Question

Find the zeroes of the given polynomial :-

(i) 3x^2-x-4.
(ii) 4u^2+8u.

Quick answer.

Answer 1

heya...!!!!

--------(1)

3x² - x - 4

=> 3x² - 4x + 3x - 4

=> x(3x - 4) + 1(3x - 4)

=> (x + 1)(3x - 4)

=> x = -1 , x = 4/3

-------(2)

4u² + 8u

=> 4u(u + 2) = 0

=> u = 0 ,u = -2

Answer 2

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we can do it by factorization method

=> 3x² - x - 4
=> 3x² -4x+3x-4
=> x( 3x - 4). + 1 (3x -4 )
=> (3x-4) (x+1)
=> x=4/3. x = -1

hece zeroes of first polynomial is 4/3 and -1

same as first we will do the second ione
ii ). = 4u² +8
= 4u(u+2)
= (4u+0) (u+2)
u = 0 => -2

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