Q. 10

# Using factor theo

Let, f (y) = y3-7y+ 6

The constant term in f (y) is equal to + 6 and factors of + 6 are ,

Putting y = 1 in f (y), we have

f (1) = (1)3 – 7 (1) + 6

= 1 – 7 + 6

= 0

Therefore, (y - 1) is a factor of f (y).

Similarly, (y - 2) and (y + 3) are the factors of f (y).

Since, f (y) is a polynomial of degree 3. So, it cannot have more than three linear factors.

Therefore, f (y) = k (y – 1) (y - 2) (y + 3)

y3-7y+ 6 = k (y – 1) (y - 2) (y + 3)

Putting x = 0 on both sides, we get

0 – 0 + 6 = k (0 – 1) (0 - 2) (0 + 3)

6 = 6k

k = 1

Putting k = 1 in f (y) = k (y – 1) (y - 2) (y + 3), we get

f (y) = (y – 1) (y - 2) (y + 3)

Hence,

y3-7y+ 6 = (y – 1) (y - 2) (y + 3)

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