find the zeroes of polynomial p(x)=6x2-4+10x
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6x² - 4 + 10x
6x² + 10x - 4 = 0
3x² + 5x - 2 = 0
3x² + (6 - 1)x - 2 = 0
3x² + 6x - x - 2 = 0
3x ( x + 2) - 1( x + 2) = 0
( 3x - 1) (x + 2) = 0
3x = 1 and x = -2
x = 1/3 and x = -2
Thus 1/3 and (-2) are the zeroes of polynomial 6x² + 10x - 4.
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6x² + 10x - 4 = 0
3x² + 5x - 2 = 0
3x² + (6 - 1)x - 2 = 0
3x² + 6x - x - 2 = 0
3x ( x + 2) - 1( x + 2) = 0
( 3x - 1) (x + 2) = 0
3x = 1 and x = -2
x = 1/3 and x = -2
Thus 1/3 and (-2) are the zeroes of polynomial 6x² + 10x - 4.
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Answered by
2
Answer:
Step-by-step explanation:
6x²+10x-4
6x²+12x-2x-4
6x(x+2)-2(x+2)
(6x-2)(x+2)
By equating factors with zero
6x-2=0
x=2/6=1/3
x+2=0
x=-2
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