Q. 9

# Using factor theo

Let, f (x) = x3-23x2+142x-120

The factors of the constant term – 120 are

Putting x = 1, we have

f (1) = (1)3 – 23 (1)2 + 142 (1) – 120

= 1 – 23 + 142 – 120

= 0

So, (x – 1) is a factor of f (x)

Let us now divide

f (x) = x3-23x2+142x-120 by (x - 1) to get the other factors of f (x)

Using long division method, we get

x3-23x2+142x-120 = (x – 1) (x2 – 22x + 120)

x2 – 22x + 120 = x2 – 10x – 12x + 120

= x (x – 10) – 12 (x – 10)

Hence, x3-23x2+142x-120 = (x – 1) (x - 10) (x - 12)

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