P(x)=x2+3x-4then find zeros
Nikki57:
which class question?
Ssc
ok
It's x square not x2
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P ( x)= x^2 +3 x - 4
=x^2 + 4x -x - 4
=x ( x + 4)- 1 ( x + 4)
=( x - 1)( x + 4 )
now put p ( x )= 0
then ,
( x -1) (x + 4 ) = 0
x = 1 , - 4
hence 1 and - 4 are zeroes of p ( x )
=x^2 + 4x -x - 4
=x ( x + 4)- 1 ( x + 4)
=( x - 1)( x + 4 )
now put p ( x )= 0
then ,
( x -1) (x + 4 ) = 0
x = 1 , - 4
hence 1 and - 4 are zeroes of p ( x )
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