6x^2-3-7x find the zeroes of these polynomial
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Given polynomial : 6x² - 7x - 3
6x² - 7x - 3 = 0
⇒ 6x² + 2x - 9x - 3 = 0
⇒ 2x(3x+1) - 3(3x+1) = 0
⇒ (3x+1) (2x-3) = 0
⇒ 3x+1 = 0 or 2x-3 = 0
⇒ 3x = -1 or 2x = 3
⇒ x = -1/3 or x = 3/2
∴ The zeroes of the polynomial are -1/3 and 3/2
6x² - 7x - 3 = 0
⇒ 6x² + 2x - 9x - 3 = 0
⇒ 2x(3x+1) - 3(3x+1) = 0
⇒ (3x+1) (2x-3) = 0
⇒ 3x+1 = 0 or 2x-3 = 0
⇒ 3x = -1 or 2x = 3
⇒ x = -1/3 or x = 3/2
∴ The zeroes of the polynomial are -1/3 and 3/2
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