help solve this equation to find the extreme values y=x^3+x^2-8x+5?

y = x³ + x² - 8x + 5
y' = 3x² + 2x - 8
y" = 6x + 2

When y' = 0 :
3x² + 2x - 8 = 0
(x + 2)(3x - 4) = 0
x = -2 or x = 4/3

When x = -2 :
y = (-2)³ + (-2)² - 8(-2) + 5 = 17
y' = 0
y" = 6(-2) + 2 = -10 < 0
Hence, relative maximum value of y = 17 at x = -2

When x = 4/3 :
y = (4/3)³ + (4/3)² - 8(4/3) + 5 = -41/27
y' = 0
y" = 6(4/3) + 2 = 10 > 0
Hence, relative minimum value of y = -41/27 at x = 4/3

y = x^3 + x^2 - 8x + 5
take derivative and equate it to zero, to find critical points
y ' = 3x^2 + 2x - 8
=> 3x^2 + 2x - 8 = 0
=> 3x^2 + 6x - 4x - 8 = 0
=> 3x(x + 2) - 4(x + 2) = 0
=> (3x - 4)(x + 2) = 0
=> x = -2 and 4/3 are critical points
These two critical points divide the graph into three intervals
(-∞, -2), (-2, 4/3) and (4/3,∞)
in the range (-∞, -2) take x = -3
y ' (-3) = ((3(-3) - 4)(-3 + 2) = 13 >0
in the interval (-2, 4/3), take x = 0
y ' (0) = -4*2 = -8 <0
in the interval (4/3, ∞), take x = 2
y ' (2) = (3*2 - 4)(2 + 2) = 8 > 0
therefore x = -2 is local maximum and x = 4/3 is local minimum
f(-2) = -8 + 4 + 16 + 5 = 17 ---------maximum
f(4/3) = (64/27) + (16/9) - (32/3) + 5
= 1/27[64 + 48 - 288 + 135)
= 1/27(-41)
= -41/27 ---------minimum