
Ques. Find the value of the f (2) and f (-3) in the given Polynomial equation f(x) =2x³-13x²+17x+12. [ 2 Marks ]
Ans. - We have f(x) =2x³-13x²+17x+12
f(2)=2x(2)³-13x(2)²x17x2+12
=2x8-13x4+34+12
=16-52+34+12
=10
f(-3)=2x(3)³-13x(-3)²+17x(-3)+12
=2x27-13x9+17x-3+12
= 54-117-51+12
=-210
Ques. Find the value of the polynomial 5x-4x²+3 when x=0. [1 Marks ]
Ans. Let f(x) = 5x-4x²+3
Now we put 0 in place of x,
we will get f(0) = 5x (0)-4x(0)²+3
=0-0+3
=3
Ques. y3 - y2 + y - 1 = 0 is a cubic polynomial equation. Find the roots of it. [ 3 Marks]
Ans. Given equation is y3 - y2 + y - 1 = 0.
Now using the trial and error method, start putting the value of x.
If y = - 1 then,
(- 1)3 - (- 1)2 - 1 - 1 = 0
-1-1-1-1 = 0
- 4 = 0, which is not possible.
Thus, -1 is not a root of this equation.
If y = 1 , then,
13 - 12 + 1 -1 = 0
0 = 0, which is true.
Therefore, one of the roots is 1.
y = 1
Therefore, the roots of this equation are y = 1.
Ques. Find the value of each of the following polynomials at the indicated value of variables:
(i) p(x) = 5x2 – 3x + 7 at x=1.
(ii) q(y)=3y3 – 4y + \(?\)11 at y=2. [2 Marks ]
Ans. (i) p(x) = 5x2 – 3x + 7
The value of the polynomial p(x) at x = 1 is,
p(1) = 5(1)2 – 3(1) + 7
= 5 – 3 + 7
= 9
(ii) q(y) = 3y3 – 4y + 11
The value of the polynomial q(y) at y = 2 is,
q(2) = 3(2)3 – 4(2) + 11
=24 – 8 + 11
=16+ 11
= 27
Ques. Check whether –2 and 2 are zeroes of the polynomial x + 2. [1 Marks]
Ans. Let p(x) = x + 2.
Then p(2) = 2 + 2 = 4,
p(–2) = –2 + 2 = 0
Thus, –2 is a zero of the polynomial equation, x + 2, but 2 is not.
Ques. Find a zero of the polynomial p(x) = 2x + 1. [1 Marks]
Ans. Finding the zero of the polynomial p(x), means
p(x) = 0
Therefore, 2x + 1 = 0, gives
x = –1/2
So, – 1/2 is a zero of the polynomial 2x+1.
Ques. Divide the polynomial equation 3x4 – 4x3 – 3x – 1 by x – 1. [3 Marks]
Ans. By long division, we get,
Here, the remainder is –5.
Now, the zero of x – 1 is 1.
So, by substituting x = 1 in p(x), we get
p(1)= 3(1)4 – 4(1)3 – 3(1) – 1
= 3 – 4 – 3 – 1
= – 5, which is the remainder.
Ques. Find the remainder obtained on dividing p(x) = x3 + 1 by x + 1. [2 Marks ]
Ans. By long division,
So, we get the remainder 0.
Here p(x) = x3 + 1, and the root of x + 1 = 0 is x= -1.
Now, p(–1) = (–1)3 + 1 = -1 + 1
= 0,
which is equal to the remainder.
Ques. Check whether the polynomial q(t) = 4t3 + 4t2 – t – 1 is a multiple of 2t + 1. [2 Marks]
Ans. Now, q(t) will be a multiple of 2t + 1 only, if q(t) is divisible by 2t + 1 and has a remainder zero.
Now, taking 2t + 1 = 0, we have t = – 1/2 .
Substituting value of t,
q(-1/2) = 4(-1/2)3 + 4(-1/2)2 - (-1/2) -1.
So the remainder obtained on dividing q(t) by 2t + 1 is 0.
So, 2t + 1 is a factor of the given polynomial q(t),which means q(t) is a multiple of 2t + 1.
Ques. Factorise x3 – 23x2 + 142x – 120. [3 Marks]
Ans. Let p(x) = x3 – 23x2 + 142x – 120.
We shall now look for all the factors of –120. Some of these are ±1, ± 2, ± 3, ± 4, ± 5, ± 6,± 8,± 10, ± 12, ± 15, ± 20, ± 24, ± 30, ± 60.
By trial, we find that p(1) = 0.
So x – 1 is a factor of p(x).
Now we see that x3 - 23x2 + 142x - 120
= x3 – x2 – 22x2 + 22x + 120x - 120.
= x2 (x - 1) - 22x(x–1) + 120(x–1)
=(x–1)(x2 - 22x + 120)
Now x2 - 22x + 120 can be factorised either by splitting the middle term or by using
the Factor theorem. By middle term splitting method, we have
x2 - 22x + 120 = x2 - 12x - 10x + 120
So,
= x(x – 12) – 10(x – 12)
= (x – 12) (x – 10)
x3 - 23x2 - 142x - 120
= (x–1)(x–10)(x–12)
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