Math help?

f(x) = 6x³ - 17x² - 5x + 6
f(3) = 6(3)³ - 17(3)² - 5(3) + 6 = 0
According to Factor Theorem, (x - 3) is a factor of f(x).

f(x) = 6x³ - 17x² - 5x + 6
= (6x³ - 18x²) + 18x ²- 17x² - 5x + 6
= (6x³ - 18x²) + x² - 5x + 6
= (6x³ - 18x²) + (x² - 3x) + 3x - 5x + 6
= (6x³ - 18x²) + (x² - 3x) - 2x + 6
= (6x³ - 18x²) + (x² - 3x) - (2x - 6)
= 6x²(x - 3) + x(x - 3) - 2(x - 3)
= (x - 3)(6x² + x - 2)
= (x - 3)(2x - 1)(3x + 2)

(x-3)(2x-1)(3x+2)

f(x) = 6x³-17x²-5x+6
f(3) = 0
factor is x = 3
x = 3 6 -17 -5 6
0 18 3 -6
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6 1 -2 0
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6x²+x-2 = 0
6x²-3x+4x-2 = 0
3x(2x-1(+2(2x-1) = 0
(2x-1)(3x+2) = 0
X = 1/2, x = -2/3

y = 6x³ - 17x² - 5x + 6 => using the rational root theorem the possible rational roots are:

±[(1, 2, 3, 6)/(1, 2, 3, 6)] => by inspection we find that x = 3 is a root and x - 3 is a factor, then by using the synthetic division:

(6x³ - 17x² - 5x + 6)/(x - 3) :

..... 6 ...... -17 ...... -5 ..... 6 .....
3... ↓ ....... 18 ........ 3 .... -6
..... 6 ........ 1 ........ -2 ..... 0 => Remainder
= 6x² + x - 2 => factor the quadratic: AC = 6*-2 = -12, look for 2 numbers that multiply to -12, add to 1=>(-3 and 4):
= 6x² - 3x + 4x - 2 => factor by grouping:
= 3x(2x - 1) + 2(2x - 1)
= (3x + 2)(2x - 1)
hence:
6x³ - 17x² - 5x + 6 = (3x + 2)(2x - 1)(x - 3)

I hope this helps.