Question

Find the zero of the following polynomials :

(i) P(x) = 2x – 1 (ii) p(x) = 2x – 5 (iii) p(x) = (x + 2)(x – 3)​

Answer 1

Solution

Given :-

1. P(x) = 2x – 1
2. p(x) = 2x – 5
3. p(x) = (x + 2)(x – 3)

Find :-

• Zeroes of given equation

Explanation

(1).

==> p(x) = 2x - 1 = 0

==> 2x = 1

==> x = 1/2

[ Ans]

__________________________

(2).

==> p(x) = 2x - 5 = 0

==> 2x = 5

==> x = 5/2

[ Ans]

___________________________

(3).

==> p(x) = (x+2)(x-3) = 0

==> (x+2) = 0. Or, ( x - 3) = 0

==> x = -2. Or. x = 3

[Ans]

_____________________

Answer 2

QUESTION

Find the zero of the following polynomials :

(i) p(x) = 2x - 1

(ii) p(x) = 2x - 5

(iii) p(x) = (x + 2)(x - 3)​

SOLUTION

Note :- To find the zero(s) of a polynomial, we must equate it with 0.

(i) p(x) = 2x - 1

Equating with 0 :-

$\sf{\implies 2x -1= 0}$

$\sf{\implies 2x = 1}$

$\sf{\implies x=\dfrac{1}{2} }$

Hence, the zero of this polynomial is 1/2.

$\rule{199}2$

(ii) p(x) = 2x - 5

Equating with 0 :-

$\sf{\implies 2x-5=0}\\\\\sf{\implies 2x=5}\\\\\sf{\implies x=\dfrac{5}{2} }$

Hence, the zero of this polynomial is 5/2.

$\rule{199}2$

(iii) p(x) = (x + 2)(x - 3)

$\sf{\implies p(x)=x^2-3x+2x-6}\\\\\sf{\implies p(x)=x^2-x-6}$

Equating with 0 :-

$\sf{\implies x^2-x-6=0}\\\\\sf{\implies x^2-3x+2x-6=0}\\\\\sf{\implies x(x-3)+2(x-3)=0}\\\\\sf{\implies (x-3)(x+2)=0}\\\\\sf{\implies (x-3)=0\:\: or \:\: (x+2)=0}\\\\\sf{\implies x=3 \:\: or \:\: x=-2}$

Hence, the zeros of this polynomial are 3 and -2.