Final answer:
To find the zeros of the polynomial x²+1.2x10-2x-6.0×10-3=0, we can solve it using the quadratic formula. The quadratic formula is given by: x = (-b ± √(b²-4ac)) / (2a). In this case, a = 1, b = 1.2x10-2, and c = -6.0x10-3. Substituting these values into the quadratic formula and simplifying, we get two possible values for x: x = -0.1918 and x = 0.3818.
Step-by-step explanation:
To find the zeros of the polynomial x²+1.2x10-2x-6.0×10-3=0, we can solve it using the quadratic formula. The quadratic formula is given by: x = (-b ± √(b²-4ac)) / (2a). In this case, a = 1, b = 1.2x10-2, and c = -6.0x10-3. Substituting these values into the quadratic formula and simplifying, we get two possible values for x: x = -0.1918 and x = 0.3818.
To find all zeros of the polynomial, we need to solve the quadratic equation provided. The equation appears to have a typographical error, but the general form given is ax² + bx + c = 0. Let's correct the equation to x² + 1.2 × 10-2x - 6.0 × 10-3 = 0.
Using the quadratic formula, x = ∛/([-b ± √{b² - 4ac}]) / (2a), we can plug in the values: