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Q.


The factors of x³ – 3x² – 9x – 5 are:

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a

(x + 2) (x + 1) (x - 5)

b

(x + 1) (x + 1) (x - 5)

c

(x + 1) (x + 6) (x - 5)

d

(x + 1) (x + 1) (x - 4) 

answer is B.

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Detailed Solution

Given: x³ – 3x² – 9x – 5
First, Let us find the factors of a constant term.
Secondly, let us check out at which factor the given polynomial becomes zero by using the hit and trial method and get one factor of the given polynomial. Then, we shall further divide the polynomial by this factor to get a quotient of the second degree.
Finally, we shall factorize the second-degree polynomial by middle term splitting to get required factors.
Let p(x) = x³ – 3x² – 9x – 5
Here, the constant term is 5.
Now, let us first find the factors of 5:
Factors of 5 are ±1 and ± 5.
Using hit and trial method, let us substitute x = - 1in p(x) we get:
p(x)= x 3 −2 x 2 −x+2p(5)  
= (−1) 3 −3 (−1) 2 −9(−1)−5  
=−1−3+9−5  
=−4+4  
=0  
Therefore, (x + 1) is the factor of  p(x).
Now, by dividing p(x) by (x + 1), we get:
 Question ImageThen p(x) = (x + 1) (x² – 4x - 5).
Here quotient is (x² – 4x - 5).
Then by using the middle term splitting in the quotient, we get:
=(x+1)( x 2 −5x+x−5)  
=(x+1)x(x−5)+1(x−5)  
=(x+1)(x−5)(x+1)  
Hence , the required factorize of p(x) is  (x + 1) (x + 1) (x - 5).
Therefore, the correct option is 2.
 
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