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Sullivan Nicholson
Mathematics
10 June, 16:59
Factorise:-
x^3 - 6x^2 + 11x - 6
+5
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2
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Ciara Miller
10 June, 17:04
0
(x - 1) (x - 2) (x - 3).
Step-by-step explanation:
f (x) = x^3 - 6x^2 + 11x - 6
f (1) = 1 - 6 + 11 - 6 = 12 - 12
= 0 so by the Factor Theorem (x - 1) is a factor.
Also, by the Rational Root theorem, as the last term is - 6 and the leading coefficient is 1 some of - 1, 2,-2, 3 - 3, 6,-6 might be zeroes of the function.
f (-1) = - 1 + 6 - 11 - 6 = - 12 so - 1 is not a zero and (x + 1) is not a factor.
f (2) = 8 - 24 + 22 - 6
= 30 - 30 = 0 so (x - 2) is also a factor)
Since the last term is - 6 the last factor must be (x - 3)
=Checking: f (3) = 27 - 6 (9) + 33 - 6
= 27 - 54 + 33 - 6
= 60-60 = 0.
So the factors are (x - 1) (x - 2) (x - 3).
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Logan Cisneros
10 June, 17:26
0
(x - 1) (x - 2) (x - 3)
Step-by-step explanation:
Note the sum of the coefficients
1 - 6 + 11 - 6 = 0
hence x = 1 is a root and (x - 1) is a factor
dividing x³ - 6x² + 11x - 6 by (x - 1) gives
(x - 1) (x² - 5x + 6)
To factor the quadratic
Consider the factors of + 6 which sum to give - 5
The factors are - 2 and - 3, since
- 2 * - 3 = 6 and - 2 - 3 = - 5, hence
x² - 5x + 6 = (x - 2) (x - 3) and
x³ - 6x² + 11x - 6 = (x - 1) (x - 2) (x - 3)
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