Question

find zeros of the polynomial given (1) p(x)=x-5 ,q(x)=x-4,h(x)=6x-1,p(x)=ax+b,r(x)=x²+3x,l(x)=x²+2x+1

Answer 1

heya...!

Good evening dear,

ur answer goes like this

(1) p(x)= x - 5


=>x - 5 = 0

=>x = 5


>>>>>>>>>>>>>>



(2) q (x) = x - 4

=>x - 4 = 0

=>x = 4



>>>>>>>>>>>

(3) h(x) = 6x - 1

=>6x - 1 = 0

=>6x = 1

=>x =1/6


>>>>>>>>>>>>>>

(4) p(x) = ax + b

=>ax + b = 0

=>ax = - b

=>x = -b/a

>>>>>>>>>>>>>

(5) r(x) = x² + 3x

=>x² + 3x = 0

=>x(x + 3) = 0


>>>>>>>>>>>>>>

=>x = 0 or x = -3

(6) l(x) x² + 2x + 1

=>x² + 2x + 1 = 0

=>x² + x + x + 1 = 0

=>x(x + 1) + 1(x + 1) = 0

=>(x + 1) (x + 1) = 0

=>x = -1 or -1




Regards

@EVERkshitijEST

Answer 2

Hello dear ✌✌☺
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your question is Find the zeroes of the polynomial .

Given :-
1). P (x) = x - 5

=> x - 5 = 0
=>x = 0 + 5
=>x = 5
so, the zero of this given polynomial is 5

2). q (x) = x - 4

=> x - 4 = 0
=>x = 0 + 4
=>x = 4
so, the zero of this given polynomial is 4

3). h (x) = 6x - 1

=> 6x - 1 = 0
=> 6x = 0 + 1 = 1
=> x = 1/6
so, the zero of given polynomial is 1/6

4). P (x) = ax + b

=> ax + b = 0
=> ax = 0 - b = -b
=>x = -b/a
so, the zero of given polynomial is -b/a

5). r (x) = x² + 3x

=> x² + 3x = 0
=>x ( x + 3 ) = 0
so, the zero of given polynomial is x (x + 3 ) = 0

6). l (x) = x² + 2x + 1

=> x² + 2x + 1
=> x² + x + x + 1
=> x ( x + 1 ) + 1 ( x + 1 )
=> (x + 1 ) (x +1)

here, l (x) = 0
=> x + 1 = 0 or x + 1 = 0
=>x = 0 - 1 or x = 0 - 1
=> X = -1 or x = -1
so, the zeroes of this given polynomial are -1 and -1

_______________________________

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@isharoy688