Find the zeroes of polyonimal x2-x-12 by graph
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x2-x-12
x2-4x+3x-12
x(x-4)+3(x-4)
(x+3)(x-4)
x+3=0,x-4=0
x=-3,x=4
x2-4x+3x-12
x(x-4)+3(x-4)
(x+3)(x-4)
x+3=0,x-4=0
x=-3,x=4
Answered by
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x²-x-12
x²-4x+3x-12
x(x-4) +3(x-4)
(x-4) (x+3)
Zeroes are....
x-4=0
x=0+4
x=4
x+3=0
x=0-3..
x= - 3
ZEROES ARE..... 4,-3
x²-4x+3x-12
x(x-4) +3(x-4)
(x-4) (x+3)
Zeroes are....
x-4=0
x=0+4
x=4
x+3=0
x=0-3..
x= - 3
ZEROES ARE..... 4,-3
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