Let’s start with the concept of “Find all the zeros of the polynomial (2x4-3x3-5x2+9x-3), it is given that two of its zeros are √3 and -√3.
Find all the zeros of the polynomial p(x)= (2x4-3x3-5x2+9x-3), it is given that two of its zeros are √3 and -√3.
Solution: √3 and -√3 are zeros of polynomial P(x) = 2x4-3x3-5x2+9x-3.
x = √3 or x = -√3
x -√3 = 0 or x + √3 =0
(x -√3 )(x +√3 ) = 0 x 0
(x)2 – (√3 )2 = 0
x2 – 3 = 0
(x2 – 3) is factor of P(x) = 2x4-3x3-5x2+9x-3.
(x2 – 3) is completely divisible by P(x) = 2x4-3x3-5x2+9x-3.
When (2x4-3x3-5x2+9x-3) is divided by (x2-3) to get (2x2-3x+1) as a quotient and 0 as a remainder.
Factorise q(x) = (2x2-3x+1)
q(x) = 2x2-3x+1
0 = 2x2-2x-x+1
0 = 2x(x-1) -1(x-1)
0 = (2x-1) (x-1)
2x-1 = 0 or x-1 = 0
X = ½ or x = 1
Hence, all zeros of polynomial P(x) = 2x4-3x3-5x2+9x-3 are √3 , -√3 , 1 and ½
Some important identity of Polynomial
- (a+b)2 = a2 + b2 + 2ab
- (a-b)2 = a2 + b2 – 2ab
- a2 – b2 = (a+b)(a-b)
- (a+b+c)2 = a2+b2+c2+2ab+2bc+2ca
- (a+b)3 = a3+b3+3a2b+3ab2
- (a-b)3 = a3+b3-3a2b+3ab2
- a3+b3 = (a+b)(a2+b2-2ab)
- a3-b3 = (a+b)(a2+b2+2ab)
An expression of the form p(x) = a0+a1x+a2x2+……+anxn, where an is not equal to zero, is called a polynomial in x of degree n.
Degree of polynomials
If P(X) is a polynomial in x, the highest power of x in P(x) is called the degree of the polynomial P(x).
Ex: The degree of polynomial P(X) = 2x3 + 5x2 -7 is 3 because the degree of a polynomial is the highest power of polynomial.
Zero of polynomial
If the value of P(x) at x = K is zero then K is called a zero of the polynomial P(x).
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