We have been asked to prove that, (x−1) is a factor of x3−5x2−x+5 hence factories completely
Book – ML Aggarwal, Avichal Publication, Factorisation, chapter 6, Question no 11
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Let us assume, x−1=0⇒x=1
Now, substitute the value of x=1 in f(x),
f(1)=13−(5×12)−1+5
=1−5−1+5
=−6+6
=0
∴ by factor theorem,
(x−1) is a factor of f(x)
Now dividing f(x) by (x−1), we get
x
3
−5x
2
−x+5=(x−1)(x
2
−4x−5)
=(x−1)(x
2
−5x+x−5)
=(x−1){x(x−5)+1(x−5)}
=(x−1)(x+1)(x−5)