Using factor theorem, factorize each of the following polynomial:y3 -

1.	Using factor theorem, factorize each of the following polynomial:y3 - 7y + 6
Answer» Let, f (y) = y³ - 7y + 6 The constant term in f (y) is equal to + 6 and factors of \(\pm\) 1, \(\pm\) 2, \(\pm\) 3 and \(\pm\) 6 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)

Using factor theorem, factorize each of the following polynomial:y3 - 7y + 6

Answer»

Let, f (y) = y³ - 7y + 6

The constant term in f (y) is equal to + 6 and factors of \(\pm\) 1, \(\pm\) 2, \(\pm\) 3 and \(\pm\) 6

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)