find the zeroes of the polynomial x2
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Answered by
1
Step-by-step explanation:
Equation: x²-7x+12
=x²+3x-4x+12
=x(x+3)-4(x+3)
=(x-4)(x+3)
(i)x-4=0
= x = 4
(ii)x+3=0
= x = -3
So, the two zeroes of the given polynomial are 4 and - 3.
HOPE IT HELPS!
Answered by
0
Answer:
x2-7x+12 =0
a=1 b=-7 c=12
a×c= 12
x2-4x-3x+12=0
x(x-4)-3(x-4) =0
(x-3) (x-4)= 0
x=3 ,x=4
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