Solving Polynomials
Introduction
To solve a polynomial, we put that polynomial equals zero (0) and find the value of the variable in it. Values of the variable are called zeroes or roots of the polynomial which depend on the degree of the polynomial. if the degree of the polynomial is 1 then there will be one zero or root and if the degree is 2 then there will be two zeroes or roots.

- Solving a linear polynomial
- Solving a quadratic polynomial
- Solving a cubic polynomial
Solving a Linear Polynomial
In the linear polynomial, the degree is always 1 so there will be one and only one zero. if the linear polynomial is p(x) then to find the zero of the polynomial p(x) we have to solve the equation p(x) = 0.
Examples
Example тАУ 1) find the zero of the polynomial p(x) = 3x + 2.
Solution тАУ let p(x) = 0
3x + 2 = 0
3x = – 2
x = -2/3
x = -2/3 is the zero of the polynomial p(x) = 3x + 2. Ans.
Example тАУ 2) find the zero of the polynomial p(y) = 2y тАУ 6.
Solution тАУ let p(y) = 0
2y тАУ 6 = 0
2y = 6
y = 6/2
y = 3
Zero of the polynomial p(y) = 2y тАУ 6 is y = 3. Ans.
Note тАУ Standard form of the linear polynomial is ax + b = 0 where a тЙа 0 so the zero will be x = -b/a. we can also find the zero by comparing it.
Solving a Quadratic Polynomial
In the quadratic polynomial, the degree is 2 so there will be two zeroes.
Examples
Example тАУ 1) find the zeroes of the polynomial x2 тАУ 3x.
Solution тАУ let p(x) = x2 тАУ 3x
Now P(x) = 0
x2 тАУ 3x = 0
x(x тАУ 3) = 0
x = 0 and x тАУ 3 = 0
x = 0 and x = 3
The two zeroes are x = 0 and x = 3. Ans.
Example тАУ 2) find the zeroes of the polynomial 6x2 + 5x тАУ 6.
Solution тАУ let p(x) = 6x2 + 5x тАУ 6
Now p(x) = 0
6x2 + 5x тАУ 6 = 0
6x2 + 9x тАУ 4x тАУ 6 = 0 (By factorization method) [9тИТ4 = 5 and 9тип(-4) = -36]
3x(2x+3) тАУ 2(2x+3) = 0
(2x + 3)(3x тАУ 2) = 0
2x + 3 = 0 and 3x тАУ 2 = 0
2x = -3 and 3x = 2
x = -3/2 and x = 2/3
These are the zeroes of the given polynomial. Ans.
Solving a Cubic Polynomial
In the cubic polynomial, the degree is 3 so there will be three zeroes. To solve the cubic polynomial first we have to arrange the polynomial in descending order then we solve it by taking common terms or by factorization.
Examples
Example тАУ 1) find the zeroes of the polynomial x3 + 2x2 тАУ x тАУ 2.
Solution тАУ let p(x) = x3 + 2x2 тАУ x тАУ 2
Now p(x) = 0
x3 + 2x2 тАУ x тАУ 2 = 0
x2(x + 2) тАУ 1(x + 2) = 0
(x + 2)(x2 тАУ 1) = 0
x + 2 = 0 and x2 тАУ 1 = 0
x = -2 and x2 = 1
x = ┬▒тИЪ1
x = ┬▒1
So, the zeroes are x = -2, x = +1, and x = -1. Ans.
Note тАУ We can check the answer by putting the value of each zero in the given polynomial because, at each value of zero, the value of the polynomial is 0.
Example тАУ 2) find the zeroes of the polynomial x3 + 6x2 + 11x + 6.
Solution тАУ let p(x) = x3 + 6x2 + 11x + 6
There is not any common term in it.
So, the factors of the constant term 6 = ┬▒1, ┬▒2, ┬▒3 and ┬▒6
Now at x = +1
p(1) = (1)3 + 6(1)2 + 11(1) + 6
p(1) = 1 + 6 + 11 + 6
p(1) = 24
тИ╡ p(1) тЙа 0 so x = +1 is not the zero of this polynomial.
Here, all the terms in the polynomial are positive so we shall take only negative factors.
Now at x = – 1
p(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6
p(-1) = -1 + 6 тАУ 11 + 6
p(-1) = 0
тИ╡ p(-1) = 0 so x = -1 is the zero of this polynomial.
Now at x = -2
p(-2) = (-2)3 + 6(-2)2 + 11(-2) + 6
p(-2) = – 8 + 6(4) тАУ 22 + 6
p(-2) = – 8 + 24 тАУ 22 + 6
p(-2) = 0
тИ╡ p(-2) = 0 so x = -2 is the zero of this polynomial.
Now at x = -3
p(-3) = (-3)3 + 6(-3)2 + 11(-3) + 6
p(-3) = -27 + 6(9) тАУ 33 + 6
p(-3) = -27 + 54 тАУ 33 + 6
p(-3) = 0
тИ╡ p(-3) = 0 so x = -3 is the zero of this polynomial.
тИ╡ The degree of the given polynomial is 3 so there will be only three zeroes.
So, the zeroes of the given polynomial are x = -1, x = -2, and x = -3. Ans.