1.) x²(x-9)(2x +1) = 0
It means x² = 0 or (x-9) = 0 or (2x +1) =0. Solving each.
x = 0, x = 0 +9, x = 9 2x +1 = 0 2x = 0 -1 2x = - 1, x = -1/2
x = 0 or 9 or -1/2
2) (x + 4)(x² - x +3) = 0
It means (x+4) = 0 or (x² - x +3) = 0
for: x + 4 =0
x = 0-4
x = -4
for: x² - x +3 = 0. This is a quadratic equation. We would use quadratic formula:
Comparing to ax^2 + bx + c = 0, and using the quadratic formula:
x = (-b + sqrt(b^2 - 4ac) / 2a
a = 1, b = -1, c = 3.
x = (- -1 + Square Root ( -1 ^ 2 - 4 * 1 * 3))
/ (2 *1))
x = (1 + Square Root (1 - 12) ) / 2 Imaginary roots.
x = -4 only real solution
3) (x+8)(x-7)(x²-2x+5) =0
x+8 = 0 or x-7 = 0 or x²-2x+5 = 0
x = 0-8 x= 0 +7
x = -8 x = 7
x²-2x+5 = 0
x = (-b + sqrt(b^2 - 4ac) / 2a
a = 1, b = -2, c = 5.
x = (- -2 + Square Root ( -2 ^ 2 - 4 * 1 * 5))
/ (2 *1)
x = (1 + Square Root (4 - 20) ) / 2 Imaginary roots.
x = -8, 7 the only real solutions
