algebra problem please help?!?!?

x^3 - x^2 - 18x + 18 = 0

can be factored by grouping

x^2(x - 1) - 18(x - 1) = 0

(x - 1)(x^2 - 18) = 0

x = 1

x^2 = 18
x = ± 3√2

Check

QED

x³ – x² – 18x + 18 = 0
(x³ – x²) – (18x – 18) = 0
x²(x – 1) – 18(x – 1) = 0
(x – 1)(x² – 18) = 0
(x – 1) = 0
x = 1
(x² – 18) = 0
x² = 18
x = ± √18
x = ± 3√2

x^3 - x^2 - 18x + 18 = 0
(x^3 - x^2) - (18x - 18) = 0
x^2(x - 1) - 18(x - 1) = 0
(x - 1)(x^2 - 18) = 0

x - 1 = 0
x = 1

x^2 - 18 = 0
x^2 = 18
x = ±√18
x = ±√(2 * 3^2)
x = ±3√2

∴ x = 1, ±3√2

f ( 1 ) = 1 - 1 - 18 + 18 = 0

Thus x - 1 is a factor

Find other factors by synthetic division :-

1 | 1______-1______-18______18
_ |________1_______0______- 18
_ | 1______ 0______- 18______ 0

( x - 1 ) ( x ² - 18 ) = 0

x = 1 , x = ± √18

x = 1 , x = ± 3 √ 2

x^3 - x^2 - 18x + 18 = 0. Factor by grouping. Group the first two and last two terms separately.

First group: x^3 - x^2. Factor out x^2 = x^2(x - 1). Come back to this later.

Second group: 18x + 18. Factor out 18 = 18(x - 1). Put the two on one line.

x^2(x - 1) - 18(x - 1) = 0. Note that you have (x - 1) that can be factored out. This will leave x^2 - 18 left.

(x - 1)(x^2 - 18) = 0. Set each binomial = to 0.

x - 1 = 0. x = 1. Answer.

x^2 - 18 = 0. x^2 = 18. x = +/- sq rt 18. Sq rt 18 = (sq rt 9)(sq rt 2) = 3(sq rt 2). Thus,

x = + 3(sq rt 2) and - 3(sq rt 2). Answer.

Answers: x = 1, 3(sq rt 2), - 3(sq rt 2).

solve by grouping

x^3-x^2-18x+18

x^2(x-1)-18(x-1)
x^2-18 x-1

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