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

1.	Using factor theorem, factorize each of the following polynomial:y3 - 2y2 - 29y - 42
Answer» Let, f (y) = y³ - 2y² - 29y - 42 The factors of the constant term – 42 are \(\pm\) 1, \(\pm\) 2, \(\pm\) 3, \(\pm\) 6, \(\pm\) 7, \(\pm\) 14, \(\pm\) 21 and \(\pm\) 42 Putting y = - 2, we have f (-2) = (-2)³ – 2 (-2)² – 29 (-2) - 42 = - 8 – 8 + 58 - 42 = 0 So, (y + 2) is a factor of f (y) Let us now divide f (y) = y³ - 2y² - 29y - 42 by (y + 2) to get the other factors of f (x) Using long division method, we get y³ - 2y²- 29y - 42 = (y + 2) (y² – 4y – 21) y² – 4y - 21 = y² – 7y + 3y - 21 = y (y – 7) + 3 (y – 7) = (y – 7) (y + 3) Hence, y³ - 2y² - 29y - 42 = (y + 2) (y - 7) (y + 3)

Using factor theorem, factorize each of the following polynomial:y3 - 2y2 - 29y - 42

Answer»

Let, f (y) = y³ - 2y² - 29y - 42

The factors of the constant term – 42 are \(\pm\) 1, \(\pm\) 2, \(\pm\) 3, \(\pm\) 6, \(\pm\) 7, \(\pm\) 14, \(\pm\) 21 and \(\pm\) 42

Putting y = - 2, we have

f (-2) = (-2)³ – 2 (-2)² – 29 (-2) - 42

= - 8 – 8 + 58 - 42

= 0

So,

(y + 2) is a factor of f (y)

Let us now divide

f (y) = y³ - 2y² - 29y - 42 by (y + 2) to get the other factors of f (x)

Using long division method, we get

y³ - 2y²- 29y - 42 = (y + 2) (y² – 4y – 21)

y² – 4y - 21 = y² – 7y + 3y - 21

= y (y – 7) + 3 (y – 7)

= (y – 7) (y + 3)

Hence,

y³ - 2y² - 29y - 42 = (y + 2) (y - 7) (y + 3)