Find the zeroes of the polynomial2xsq+7/2x+3/4 by factorisation method and verify the relation between the zeroes and the coefficients of the polynomial
Answers
Given polynomial,
- p(x) = 2x² + 7/2x + 3/4
We have to find the zeroes of the given polynomial p(x) and verify the relation between the zeroes and the coefficients of the polynomial.
Since, the polynomial has its some terms in fraction, So let us convert it into integers only. By multiplying the whole polynomial by 4.
⇒ p(x) = 8x² + 14x + 3
⇒ p(x) = 8x² + 2x + 12x + 3
⇒ p(x) = 2x(4x + 1) + 3(4x + 1)
⇒ p(x) = (4x + 1)(2x + 3)
Hence,
- x = -1/4 , -3/2
Let's verify the relation now,
We know,
⇒ Sum of zeroes = - ( coefficient of x) / (coefficient of x² )
⇒ -1/4 + (-3/2) = - (14 / 8)
⇒ -1/4 - 3/2 = -7/4
⇒ -(1 + 6)/4 = -7/4
⇒ -7/4 = -7/4
Similarly,
⇒ Product of zeroes = ( constant term ) / (coefficient of x² )
⇒ -1/4 × -3/2 = 3 / 8
⇒ 3/8 = 3/8
Hence, Verified.
Given :
- the polynomial2xsq+7/2x+3/4
To Find :
- Find the polynomial
- verify the relation between the zeroes and the coefficients of the polynomial
Solution :
Sum of zeros = a + b
Sum of zeros = - 3 / 2 + - 1 / 4
Sum of zeros = - 7 / 4
Product of zeros = a × b
Product of zeros = - 3 / 2 × - 1 / 4
Product of zeros = 3/8
