Second degree polynomial
Introduction :
The maximum power of the variable in a given polynomial is called the degree of the polynomial. The degree of the polynomial 5x2 + 13x – 9 in x is 2. We say that it is a second- degree of polynomial or a polynomial of degree 2. Quadratic polynomial is also called as second degree polynomial.
The basic form of a second degree polynomial is given by
ax2 + bx + c
Where, a is the coefficient of x2, b is the coefficient of x, c is the constant and x is the variable.
Three methods are used to solve the quadratic polynomial equation.
1) Factoring method
2) Quadratic formula method
3) Completing the square method
Second degree polynomial problems:
Example problem 1:
Find whether the given equation is a second degree polynomial or not.
(x - 2) (x - 3) = 0
Solution:
Multiply two expressions in the left hand side,
(x - 2) (x - 3) = 0
x(x-3) -2(x-3)=0
x2 -3x -2x + 2(3) = 0
x2 -5x +6 = 0
So, the highest exponent of variable x is 2.
So, the given Solving Polynomial Equations is second degree polynomial equation.
Example problem 2:
Solve the quadratic equation by factoring: (x + 3) (x + 5) = 1 – x
Solution:
(x + 3) (x + 5) = 1 – x
x2 + 8x + 15 – 1 + x = 0
x2 + 9x + 14 = 0
Here a = coefficient of x2 = 1
b = coefficient of x = 9
c = constant term = 14
We find a × c = 1 × 14 = 14 = 7 × 2, 7 + 2 = 9 = b. Hence factorization by splitting the middle term and grouping as follows:
x2 + 9x + 14 = 0
(x + 2) (x + 7) = 0
x = –2 x = –7
The solution set = {–2, –7}.
Additional second degree polynomial problems:
Example problem 3:
Factorize 2x2+ 7x + 3.
Solution:
Here a = coefficient of x2 = 2
b = coefficient of x = 7
c = constant term = 3
We find a × c = 2 × 3 = 6 = 6 × 1, 6 + 1 = 7 = b.
Hence factorization by splitting the middle term and grouping as follows:
2x2 + 7x + 3 = 2x2 + (6 + 1)x + 3
= 2x2 + 6x + x + 3
= 2x(x + 3) + (1)(x+3)
= (2x+1) (x+3).
Example problem 4:
Solve the equation by completing the square x2 + 6x – 7 = 0.
Solution: x2 + 6x = 7. The term to be added =(6/2)2=9
Adding 9 on both sides we get
x2 + 6x + 9 = 7 + 9
(x + 3)2 = 16
Taking square root on both sides x + 3 = ± 16 = ± 4
x + 3 = 4 x + 3 = –4
x=1 x=-7
So, the answer is x=1 (or) -7.
The maximum power of the variable in a given polynomial is called the degree of the polynomial. The degree of the polynomial 5x2 + 13x – 9 in x is 2. We say that it is a second- degree of polynomial or a polynomial of degree 2. Quadratic polynomial is also called as second degree polynomial.
The basic form of a second degree polynomial is given by
ax2 + bx + c
Where, a is the coefficient of x2, b is the coefficient of x, c is the constant and x is the variable.
Three methods are used to solve the quadratic polynomial equation.
1) Factoring method
2) Quadratic formula method
3) Completing the square method
Second degree polynomial problems:
Example problem 1:
Find whether the given equation is a second degree polynomial or not.
(x - 2) (x - 3) = 0
Solution:
Multiply two expressions in the left hand side,
(x - 2) (x - 3) = 0
x(x-3) -2(x-3)=0
x2 -3x -2x + 2(3) = 0
x2 -5x +6 = 0
So, the highest exponent of variable x is 2.
So, the given Solving Polynomial Equations is second degree polynomial equation.
Example problem 2:
Solve the quadratic equation by factoring: (x + 3) (x + 5) = 1 – x
Solution:
(x + 3) (x + 5) = 1 – x
x2 + 8x + 15 – 1 + x = 0
x2 + 9x + 14 = 0
Here a = coefficient of x2 = 1
b = coefficient of x = 9
c = constant term = 14
We find a × c = 1 × 14 = 14 = 7 × 2, 7 + 2 = 9 = b. Hence factorization by splitting the middle term and grouping as follows:
x2 + 9x + 14 = 0
(x + 2) (x + 7) = 0
x = –2 x = –7
The solution set = {–2, –7}.
Additional second degree polynomial problems:
Example problem 3:
Factorize 2x2+ 7x + 3.
Solution:
Here a = coefficient of x2 = 2
b = coefficient of x = 7
c = constant term = 3
We find a × c = 2 × 3 = 6 = 6 × 1, 6 + 1 = 7 = b.
Hence factorization by splitting the middle term and grouping as follows:
2x2 + 7x + 3 = 2x2 + (6 + 1)x + 3
= 2x2 + 6x + x + 3
= 2x(x + 3) + (1)(x+3)
= (2x+1) (x+3).
Example problem 4:
Solve the equation by completing the square x2 + 6x – 7 = 0.
Solution: x2 + 6x = 7. The term to be added =(6/2)2=9
Adding 9 on both sides we get
x2 + 6x + 9 = 7 + 9
(x + 3)2 = 16
Taking square root on both sides x + 3 = ± 16 = ± 4
x + 3 = 4 x + 3 = –4
x=1 x=-7
So, the answer is x=1 (or) -7.