find zeros of polynomial x^2+4x-192?
Answers
Answered by
19
First factorize 192
x^2+4x-192=x^2-12x+16x-192
=x(x-12)+16(x-12)
=(x-12)(x+16)
x=12 or x=-16
The zeroes are 12 & -16
Answered by
28
⇒x² + 4x -192 = 0
⇒x² + 16x - 12x - 192 = 0
⇒ x ( x + 16 ) -12 ( x + 16 ) = 0
⇒ ( x + 16 ) ( x - 12 )
⇒ There fore..
⇒ X = - 16 and X = 12
⇒x² + 16x - 12x - 192 = 0
⇒ x ( x + 16 ) -12 ( x + 16 ) = 0
⇒ ( x + 16 ) ( x - 12 )
⇒ There fore..
⇒ X = - 16 and X = 12
Similar questions