Factorise x3-3x2-9x-5
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Posted by Nikhil Sathya 5 years, 2 months ago
- 3 answers
Gaurav Seth 5 years, 2 months ago
X³ - 3x² - 9x - 5
= x³ + x² - 4x² - 4x - 5x - 5
= x²( x + 1) - 4x ( x + 1) - 5(x + 1)
= (x + 1)(x² - 4x - 5)
= (x + 1)(x² -5x + x - 5)
= (x + 1)(x - 5) (x + 1)
hence, (x +1), (x -5) and (x +1) are the factors of given polynomial .
OR
x3 - 3x2 - 9x - 5
Let p(x) = x3 - 3x2 - 9x - 5
By trial, we find that
p(- 1) = (- 1)3 - 3(- 1)2- 9(- 1) -5
= - 1 - 3 + 9 - 5 = 0
∴ By Factor Theorem, x - (- 1), i.e., (x + 1) is a factor of p(x).
Now,
x3 - 3x2 - 9x - 5
= x2(x + 1) - 4x(x + 1) - 5(x + 1)
= (x + 1)(x2- 4x - 5)
= (x + 1)(x2 - 5x + x - 5)
= (x+ 1){x(x - 5) + 1 (x - 5)}
= (x + 1)(x - 5)(x + 1).
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Ashish Patel 3 years, 8 months ago
0Thank You