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Question

Factorise the expression x3−6x2+11x−6.

A
(x+1)(x+2)(x+3)
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B
(x1)(x+2)(x+3)
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C
(x+1)(x+2)(x3)
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D
(x1)(x2)(x3)
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Solution

The correct option is D (x1)(x2)(x3)
Let us assume that all factors are integers. Product of the factors is equal to the constant term (-6). Possible roots are the factors of -6. Hence, possible factors are:
±1,±2,±3,±6

For x = 1, the value of the given expression becomes:p(1)=136×12+11×16=0

By factor theorem, (x+1) is a factor of x36x2+11x6.

Performing long division as shown below:

x25x+6x1)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x36x2+11x6 x3x2 + ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 5x2+11x6 5x2+5x + ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 6x6 6x6 ––––––––––– 0 –––––––––––

We get the quotient:
q(x)=x25x+6=x22x3x+6=x(x2)3(x2)=(x3)(x2)

Polynomial p(x) can be written as:p(x)=(x1)q(x)p(x)=(x1)(x2)(x3)

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