Factor Theorem-8

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Factorize x3 - 7x + 6, using factor theorem

Solution: For factorizing the polynomial f(x) = x3 - 7x + 6, we have to first find one value of x,

for which f(x) = 0 by hit and trial method.

For this we see the constant term + 6 and consider its factors ± 1, ± 2, ± 3, ± 6.

Putting these factors as the values of x, one by one in f(x), we find the value of x for which f(x) = 0.

That gives us the first factor of f(x). The rest we find either by continuing this process or by long division

by the factor that we have got.

On putting, x = 1 in f(x) = x3 - 7x + 6, we get

f(i) = 1 – 7 + 6 = 0 => (x - 1) is a factor of x3 - 7x + 6.

On putting, x = 2 in f(x), we get

f(2) = 8 – 14 + 6 = 0 => (x - 2) is a factor of x3 - 7x + 6

On putting, x = -3 in f(x), we get

f( -3) = - 27 + 21 + 6 = 0

=>(x + 3) is a factor of x3 - 7x + 6.

Since the highest power of x in the expression is 3, there cannot be more than three factors.

Therefore, x3 - 7x + 6 = (x - 1) (x - 2) (x + 3)

Alternative method:

After getting one factor, say x – 1, we divide x3 - 7x + 6 by

x – 1 and we get the quotient as x2 + x – 6

which can be easily factorized as (x + 3) (x - 2)

Hence, x3 - 7x + 6 = (x - 1) (x + 3) (x - 2)

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