Factorise
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2x³ - 3x² - 17x + 30 =
note that x = 2 is a zero of the polynomial:
2(2)³ - 3(2)² - 17(2) + 30 =
2(8) - 3(4) - 34 + 30 =
16 - 12 - 4 = 0
then the polynomial is divisible by (x - 2):
let's verify this by synthetic division:
...2 | 2...- 3.....-17....+30
......|
......| .....+4.....+2.....-30
-----------------------------------
.......2....+1.....-15.......0
we get:
(2x³ - 3x² - 17x + 30) /(x - 2) = 2x² + (1)x - 15
therefore:
(2x³ - 3x² - 17x + 30) = (x - 2)(2x² + x - 15)
let's now factor the quadratic replacing x with 6x - 5x:
2x² + x - 15 = 2x² + 6x - 5x - 15 =
(factoring by grouping)
2x(x + 3) - 5(x + 3) =
(factoring the common term (x + 3))
(x + 3)(2x - 5)
thus the answer is:
(2x³ - 3x² - 17x + 30) = (x - 2)(x + 3)(2x - 5)
note that x = 2 is a zero of the polynomial:
2(2)³ - 3(2)² - 17(2) + 30 =
2(8) - 3(4) - 34 + 30 =
16 - 12 - 4 = 0
then the polynomial is divisible by (x - 2):
let's verify this by synthetic division:
...2 | 2...- 3.....-17....+30
......|
......| .....+4.....+2.....-30
-----------------------------------
.......2....+1.....-15.......0
we get:
(2x³ - 3x² - 17x + 30) /(x - 2) = 2x² + (1)x - 15
therefore:
(2x³ - 3x² - 17x + 30) = (x - 2)(2x² + x - 15)
let's now factor the quadratic replacing x with 6x - 5x:
2x² + x - 15 = 2x² + 6x - 5x - 15 =
(factoring by grouping)
2x(x + 3) - 5(x + 3) =
(factoring the common term (x + 3))
(x + 3)(2x - 5)
thus the answer is:
(2x³ - 3x² - 17x + 30) = (x - 2)(x + 3)(2x - 5)
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