Explanation:
To find the zeros of the polynomial p(x) = (2x^2 - 9x + 7)(x - 2), we set p(x) equal to zero and solve for x:
(2x^2 - 9x + 7)(x - 2) = 0
This equation will be satisfied if either of the factors on the left side equals zero. So we set each factor equal to zero and solve for x separately.
Setting 2x^2 - 9x + 7 = 0:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 2, b = -9, and c = 7.
Plugging these values into the quadratic formula, we have:
x = (-(-9) ± √((-9)^2 - 4 * 2 * 7)) / (2 * 2)
x = (9 ± √(81 - 56)) / 4
x = (9 ± √25) / 4
x = (9 ± 5) / 4
So we have two solutions:
x1 = (9 + 5) / 4 = 14 / 4 = 3.5
x2 = (9 - 5) / 4 = 4 / 4 = 1
Setting x - 2 = 0:
x = 2
Therefore, the zeros of the polynomial p(x) = (2x^2 - 9x + 7)(x - 2) are x = 3.5, 1, and 2.